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For this Question: Given an integer N, your task is to print all the even integer from 1 to N.<b>User task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the functions <b>For_Loop()</b> that take the integer n as a parameter.
</b>Constraints:</b>
1 <= n <= 100
<b>Note:</b>
<i>But there is a catch here, given user function has already code in it which may or may not be correct, now you need to figure out these and correct them if it is required</i>Print all the even numbers from 1 to n. (print all the numbers in the same line, space-separated)Sample Input:-
5
Sample Output:-
2 4
Sample Input:-
6
Sample Output:-
2 4 6, I have written this Solution Code: public static void For_Loop(int n){
for(int i=2;i<=n;i+=2){
System.out.print(i+" ");
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: The cost of stock on each day is given in an array A[] of size N. If you can only perform at most two transactions what is the maximum profit you can gain.
Note:- The second transaction can only start after the first one is complete (Sell- >buy- >sell- >buy).The first line of input contains a single integer N. The next line of input contains N space-separated integers depicting the values of A[].
Constraints:-
2 <= N <= 10000
1 <= A[i] <= 1000000000Print the maximum profit gain in at most two transactions.Sample Input:-
5
3 5 2 8 3
Sample Output:-
8
Explanation:-
Buy at index 1, sell at index 2, Buy at index 3, sell at index 8.
Sample Input:-
4
1 2 4 5
Sample Output:-
4, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int n = Integer.parseInt(br.readLine());
String str1 = br.readLine();
String[] str2 = str1.split(" ");
long[] arr = new long[n];
for(int i = 0; i < n; ++i) {
arr[i] = Long.parseLong(str2[i]);
}
System.out.print(maxProfit(arr, n));
}
static long maxProfit(long price[], int n)
{
long profit[] = new long[n];
long max_price = price[n - 1];
for (int i = n - 2; i >= 0; i--) {
if (price[i] > max_price)
max_price = price[i];
profit[i] = Math.max(profit[i + 1],
max_price - price[i]);
}
long min_price = price[0];
for (int i = 1; i < n; i++) {
if (price[i] < min_price)
min_price = price[i];
profit[i] = Math.max(
profit[i - 1],
profit[i] + (price[i] - min_price));
}
long result = profit[n - 1];
return result;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: The cost of stock on each day is given in an array A[] of size N. If you can only perform at most two transactions what is the maximum profit you can gain.
Note:- The second transaction can only start after the first one is complete (Sell- >buy- >sell- >buy).The first line of input contains a single integer N. The next line of input contains N space-separated integers depicting the values of A[].
Constraints:-
2 <= N <= 10000
1 <= A[i] <= 1000000000Print the maximum profit gain in at most two transactions.Sample Input:-
5
3 5 2 8 3
Sample Output:-
8
Explanation:-
Buy at index 1, sell at index 2, Buy at index 3, sell at index 8.
Sample Input:-
4
1 2 4 5
Sample Output:-
4, I have written this Solution Code:
#include <bits/stdc++.h>
using namespace std;
#define MEM(a, b) memset(a, (b), sizeof(a))
#define FOREACH(it, l) for (auto it = l.begin(); it != l.end(); it++)
#define IN(A, B, C) assert( B <= A && A <= C)
#define MP make_pair
#define FOR(i,a) for(int i=0;i<a;i++)
#define FOR1(i,j,a) for(int i=j;i<a;i++)
#define EB emplace_back
#define INF (int)1e9
#define EPS 1e-9
#define PI 3.1415926535897932384626433832795
#define max1 1000001
#define MOD 1000000007
#define read(type) readInt<type>()
#define out(x) cout<<x<<'\n'
#define out1(x) cout<<x<<" "
#define END cout<<'\n'
#define int long long
#define sz(v) ((int)(v).size())
#define all(v) (v).begin(), (v).end()
void fast(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
signed main(){
int n;
cin>>n;
int a[n];
FOR(i,n){cin>>a[i];}
int left[n],right[n+1];
right[n]=0;
for(int i=0;i<n;i++){
left[i]=0;
right[i]=0;
}
int m = a[n-1];
for(int i=n-2;i>=0;i--){
m=max(m,a[i]);
right[i]=max(m-a[i],right[i+1]);
}
m = a[0];
for(int i=1;i<n;i++){
m=min(m,a[i]);
left[i]=max(a[i]-m,left[i-1]);
}
int ans=0;
for(int i=0;i<n;i++){
ans=max(ans,left[i]+right[i+1]);
}
out(ans);
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Given a series and an integer N, your task is to print the sum of it's first N terms.
Series:- 3, 5, 9, 17, 33.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>SeriesSum()</b> that takes the integer N as parameter.
<b>Constraints:</b>
1 <= N <= 20Return the sum of N terms.Sample Input:-
3
Sample Output:-
17
Sample Input:-
2
Sample Output:-
8, I have written this Solution Code:
int SeriesSum(int N){
int n=1;
int ans=3;
int res=3;
for(int i=1;i<N;i++){
n*=2;
res+=n;
ans+=res;
}
return ans;
}
, In this Programming Language: C, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Given a series and an integer N, your task is to print the sum of it's first N terms.
Series:- 3, 5, 9, 17, 33.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>SeriesSum()</b> that takes the integer N as parameter.
<b>Constraints:</b>
1 <= N <= 20Return the sum of N terms.Sample Input:-
3
Sample Output:-
17
Sample Input:-
2
Sample Output:-
8, I have written this Solution Code:
int SeriesSum(int N){
int n=1;
int ans=3;
int res=3;
for(int i=1;i<N;i++){
n*=2;
res+=n;
ans+=res;
}
return ans;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Given a series and an integer N, your task is to print the sum of it's first N terms.
Series:- 3, 5, 9, 17, 33.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>SeriesSum()</b> that takes the integer N as parameter.
<b>Constraints:</b>
1 <= N <= 20Return the sum of N terms.Sample Input:-
3
Sample Output:-
17
Sample Input:-
2
Sample Output:-
8, I have written this Solution Code: static int SeriesSum(int N){
int n=1;
int ans=3;
int res=3;
for(int i=1;i<N;i++){
n*=2;
res+=n;
ans+=res;
}
return ans;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Given a series and an integer N, your task is to print the sum of it's first N terms.
Series:- 3, 5, 9, 17, 33.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>SeriesSum()</b> that takes the integer N as parameter.
<b>Constraints:</b>
1 <= N <= 20Return the sum of N terms.Sample Input:-
3
Sample Output:-
17
Sample Input:-
2
Sample Output:-
8, I have written this Solution Code: def SeriesSum(N):
n=1
res=3
ans=3
for i in range (1,N):
n=n*2
res=res+n
ans=ans+res
return ans
, In this Programming Language: Python, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Given a natural number N, your task is to print all the digits of the number in English words. The words have to separate by space and in lowercase English letters.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Print_Digit()</b> that takes integer N as a parameter.
<b>Constraints:-</b>
1 ≤ N ≤ 10<sup>7</sup>Print the digits of the number as shown in the example.
<b>Note:-</b>
Print all digits in lowercase English lettersSample Input:-
1024
Sample Output:-
one zero two four
Sample Input:-
2
Sample Output:-
two, I have written this Solution Code: def Print_Digit(n):
dc = {1: "one", 2: "two", 3: "three", 4: "four",
5: "five", 6: "six", 7: "seven", 8: "eight", 9: "nine", 0: "zero"}
final_list = []
while (n > 0):
final_list.append(dc[int(n%10)])
n = int(n / 10)
for val in final_list[::-1]:
print(val, end=' '), In this Programming Language: Python, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Given a natural number N, your task is to print all the digits of the number in English words. The words have to separate by space and in lowercase English letters.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Print_Digit()</b> that takes integer N as a parameter.
<b>Constraints:-</b>
1 ≤ N ≤ 10<sup>7</sup>Print the digits of the number as shown in the example.
<b>Note:-</b>
Print all digits in lowercase English lettersSample Input:-
1024
Sample Output:-
one zero two four
Sample Input:-
2
Sample Output:-
two, I have written this Solution Code: class Solution {
public static void Print_Digits(int N){
if(N==0){return;}
Print_Digits(N/10);
int x=N%10;
if(x==1){System.out.print("one ");}
else if(x==2){System.out.print("two ");}
else if(x==3){System.out.print("three ");}
else if(x==4){System.out.print("four ");}
else if(x==5){System.out.print("five ");}
else if(x==6){System.out.print("six ");}
else if(x==7){System.out.print("seven ");}
else if(x==8){System.out.print("eight ");}
else if(x==9){System.out.print("nine ");}
else if(x==0){System.out.print("zero ");}
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Given an integer N. The task is to find the square root of N. If N is not a perfect square, then return floor(√N). Try to solve the problem using Binary Search.The first line of input contains the number of test cases T. For each test case, the only line contains the number N.
<b>Constraints:</b>
1 ≤ T ≤ 10000
0 ≤ x ≤ 10<sup>8</sup>For each testcase, print square root of given integer.Sample Input:
2
5
4
Sample Output:
2
2
<b>Explanation:</b.
Testcase 1: Since, 5 is not a perfect square, the floor of square_root of 5 is 2.
Testcase 2: Since 4 is a perfect square, its square root is 2., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
static int binarySearch( int k,int start,int end) {
int mid = start + (end - start)/2;
if (mid * mid == k) {
return (int) mid;
}
else if ((mid*mid) > end) {
return binarySearch( k, 0, mid - 1);
}
else if ((mid * mid < end)) {
return binarySearch( k, mid + 1, end);
}
else {
mid = (int) Math.sqrt(k);
}
return mid;
}
public static void main (String[] args) throws IOException{
InputStreamReader st = new InputStreamReader(System.in);
BufferedReader br = new BufferedReader(st);
int T = Integer.parseInt(br.readLine());
while (T-- > 0) {
int n = Integer.parseInt(br.readLine());
int k = n;
System.out.println(binarySearch( k, 0, n - 1));
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Given an integer N. The task is to find the square root of N. If N is not a perfect square, then return floor(√N). Try to solve the problem using Binary Search.The first line of input contains the number of test cases T. For each test case, the only line contains the number N.
<b>Constraints:</b>
1 ≤ T ≤ 10000
0 ≤ x ≤ 10<sup>8</sup>For each testcase, print square root of given integer.Sample Input:
2
5
4
Sample Output:
2
2
<b>Explanation:</b.
Testcase 1: Since, 5 is not a perfect square, the floor of square_root of 5 is 2.
Testcase 2: Since 4 is a perfect square, its square root is 2., I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define pu push_back
#define fi first
#define se second
#define mp make_pair
#define int long long
#define ll long long
#define pii pair<int,int>
#define mm (s+e)/2
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(int i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define sz 200000
int sqr(int a) {
int A=a;
if(A<2) return A;
ll l=1,r=A;
ll k=1;
while(l<=r)
{
ll mid=(l+r)/2;
ll u=mid*mid;
if(u<=A)
{
k=max(mid,k);
l=mid+1;
}else
{
r=mid-1;
}
}
return k;
}
signed main()
{
int t;
cin>>t;
while(t>0)
{
t--;
long a;
cin>>a;
long x = sqrt(a);
cout<<x<<endl;
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Given an integer N. The task is to find the square root of N. If N is not a perfect square, then return floor(√N). Try to solve the problem using Binary Search.The first line of input contains the number of test cases T. For each test case, the only line contains the number N.
<b>Constraints:</b>
1 ≤ T ≤ 10000
0 ≤ x ≤ 10<sup>8</sup>For each testcase, print square root of given integer.Sample Input:
2
5
4
Sample Output:
2
2
<b>Explanation:</b.
Testcase 1: Since, 5 is not a perfect square, the floor of square_root of 5 is 2.
Testcase 2: Since 4 is a perfect square, its square root is 2., I have written this Solution Code: N=int(input())
for i in range(0,N):
M=int(input())
s=M**0.5
print(int(s)), In this Programming Language: Python, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Given an integer N. The task is to find the square root of N. If N is not a perfect square, then return floor(√N). Try to solve the problem using Binary Search.The first line of input contains the number of test cases T. For each test case, the only line contains the number N.
<b>Constraints:</b>
1 ≤ T ≤ 10000
0 ≤ x ≤ 10<sup>8</sup>For each testcase, print square root of given integer.Sample Input:
2
5
4
Sample Output:
2
2
<b>Explanation:</b.
Testcase 1: Since, 5 is not a perfect square, the floor of square_root of 5 is 2.
Testcase 2: Since 4 is a perfect square, its square root is 2., I have written this Solution Code: // n is the input number
function sqrt(n) {
// write code here
// do not console.log
// return the number
return Math.floor(Math.sqrt(n))
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Given a year(an integer variable) as input, find if it is a leap year or not.
Note: Leap year is the year that is multiple of 4. But, if the year is a multiple of 100 and not a multiple of 400, then it is not a leap year.<b>User Task:</b>
Complete the function <b>LeapYear()</b> that takes integer n as a parameter.
<b>Constraint:</b>
1 <= n <= 5000If it is a leap year then print <b>YES</b> and if it is not a leap year, then print <b>NO</b>Sample Input:
2000
Sample Output:
YES
Sample Input:
2003
Sample Output:
NO
Sample Input:
1900
Sample Output:
NO, I have written this Solution Code: n = int(input())
if (n%4==0 and n%100!=0 or n%400==0):
print("YES")
elif n==0:
print("YES")
else:
print("NO"), In this Programming Language: Python, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Given a year(an integer variable) as input, find if it is a leap year or not.
Note: Leap year is the year that is multiple of 4. But, if the year is a multiple of 100 and not a multiple of 400, then it is not a leap year.<b>User Task:</b>
Complete the function <b>LeapYear()</b> that takes integer n as a parameter.
<b>Constraint:</b>
1 <= n <= 5000If it is a leap year then print <b>YES</b> and if it is not a leap year, then print <b>NO</b>Sample Input:
2000
Sample Output:
YES
Sample Input:
2003
Sample Output:
NO
Sample Input:
1900
Sample Output:
NO, I have written this Solution Code:
import java.util.Scanner;
class Main {
public static void main (String[] args)
{
//Capture the user's input
Scanner scanner = new Scanner(System.in);
//Storing the captured value in a variable
int n = scanner.nextInt();
LeapYear(n);
}
static void LeapYear(int year){
if(year%400==0 || (year%100 != 0 && year%4==0)){System.out.println("YES");}
else {
System.out.println("NO");}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Given a String <b>S</b>, you need to typecast this String to Long. If the typecasting is done successfully then we will print "<b>Nice Job</b>" otherwise "<b>Wrong answer</b>".User task:
Since this is a functional problem you don't have to worry about the input. You just have to complete the function <b>checkConvertion()</b>, which contains S as a parameter.
<b>Constraints:-</b>
1 <= |S| <= 15
The string will contain only numeric digits(1-9)You need to return the typecasted Long value. The driver code will print "<b>Nice Job</b>" otherwise "<b>Wrong answer</b>".
<b>Note:-</b> We are not converting this string to int here because the size of the number exceeds the int limits.Sample Input 1:-
43
Sample Output 1:-
Nice Job
Sample Input 2:-
2584563259874
Sample Output 2:-
Nice Job, I have written this Solution Code: def checkConevrtion(a):
return int(a)
, In this Programming Language: Python, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Given a String <b>S</b>, you need to typecast this String to Long. If the typecasting is done successfully then we will print "<b>Nice Job</b>" otherwise "<b>Wrong answer</b>".User task:
Since this is a functional problem you don't have to worry about the input. You just have to complete the function <b>checkConvertion()</b>, which contains S as a parameter.
<b>Constraints:-</b>
1 <= |S| <= 15
The string will contain only numeric digits(1-9)You need to return the typecasted Long value. The driver code will print "<b>Nice Job</b>" otherwise "<b>Wrong answer</b>".
<b>Note:-</b> We are not converting this string to int here because the size of the number exceeds the int limits.Sample Input 1:-
43
Sample Output 1:-
Nice Job
Sample Input 2:-
2584563259874
Sample Output 2:-
Nice Job, I have written this Solution Code: static long checkConevrtion(String S)
{
return Long.parseLong(S);
}, In this Programming Language: Java, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Given a number N for each i (1 < = i < = N), you have to print the number except :-
For each multiple of 3, print "Newton" instead of the number.
For each multiple of 5, print "School" instead of the number.
For numbers that are multiples of both 3 and 5, print "NewtonSchool" instead of the number.The first line of the input contains N.
<b>Constraints</b>
1 < = N < = 1000
Print N space separated number or Newton School according to the condition.Sample Input:-
3
Sample Output:-
1 2 Newton
Sample Input:-
5
Sample Output:-
1 2 Newton 4 School, I have written this Solution Code: n=int(input())
for i in range(1,n+1):
if i%3==0 and i%5==0:
print("NewtonSchool",end=" ")
elif i%3==0:
print("Newton",end=" ")
elif i%5==0:
print("School",end=" ")
else:
print(i,end=" "), In this Programming Language: Python, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Given a number N for each i (1 < = i < = N), you have to print the number except :-
For each multiple of 3, print "Newton" instead of the number.
For each multiple of 5, print "School" instead of the number.
For numbers that are multiples of both 3 and 5, print "NewtonSchool" instead of the number.The first line of the input contains N.
<b>Constraints</b>
1 < = N < = 1000
Print N space separated number or Newton School according to the condition.Sample Input:-
3
Sample Output:-
1 2 Newton
Sample Input:-
5
Sample Output:-
1 2 Newton 4 School, I have written this Solution Code: import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
static void NewtonSchool(int n){
for(int i=1;i<=n;i++){
if(i%3==0 && i%5==0){System.out.print("NewtonSchool ");}
else if(i%5==0){System.out.print("School ");}
else if(i%3==0){System.out.print("Newton ");}
else{System.out.print(i+" ");}
}
}
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int x= sc.nextInt();
NewtonSchool(x);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Final fight of Harry and Voldemort has began. In order to defeat Lord Voldemort, Harry must use a spell greater than the Voldemort's spell and as Voldemort loves palindromes, Harry's spell must not contain palindromic substring of length more than or equal to X.
Given X and string S, find a string R such that R is lexicographically larger than S and R does not contain any palindrome substring with length more than or equal to X. Among all possible R, you must chose lexicographically minimum.
This fight will decide the fate of Hogwarts, so please help Harry find an appropriate spell (string R) or tell Harry that no such spell exists.First line of input contains a single integer X.
Second line of input contains a string S (Lord Voldemort's spell).
Constraints:
1 <= |S| <= 100000
1 <= X <= |S|
S contains only lowercase english alphabets
Note: R must contain only lowercase english alphabets.Print a single string R denoting Harry's spell if it exists else print "Impossible".Sample Input
3
bbbbbbb
Sample Output
bbcaabb
Sample Input
2
zyz
Sample Output
Impossible
, I have written this Solution Code: #pragma GCC optimize ("Ofast")
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class T>
using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define pii pair<int,int>
/////////////
#define P 1000000007
#define N 400010
char s[N],t[N];int m,n,a[N],b[N],p[N];
bool F=0;
bool can(int x,int y)
{
x++;
if(x<y)return 1;
int le=a[x]-a[x-y]*p[y],ri=b[x]-b[x-y];
return le*p[x-y]!=ri;
}
void ff(int x,bool y)
{
if(x==n){cout<<t,F=1;return;}
for(t[x]=(y?s[x]:'a');t[x]<='z';t[x]++)
{
a[x+1]=a[x]*P+t[x],
b[x+1]=b[x]+t[x]*p[x];
if(can(x,m)&&can(x,m+1))ff(x+1,y&&(t[x]==s[x]));
if(F)return;
}
return;
}
signed main()
{
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
cin>>m>>s;
n=strlen(s);t[n]=0;
int w=n-1;
while(w>=0&&s[w]=='z')s[w]='a',w--;
if(w<0){cout<<"Impossible";return 0;}
s[w]++;
p[0]=1;for(int i=1;i<N;i++)p[i]=p[i-1]*P;
ff(0,1);
if(!F)cout<<"Impossible";
#ifdef ANIKET_GOYAL
// cout<<endl<<endl<<endl<<endl<<"Time elapsed: "<<(double)(clock()-clk)/CLOCKS_PER_SEC<<endl;
#endif
}, In this Programming Language: C++, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: English Team has now adopted a rotation policy for two of their players: Dom and Leach.
On the first day, both of them played but, from the second day onwards, Dom plays every second day, while Leach plays every third day.
For example, on:
Day 1 - Both players play,
Day 2 - Neither of them plays,
Day 3 - Only Dom plays,
Day 4 - Only Leach plays,
Day 5 - Only Dom plays,
Day 6 - Neither of them plays,
Day 7 - Both the players play.. and so on.
Find the number of days in the interval [A, B] (A and B, both inclusive) when neither Dom nor Leach plays.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>RotationPolicy()</b> that takes integers A, and B as arguments.
Constraints:-
1 <= A, B <=100000Return the number of days when neither of the two players played the game.Sample Input:-
3 8
Sample Output:-
2
Sample Input:-
1 4
Sample Output:-
1, I have written this Solution Code:
int RotationPolicy(int A, int B){
int cnt=0;
for(int i=A;i<=B;i++){
if((i-1)%2!=0 && (i-1)%3!=0){cnt++;}
}
return cnt;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: English Team has now adopted a rotation policy for two of their players: Dom and Leach.
On the first day, both of them played but, from the second day onwards, Dom plays every second day, while Leach plays every third day.
For example, on:
Day 1 - Both players play,
Day 2 - Neither of them plays,
Day 3 - Only Dom plays,
Day 4 - Only Leach plays,
Day 5 - Only Dom plays,
Day 6 - Neither of them plays,
Day 7 - Both the players play.. and so on.
Find the number of days in the interval [A, B] (A and B, both inclusive) when neither Dom nor Leach plays.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>RotationPolicy()</b> that takes integers A, and B as arguments.
Constraints:-
1 <= A, B <=100000Return the number of days when neither of the two players played the game.Sample Input:-
3 8
Sample Output:-
2
Sample Input:-
1 4
Sample Output:-
1, I have written this Solution Code:
int RotationPolicy(int A, int B){
int cnt=0;
for(int i=A;i<=B;i++){
if((i-1)%2!=0 && (i-1)%3!=0){cnt++;}
}
return cnt;
}
, In this Programming Language: C, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: English Team has now adopted a rotation policy for two of their players: Dom and Leach.
On the first day, both of them played but, from the second day onwards, Dom plays every second day, while Leach plays every third day.
For example, on:
Day 1 - Both players play,
Day 2 - Neither of them plays,
Day 3 - Only Dom plays,
Day 4 - Only Leach plays,
Day 5 - Only Dom plays,
Day 6 - Neither of them plays,
Day 7 - Both the players play.. and so on.
Find the number of days in the interval [A, B] (A and B, both inclusive) when neither Dom nor Leach plays.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>RotationPolicy()</b> that takes integers A, and B as arguments.
Constraints:-
1 <= A, B <=100000Return the number of days when neither of the two players played the game.Sample Input:-
3 8
Sample Output:-
2
Sample Input:-
1 4
Sample Output:-
1, I have written this Solution Code: static int RotationPolicy(int A, int B){
int cnt=0;
for(int i=A;i<=B;i++){
if((i-1)%2!=0 && (i-1)%3!=0){cnt++;}
}
return cnt;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: English Team has now adopted a rotation policy for two of their players: Dom and Leach.
On the first day, both of them played but, from the second day onwards, Dom plays every second day, while Leach plays every third day.
For example, on:
Day 1 - Both players play,
Day 2 - Neither of them plays,
Day 3 - Only Dom plays,
Day 4 - Only Leach plays,
Day 5 - Only Dom plays,
Day 6 - Neither of them plays,
Day 7 - Both the players play.. and so on.
Find the number of days in the interval [A, B] (A and B, both inclusive) when neither Dom nor Leach plays.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>RotationPolicy()</b> that takes integers A, and B as arguments.
Constraints:-
1 <= A, B <=100000Return the number of days when neither of the two players played the game.Sample Input:-
3 8
Sample Output:-
2
Sample Input:-
1 4
Sample Output:-
1, I have written this Solution Code:
def RotationPolicy(A, B):
cnt=0
for i in range (A,B+1):
if(i-1)%2!=0 and (i-1)%3!=0:
cnt=cnt+1
return cnt
, In this Programming Language: Python, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: English Team has now adopted a rotation policy for two of their players: Dom and Leach.
On the first day, both of them played but, from the second day onwards, Dom plays every second day, while Leach plays every third day.
For example, on:
Day 1 - Both players play,
Day 2 - Neither of them plays,
Day 3 - Only Dom plays,
Day 4 - Only Leach plays,
Day 5 - Only Dom plays,
Day 6 - Neither of them plays,
Day 7 - Both the players play.. and so on.
Find the number of days in the interval [A, B] (A and B, both inclusive) when neither Dom nor Leach plays.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>RotationPolicy()</b> that takes integers A, and B as arguments.
Constraints:-
1 <= A, B <=100000Return the number of days when neither of the two players played the game.Sample Input:-
3 8
Sample Output:-
2
Sample Input:-
1 4
Sample Output:-
1, I have written this Solution Code: function RotationPolicy(a, b) {
// write code here
// do no console.log the answer
// return the output using return keyword
let count = 0
for (let i = a; i <= b; i++) {
if((i-1)%2 !== 0 && (i-1)%3 !==0){
count++
}
}
return count
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Implement pow(X, N), which calculates x raised to the power N i.e. (X^N).
Try using a recursive approachThe first line contains T, denoting the number of test cases. Each test case contains a single line containing X, and N.
<b>Constraints:</b>
1 ≤ T ≤ 100
-10.00 ≤ X ≤10.00
-10 ≤ N ≤ 10
For each test case, you need to print the value of X^N. Print up to two places of decimal.
<b>Note:</b>
Please take care that the output can be very large but it will not exceed double the data type value.Input:
1
2.00 -2
Output:
0.25
<b>Explanation:</b>
2^(-2) = 1/2^2 = 1/4 = 0.25, I have written this Solution Code:
// X and Y are numbers
// ignore number of testcases variable
function pow(X, Y) {
// write code here
// console.log the output in a single line,like example
console.log(Math.pow(X, Y).toFixed(2))
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Implement pow(X, N), which calculates x raised to the power N i.e. (X^N).
Try using a recursive approachThe first line contains T, denoting the number of test cases. Each test case contains a single line containing X, and N.
<b>Constraints:</b>
1 ≤ T ≤ 100
-10.00 ≤ X ≤10.00
-10 ≤ N ≤ 10
For each test case, you need to print the value of X^N. Print up to two places of decimal.
<b>Note:</b>
Please take care that the output can be very large but it will not exceed double the data type value.Input:
1
2.00 -2
Output:
0.25
<b>Explanation:</b>
2^(-2) = 1/2^2 = 1/4 = 0.25, I have written this Solution Code: def power(N,K):
return ("{0:.2f}".format(N**K))
T=int(input())
for i in range(T):
X,N = map(float,input().strip().split())
print(power(X,N)), In this Programming Language: Python, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Implement pow(X, N), which calculates x raised to the power N i.e. (X^N).
Try using a recursive approachThe first line contains T, denoting the number of test cases. Each test case contains a single line containing X, and N.
<b>Constraints:</b>
1 ≤ T ≤ 100
-10.00 ≤ X ≤10.00
-10 ≤ N ≤ 10
For each test case, you need to print the value of X^N. Print up to two places of decimal.
<b>Note:</b>
Please take care that the output can be very large but it will not exceed double the data type value.Input:
1
2.00 -2
Output:
0.25
<b>Explanation:</b>
2^(-2) = 1/2^2 = 1/4 = 0.25, I have written this Solution Code: import java.util.*;
import java.io.*;
import java.lang.*;
class Main
{
public static void main (String[] args)throws Exception {
BufferedReader read = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(read.readLine());
while(t-- > 0)
{
String str[] = read.readLine().trim().split(" ");
double X = Double.parseDouble(str[0]);
int N = Integer.parseInt(str[1]);
System.out.println(String.format("%.2f", myPow(X, N)));
}
}
public static double myPow(double x, int n) {
if (n == Integer.MIN_VALUE)
n = - (Integer.MAX_VALUE - 1);
if (n == 0)
return 1.0;
else if (n < 0)
return 1 / myPow(x, -n);
else if (n % 2 == 1)
return x * myPow(x, n - 1);
else {
double sqrt = myPow(x, n / 2);
return sqrt * sqrt;
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Implement pow(X, N), which calculates x raised to the power N i.e. (X^N).
Try using a recursive approachThe first line contains T, denoting the number of test cases. Each test case contains a single line containing X, and N.
<b>Constraints:</b>
1 ≤ T ≤ 100
-10.00 ≤ X ≤10.00
-10 ≤ N ≤ 10
For each test case, you need to print the value of X^N. Print up to two places of decimal.
<b>Note:</b>
Please take care that the output can be very large but it will not exceed double the data type value.Input:
1
2.00 -2
Output:
0.25
<b>Explanation:</b>
2^(-2) = 1/2^2 = 1/4 = 0.25, I have written this Solution Code: #include "bits/stdc++.h"
using namespace std;
#define ld long double
#define int long long int
#define speed ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#define endl '\n'
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
ld power(ld x, ld n){
if(n == 0)
return 1;
else
return x*power(x, n-1);
}
signed main() {
speed;
int t; cin >> t;
while(t--){
double x; int n;
cin >> x >> n;
if(n < 0)
x = 1.0/x, n *= -1;
cout << setprecision(2) << fixed << power(x, n) << endl;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Rachel got a string as a gift from Ross. She wants to exchange this string with another string of same length. A string with all the same characters is fashionable string. Find Rachel lexicographically minimum fashionable string but make sure that the new string has at least one character common with the string Ross gave her otherwise he will get upset.Input contains a single string S, denoting the string Ross gave.
Constraints:
1 <= |S| <= 10000
S contains only lowercase english letters.Print the new string Rachel gets.Sample Input
bccde
Sample Output
bbbbb
Explanation: "bbbbb" is lexicographically minimum fashionable string which has atleast one character common with the initial string.
Sample Input
a
Sample Output
a, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String s = br.readLine().trim();
char n[] = s.toCharArray();
int posMin = 0;
for(int i=0;i<n.length;i++){
if(n[posMin]>n[i])
posMin = i;
}
for(int i=0;i<n.length;i++)
n[i] = n[posMin];
System.out.print(String.valueOf(n));
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Rachel got a string as a gift from Ross. She wants to exchange this string with another string of same length. A string with all the same characters is fashionable string. Find Rachel lexicographically minimum fashionable string but make sure that the new string has at least one character common with the string Ross gave her otherwise he will get upset.Input contains a single string S, denoting the string Ross gave.
Constraints:
1 <= |S| <= 10000
S contains only lowercase english letters.Print the new string Rachel gets.Sample Input
bccde
Sample Output
bbbbb
Explanation: "bbbbb" is lexicographically minimum fashionable string which has atleast one character common with the initial string.
Sample Input
a
Sample Output
a, I have written this Solution Code: #pragma GCC optimize ("Ofast")
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class T>
using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
template<class C> void mini(C&a4, C b4){a4=min(a4,b4);}
typedef unsigned long long ull;
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define mod 1000000007ll
#define pii pair<int,int>
/////////////
signed main(){
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
string s;
cin>>s;
char c='z';
for(auto r:s)
c=min(c,r);
for(int i=0;i<s.length();++i)
cout<<c;
#ifdef ANIKET_GOYAL
// cout<<endl<<endl<<endl<<endl<<"Time elapsed: "<<(double)(clock()-clk)/CLOCKS_PER_SEC<<endl;
#endif
}, In this Programming Language: C++, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Rachel got a string as a gift from Ross. She wants to exchange this string with another string of same length. A string with all the same characters is fashionable string. Find Rachel lexicographically minimum fashionable string but make sure that the new string has at least one character common with the string Ross gave her otherwise he will get upset.Input contains a single string S, denoting the string Ross gave.
Constraints:
1 <= |S| <= 10000
S contains only lowercase english letters.Print the new string Rachel gets.Sample Input
bccde
Sample Output
bbbbb
Explanation: "bbbbb" is lexicographically minimum fashionable string which has atleast one character common with the initial string.
Sample Input
a
Sample Output
a, I have written this Solution Code: s=input()
l=len(s)
k=min(s)
m=k*l
print(m), In this Programming Language: Python, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Given an array of N Books B[], where B[i] denotes the number of pages P in i'th book. Now you need to find the maximum number of distinct books (having a different number of pages) after removing K books from the array.The input line contains T, denoting the number of testcases. Each testcase contains two lines. First line contains N, number of books and K, number of books you need to remove. Second line contains N positive integers denoting the number of pages in ith book.
Constraints:
1 <= T <= 100
1 <= N <= 10^4
1 <= K <= N
1 <= B[i] <= 10^4You need to print the maximum number of distinct books.Sample Input:
2
7 3
5 7 5 5 1 2 2
6 4
1 2 3 4 5 6
Sample Output:
4
2
Explanation:
Testcase 1:
Remove 2 occurrences of books having 5 pages and 1 occurrence of book having 2 pages., I have written this Solution Code: import java.util.*;
import java.io.*;
class Main{
static class Reader {
final private int BUFFER_SIZE = 1 << 16;
private DataInputStream din;
private byte[] buffer;
private int bufferPointer, bytesRead;
public Reader()
{
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public Reader(String file_name) throws IOException
{
din = new DataInputStream(
new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public String readLine() throws IOException
{
byte[] buf = new byte[64];
int cnt = 0, c;
while ((c = read()) != -1) {
if (c == '\n') {
if (cnt != 0) {
break;
}
else {
continue;
}
}
buf[cnt++] = (byte)c;
}
return new String(buf, 0, cnt);
}
public int nextInt() throws IOException
{
int ret = 0;
byte c = read();
while (c <= ' ') {
c = read();
}
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public long nextLong() throws IOException
{
long ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public double nextDouble() throws IOException
{
double ret = 0, div = 1;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (c == '.') {
while ((c = read()) >= '0' && c <= '9') {
ret += (c - '0') / (div *= 10);
}
}
if (neg)
return -ret;
return ret;
}
private void fillBuffer() throws IOException
{
bytesRead = din.read(buffer, bufferPointer = 0,
BUFFER_SIZE);
if (bytesRead == -1)
buffer[0] = -1;
}
private byte read() throws IOException
{
if (bufferPointer == bytesRead)
fillBuffer();
return buffer[bufferPointer++];
}
public void close() throws IOException
{
if (din == null)
return;
din.close();
}
}
public static void main(String[] args) throws IOException {
Reader sc = new Reader();
int t = sc.nextInt();
while (t-->0) {
HashMap<Integer,Integer> map = new HashMap<>();
int n = sc.nextInt();
int k = sc.nextInt();
for (int i = 0; i < n; i++) {
int v = sc.nextInt();
map.put(v, map.getOrDefault(v, 0)+1);
}
int maxRemove = 0;
for (Integer integer : map.keySet()) {
maxRemove += map.get(integer) -1;
}
if (k<=maxRemove) {
System.out.println(map.size());
}else{
System.out.println(map.size() - (k-maxRemove));
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Given an array of N Books B[], where B[i] denotes the number of pages P in i'th book. Now you need to find the maximum number of distinct books (having a different number of pages) after removing K books from the array.The input line contains T, denoting the number of testcases. Each testcase contains two lines. First line contains N, number of books and K, number of books you need to remove. Second line contains N positive integers denoting the number of pages in ith book.
Constraints:
1 <= T <= 100
1 <= N <= 10^4
1 <= K <= N
1 <= B[i] <= 10^4You need to print the maximum number of distinct books.Sample Input:
2
7 3
5 7 5 5 1 2 2
6 4
1 2 3 4 5 6
Sample Output:
4
2
Explanation:
Testcase 1:
Remove 2 occurrences of books having 5 pages and 1 occurrence of book having 2 pages., I have written this Solution Code: t=int(input())
for m in range(0,t):
n,k=input().split()
n=int(n)
k=int(k)
arr=input().split()
hm={}
temp=k
dictinct=0
for i in range(0,n):
arr[i]=int(arr[i])
if(arr[i] not in hm):
hm[arr[i]]=1
else:
k-=1
dictinct+=1
if(k>0):
print (len(hm)-k)
else:
print (len(hm)), In this Programming Language: Python, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Given an array of N Books B[], where B[i] denotes the number of pages P in i'th book. Now you need to find the maximum number of distinct books (having a different number of pages) after removing K books from the array.The input line contains T, denoting the number of testcases. Each testcase contains two lines. First line contains N, number of books and K, number of books you need to remove. Second line contains N positive integers denoting the number of pages in ith book.
Constraints:
1 <= T <= 100
1 <= N <= 10^4
1 <= K <= N
1 <= B[i] <= 10^4You need to print the maximum number of distinct books.Sample Input:
2
7 3
5 7 5 5 1 2 2
6 4
1 2 3 4 5 6
Sample Output:
4
2
Explanation:
Testcase 1:
Remove 2 occurrences of books having 5 pages and 1 occurrence of book having 2 pages., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define max1 1000001
int a[max1];
int main(){
int t;
cin>>t;
while(t--){
int n,k;
cin>>n>>k;
int arr;
map<int,int> m;
int cnt=0;
for(int i=0;i<n;i++){
cin>>arr;
if(m.find(arr)!=m.end()){cnt++;}
m[arr]++;
}
if(k<=cnt){cout<<m.size()<<endl;}
else{cout<<m.size()-(k-cnt)<<endl;}
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Sara loves pattern, so this time she wishes to draw a pattern as:-
*****
****
***
**
*
Since Sara does not know how to code, help her to draw this pattern.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Triangle()</b> that takes no parameters.Print the pattern as shown in the example.Sample Output:-
*****
****
***
**
*, I have written this Solution Code: def Triangle():
for i in range(5):
for j in range(5-i):
print("*", end='')
print()
, In this Programming Language: Python, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Sara loves pattern, so this time she wishes to draw a pattern as:-
*****
****
***
**
*
Since Sara does not know how to code, help her to draw this pattern.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Triangle()</b> that takes no parameters.Print the pattern as shown in the example.Sample Output:-
*****
****
***
**
*, I have written this Solution Code: static void Triangle(){
System.out.println("*****");
System.out.println("****");
System.out.println("***");
System.out.println("**");
System.out.println("*");
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Given the heads of two singly linked lists head1 and head2, return the node at which the two lists intersect. If the two linked lists have no intersection at all, return null
For example, the following two linked lists begin to intersect at node c1<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>intersection()</b> that takes the head node of both lists as a parameter.
The first line of the test case contains:
<ul>
<li>the size of the 1st linked list</li>
<li>size of 2nd linked list</li>
<li>the size of the common part of two linked lists</li>
</ul>
The rest of the test case contains:
<ul>
<li>elements of 1st linked list</li>
<li>elements of 2nd linked list</li>
<li>common part of two linked lists</li>
</ul>Return the node of intersectionSample Input:-
4 3 4
1 2 3 4
5 6 7
9 10 11 12
Sample Output:-
9
Explanation:
1 -> 2 -> 3 -> 4
      |
       9 -> 10 -> 11 -> 12
      |
  5 -> 6 -> 7, I have written this Solution Code:
public static Node intersection(Node head1,Node head2){ int aLength = getLength(head1), bLength = getLength(head2);
Node currA = head1, currB = head2;
if (bLength > aLength)
for (int i = 0; i < bLength - aLength; i++) currB = currB.next;
if (aLength > bLength)
for (int i = 0; i < aLength - bLength; i++) currA = currA.next;
while (currA != null && currB != null) {
if (currA == currB) return currA;
currA = currA.next;
currB = currB.next;
}
return null;
}
private static int getLength(Node head) {
Node curr = head;
int len = 0;
while (curr != null) {
curr = curr.next;
len++;
}
return len;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Harry is trying to find the chamber of secrets but the chamber of secrets is well protected by dark spells. To find the chamber Harry needs to find the initial position of the chamber. Harry knows the final position of the chamber is X. He also knows that in total, the chamber has been repositioned N times. During the i<sup>th</sup> repositioning, the position of the chamber is changed to (previous position * Arr[i]) % 1000000007.
Given X and Arr, help Harry find the initial position of the chamber.
Note:- The initial position of the chamber is less than 1000000007.The first line of the input contains two integers N and X.
The second line of the input contains N integers denoting Arr.
Constraints
1 <= N <= 100000
1 <= Arr[i] < 1000000007Print a single integer denoting the initial position of the chamber.Sample Input
5 480
1 4 4 3 1
Sample Output
10
Explanation: Initial position = 10
After first repositioning position = (10*1)%1000000007 = 10
After second repositioning position = (10*4)%1000000007 = 40
After third repositioning position = (40*4)%1000000007 = 160
After fourth repositioning position = (160*3)%1000000007 = 480
After fifth repositioning position = (480*1)%1000000007 = 480, I have written this Solution Code: def inv(a,b,m):
res = 1
while(b):
if b&1:
res = (res*a)%m
a = (a*a)%m
b >>= 1
return res
n,x = map(int,input().split())
a = list(map(int,input().split()))
m = 1000000007
for i in a:
x = (x*inv(i,m-2,m))%m
print(x), In this Programming Language: Python, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Harry is trying to find the chamber of secrets but the chamber of secrets is well protected by dark spells. To find the chamber Harry needs to find the initial position of the chamber. Harry knows the final position of the chamber is X. He also knows that in total, the chamber has been repositioned N times. During the i<sup>th</sup> repositioning, the position of the chamber is changed to (previous position * Arr[i]) % 1000000007.
Given X and Arr, help Harry find the initial position of the chamber.
Note:- The initial position of the chamber is less than 1000000007.The first line of the input contains two integers N and X.
The second line of the input contains N integers denoting Arr.
Constraints
1 <= N <= 100000
1 <= Arr[i] < 1000000007Print a single integer denoting the initial position of the chamber.Sample Input
5 480
1 4 4 3 1
Sample Output
10
Explanation: Initial position = 10
After first repositioning position = (10*1)%1000000007 = 10
After second repositioning position = (10*4)%1000000007 = 40
After third repositioning position = (40*4)%1000000007 = 160
After fourth repositioning position = (160*3)%1000000007 = 480
After fifth repositioning position = (480*1)%1000000007 = 480, I have written this Solution Code: #pragma GCC optimize ("Ofast")
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class T>
using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define pii pair<int,int>
/////////////
int power_mod(int a,int b,int mod){
int ans = 1;
while(b){
if(b&1)
ans = (ans*a)%mod;
b = b/2;
a = (a*a)%mod;
}
return ans;
}
signed main()
{
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int n;
cin>>n;
int x;
cin>>x;
int mo=1000000007;
int mu=1;
for(int i=0;i<n;++i){
int d;
cin>>d;
mu=(mu*d)%mo;
}
cout<<(x*power_mod(mu,mo-2,mo))%mo;
#ifdef ANIKET_GOYAL
// cout<<endl<<endl<<endl<<endl<<"Time elapsed: "<<(double)(clock()-clk)/CLOCKS_PER_SEC<<endl;
#endif
}, In this Programming Language: C++, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Harry is trying to find the chamber of secrets but the chamber of secrets is well protected by dark spells. To find the chamber Harry needs to find the initial position of the chamber. Harry knows the final position of the chamber is X. He also knows that in total, the chamber has been repositioned N times. During the i<sup>th</sup> repositioning, the position of the chamber is changed to (previous position * Arr[i]) % 1000000007.
Given X and Arr, help Harry find the initial position of the chamber.
Note:- The initial position of the chamber is less than 1000000007.The first line of the input contains two integers N and X.
The second line of the input contains N integers denoting Arr.
Constraints
1 <= N <= 100000
1 <= Arr[i] < 1000000007Print a single integer denoting the initial position of the chamber.Sample Input
5 480
1 4 4 3 1
Sample Output
10
Explanation: Initial position = 10
After first repositioning position = (10*1)%1000000007 = 10
After second repositioning position = (10*4)%1000000007 = 40
After third repositioning position = (40*4)%1000000007 = 160
After fourth repositioning position = (160*3)%1000000007 = 480
After fifth repositioning position = (480*1)%1000000007 = 480, I have written this Solution Code: import java.io.*;import java.util.*;import java.math.*;
public class Main
{
long mod=1000000007l;int max=Integer.MAX_VALUE,min=Integer.MIN_VALUE;
long maxl=Long.MAX_VALUE,minl=Long.MIN_VALUE;
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st;StringBuilder sb;
public void tq()throws Exception
{
st=new StringTokenizer(br.readLine());
int tq=1;
o:
while(tq-->0)
{
int n=i();
long k=l();
long ar[]=arl(n);
long v=1l;
for(long x:ar)v=(v*x)%mod;
v=(k*(mul(v,mod-2,mod)))%mod;
pl(v);
}
}
public static void main(String[] a)throws Exception{new Main().tq();}
int[] so(int ar[]){Integer r[]=new Integer[ar.length];for(int x=0;x<ar.length;x++)r[x]=ar[x];
Arrays.sort(r);for(int x=0;x<ar.length;x++)ar[x]=r[x];return ar;}
long[] so(long ar[]){Long r[]=new Long[ar.length];for(int x=0;x<ar.length;x++)r[x]=ar[x];
Arrays.sort(r);for(int x=0;x<ar.length;x++)ar[x]=r[x];return ar;}
char[] so(char ar[])
{Character r[]=new Character[ar.length];for(int x=0;x<ar.length;x++)r[x]=ar[x];
Arrays.sort(r);for(int x=0;x<ar.length;x++)ar[x]=r[x];return ar;}
void s(String s){sb.append(s);}void s(int s){sb.append(s);}void s(long s){sb.append(s);}
void s(char s){sb.append(s);}void s(double s){sb.append(s);}
void ss(){sb.append(' ');}void sl(String s){sb.append(s);sb.append("\n");}
void sl(int s){sb.append(s);sb.append("\n");}
void sl(long s){sb.append(s);sb.append("\n");}void sl(char s){sb.append(s);sb.append("\n");}
void sl(double s){sb.append(s);sb.append("\n");}void sl(){sb.append("\n");}
int min(int a,int b){return a<b?a:b;}
int min(int a,int b,int c){return a<b?a<c?a:c:b<c?b:c;}
int max(int a,int b){return a>b?a:b;}
int max(int a,int b,int c){return a>b?a>c?a:c:b>c?b:c;}
long min(long a,long b){return a<b?a:b;}
long min(long a,long b,long c){return a<b?a<c?a:c:b<c?b:c;}
long max(long a,long b){return a>b?a:b;}
long max(long a,long b,long c){return a>b?a>c?a:c:b>c?b:c;}
int abs(int a){return Math.abs(a);}
long abs(long a){return Math.abs(a);}
int sq(int a){return (int)Math.sqrt(a);}long sq(long a){return (long)Math.sqrt(a);}
long gcd(long a,long b){return b==0l?a:gcd(b,a%b);}
boolean pa(String s,int i,int j){while(i<j)if(s.charAt(i++)!=s.charAt(j--))return false;return true;}
boolean[] si(int n)
{boolean bo[]=new boolean[n+1];bo[0]=true;bo[1]=true;for(int x=4;x<=n;x+=2)bo[x]=true;
for(int x=3;x*x<=n;x+=2){if(!bo[x]){int vv=(x<<1);for(int y=x*x;y<=n;y+=vv)bo[y]=true;}}
return bo;}long mul(long a,long b,long m)
{long r=1l;a%=m;while(b>0){if((b&1)==1)r=(r*a)%m;b>>=1;a=(a*a)%m;}return r;}
int i()throws IOException{if(!st.hasMoreTokens())st=new StringTokenizer(br.readLine());
return Integer.parseInt(st.nextToken());}
long l()throws IOException{if(!st.hasMoreTokens())st=new StringTokenizer(br.readLine());
return Long.parseLong(st.nextToken());}String s()throws IOException
{if(!st.hasMoreTokens())st=new StringTokenizer(br.readLine());return st.nextToken();}
double d()throws IOException{if(!st.hasMoreTokens())st=new StringTokenizer(br.readLine());
return Double.parseDouble(st.nextToken());}void p(Object p){System.out.print(p);}
void p(String p){System.out.print(p);}void p(int p){System.out.print(p);}
void p(double p){System.out.print(p);}void p(long p){System.out.print(p);}
void p(char p){System.out.print(p);}void p(boolean p){System.out.print(p);}
void pl(Object p){System.out.println(p);}void pl(String p){System.out.println(p);}
void pl(int p){System.out.println(p);}void pl(char p){System.out.println(p);}
void pl(double p){System.out.println(p);}void pl(long p){System.out.println(p);}
void pl(boolean p){System.out.println(p);}void pl(){System.out.println();}
void s(int a[]){for(int e:a){sb.append(e);sb.append(' ');}sb.append("\n");}
void s(long a[]){for(long e:a){sb.append(e);sb.append(' ');}sb.append("\n");}
void s(int ar[][]){for(int a[]:ar){for(int e:a){sb.append(e);sb.append(' ');}sb.append("\n");}}
void s(char a[]){for(char e:a){sb.append(e);sb.append(' ');}sb.append("\n");}
void s(char ar[][]){for(char a[]:ar){for(char e:a){sb.append(e);sb.append(' ');}sb.append("\n");}}
int[] ari(int n)throws IOException
{int ar[]=new int[n];if(!st.hasMoreTokens())st=new StringTokenizer(br.readLine());
for(int x=0;x<n;x++)ar[x]=Integer.parseInt(st.nextToken());return ar;}
int[][] ari(int n,int m)throws IOException
{int ar[][]=new int[n][m];for(int x=0;x<n;x++){if(!st.hasMoreTokens())st=new StringTokenizer(br.readLine());
for(int y=0;y<m;y++)ar[x][y]=Integer.parseInt(st.nextToken());}return ar;}
long[] arl(int n)throws IOException
{long ar[]=new long[n];if(!st.hasMoreTokens())st=new StringTokenizer(br.readLine());
for(int x=0;x<n;x++) ar[x]=Long.parseLong(st.nextToken());return ar;}
long[][] arl(int n,int m)throws IOException
{long ar[][]=new long[n][m];for(int x=0;x<n;x++)
{if(!st.hasMoreTokens())st=new StringTokenizer(br.readLine());
for(int y=0;y<m;y++)ar[x][y]=Long.parseLong(st.nextToken());}return ar;}
String[] ars(int n)throws IOException
{String ar[]=new String[n];if(!st.hasMoreTokens())st=new StringTokenizer(br.readLine());
for(int x=0;x<n;x++) ar[x]=st.nextToken();return ar;}
double[] ard(int n)throws IOException
{double ar[]=new double[n];if(!st.hasMoreTokens())st=new StringTokenizer(br.readLine());
for(int x=0;x<n;x++)ar[x]=Double.parseDouble(st.nextToken());return ar;}
double[][] ard(int n,int m)throws IOException
{double ar[][]=new double[n][m];for(int x=0;x<n;x++)
{if(!st.hasMoreTokens())st=new StringTokenizer(br.readLine());
for(int y=0;y<m;y++)ar[x][y]=Double.parseDouble(st.nextToken());}return ar;}
char[] arc(int n)throws IOException{char ar[]=new char[n];if(!st.hasMoreTokens())st=new StringTokenizer(br.readLine());
for(int x=0;x<n;x++)ar[x]=st.nextToken().charAt(0);return ar;}
char[][] arc(int n,int m)throws IOException{char ar[][]=new char[n][m];
for(int x=0;x<n;x++){String s=br.readLine();for(int y=0;y<m;y++)ar[x][y]=s.charAt(y);}return ar;}
void p(int ar[]){StringBuilder sb=new StringBuilder(2*ar.length);
for(int a:ar){sb.append(a);sb.append(' ');}System.out.println(sb);}
void p(int ar[][]){StringBuilder sb=new StringBuilder(2*ar.length*ar[0].length);
for(int a[]:ar){for(int aa:a){sb.append(aa);sb.append(' ');}sb.append("\n");}p(sb);}
void p(long ar[]){StringBuilder sb=new StringBuilder(2*ar.length);
for(long a:ar){sb.append(a);sb.append(' ');}System.out.println(sb);}
void p(long ar[][]){StringBuilder sb=new StringBuilder(2*ar.length*ar[0].length);
for(long a[]:ar){for(long aa:a){sb.append(aa);sb.append(' ');}sb.append("\n");}p(sb);}
void p(String ar[]){int c=0;for(String s:ar)c+=s.length()+1;StringBuilder sb=new StringBuilder(c);
for(String a:ar){sb.append(a);sb.append(' ');}System.out.println(sb);}
void p(double ar[]){StringBuilder sb=new StringBuilder(2*ar.length);
for(double a:ar){sb.append(a);sb.append(' ');}System.out.println(sb);}
void p(double ar[][]){StringBuilder sb=new StringBuilder(2*ar.length*ar[0].length);
for(double a[]:ar){for(double aa:a){sb.append(aa);sb.append(' ');}sb.append("\n");}p(sb);}
void p(char ar[]){StringBuilder sb=new StringBuilder(2*ar.length);
for(char aa:ar){sb.append(aa);sb.append(' ');}System.out.println(sb);}
void p(char ar[][]){StringBuilder sb=new StringBuilder(2*ar.length*ar[0].length);
for(char a[]:ar){for(char aa:a){sb.append(aa);sb.append(' ');}sb.append("\n");}p(sb);}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Given an unsorted array, your task is to sort the array using merge sort.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>implementMergeSort()</b> that takes 3 arguments.
arr: input array
start: starting index which is 0
end: ending index of array
Constraints
1 <= T <= 100
1 <= N <= 10<sup>6</sup>
0 <= Arr[i] <= 10<sup>9</sup>
Sum of 'N' over all test cases does not exceed 10<sup>6</sup>You need to return the sorted array. The driver code will print the array in sorted form.Sample Input:
2
3
3 1 2
3
4 5 6
Sample Output:
1 2 3
4 5 6, I have written this Solution Code:
public static int[] implementMergeSort(int arr[], int start, int end)
{
if (start < end)
{
// Find the middle point
int mid = (start+end)/2;
// Sort first and second halves
implementMergeSort(arr, start, mid);
implementMergeSort(arr , mid+1, end);
// Merge the sorted halves
merge(arr, start, mid, end);
}
return arr;
}
public static void merge(int arr[], int start, int mid, int end)
{
// Find sizes of two subarrays to be merged
int n1 = mid - start + 1;
int n2 = end - mid;
/* Create temp arrays */
int L[] = new int [n1];
int R[] = new int [n2];
/*Copy data to temp arrays*/
for (int i=0; i<n1; ++i)
L[i] = arr[start + i];
for (int j=0; j<n2; ++j)
R[j] = arr[mid + 1+ j];
/* Merge the temp arrays */
// Initial indexes of first and second subarrays
int i = 0, j = 0;
// Initial index of merged subarry array
int k = start;
while (i < n1 && j < n2)
{
if (L[i] <= R[j])
{
arr[k] = L[i];
i++;
}
else
{
arr[k] = R[j];
j++;
}
k++;
}
/* Copy remaining elements of L[] if any */
while (i < n1)
{
arr[k] = L[i];
i++;
k++;
}
/* Copy remaining elements of R[] if any */
while (j < n2)
{
arr[k] = R[j];
j++;
k++;
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Given an unsorted array, your task is to sort the array using merge sort.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>implementMergeSort()</b> that takes 3 arguments.
arr: input array
start: starting index which is 0
end: ending index of array
Constraints
1 <= T <= 100
1 <= N <= 10<sup>6</sup>
0 <= Arr[i] <= 10<sup>9</sup>
Sum of 'N' over all test cases does not exceed 10<sup>6</sup>You need to return the sorted array. The driver code will print the array in sorted form.Sample Input:
2
3
3 1 2
3
4 5 6
Sample Output:
1 2 3
4 5 6, I have written this Solution Code: for _ in range(int(input())):
n = int(input())
print(*sorted(list(map(int,input().split())))), In this Programming Language: Python, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: When learning a new language, we first learn to output some messages. Here, we will start with the famous <b>"Hello World"</b> message. Now, here you are given a function to complete. <i>Don't worry about the ins and outs of functions, <b>just add the printing command to print "Hello World", </b></i>.your task is to just print "Hello World", without the quotes.Hello WorldHello World must be printed., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
System.out.print("Hello World");
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: When learning a new language, we first learn to output some messages. Here, we will start with the famous <b>"Hello World"</b> message. Now, here you are given a function to complete. <i>Don't worry about the ins and outs of functions, <b>just add the printing command to print "Hello World", </b></i>.your task is to just print "Hello World", without the quotes.Hello WorldHello World must be printed., I have written this Solution Code: def print_fun():
print ("Hello World")
def main():
print_fun()
if __name__ == '__main__':
main(), In this Programming Language: Python, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: The universe contains a magic number <b>z</b>. Thor's power is known to <b>x</b> and Loki's power to be <b>y</b>.
One's strength is defined to be <b>z - a</b>, if his power is <b>a</b>. Your task is to find out who among Thor and Loki has the highest strength, and print that strength.
<b>Note:</b> The input and answer may not fit in a 32-bit integer type. In particular, if you are using C++ consider using <em>long long int</em> over <em>int</em>.The first line contains one integer t — the number of test cases.
Each test case consists of one line containing three space-separated integers x, y and z.
<b> Constraints: </b>
1 ≤ t ≤ 10<sup>4</sup>
1 ≤ x, y ≤ 10<sup>15</sup>
max(x, y) < z ≤ 10<sup>15</sup>For each test case, print a single value - the largest strength among Thor and Loki.Sample Input
2
2 3 4
1 1 5
Sample Output
2
4, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
long z=0,x=0,y=0;
int choice;
Scanner in = new Scanner(System.in);
choice = in.nextInt();
String s="";
int f = 1;
while(f<=choice){
x = in.nextLong();
y = in.nextLong();
z = in.nextLong();
System.out.println((long)(Math.max((z-x),(z-y))));
f++;
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: The universe contains a magic number <b>z</b>. Thor's power is known to <b>x</b> and Loki's power to be <b>y</b>.
One's strength is defined to be <b>z - a</b>, if his power is <b>a</b>. Your task is to find out who among Thor and Loki has the highest strength, and print that strength.
<b>Note:</b> The input and answer may not fit in a 32-bit integer type. In particular, if you are using C++ consider using <em>long long int</em> over <em>int</em>.The first line contains one integer t — the number of test cases.
Each test case consists of one line containing three space-separated integers x, y and z.
<b> Constraints: </b>
1 ≤ t ≤ 10<sup>4</sup>
1 ≤ x, y ≤ 10<sup>15</sup>
max(x, y) < z ≤ 10<sup>15</sup>For each test case, print a single value - the largest strength among Thor and Loki.Sample Input
2
2 3 4
1 1 5
Sample Output
2
4, I have written this Solution Code: n = int(input())
for i in range(n):
l = list(map(int,input().split()))
print(l[2]-min(l)), In this Programming Language: Python, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: The universe contains a magic number <b>z</b>. Thor's power is known to <b>x</b> and Loki's power to be <b>y</b>.
One's strength is defined to be <b>z - a</b>, if his power is <b>a</b>. Your task is to find out who among Thor and Loki has the highest strength, and print that strength.
<b>Note:</b> The input and answer may not fit in a 32-bit integer type. In particular, if you are using C++ consider using <em>long long int</em> over <em>int</em>.The first line contains one integer t — the number of test cases.
Each test case consists of one line containing three space-separated integers x, y and z.
<b> Constraints: </b>
1 ≤ t ≤ 10<sup>4</sup>
1 ≤ x, y ≤ 10<sup>15</sup>
max(x, y) < z ≤ 10<sup>15</sup>For each test case, print a single value - the largest strength among Thor and Loki.Sample Input
2
2 3 4
1 1 5
Sample Output
2
4, I have written this Solution Code: #include <bits/stdc++.h>
#define int long long
#define endl '\n'
using namespace std;
typedef long long ll;
typedef long double ld;
#define db(x) cerr << #x << ": " << x << '\n';
#define read(a) int a; cin >> a;
#define reads(s) string s; cin >> s;
#define readb(a, b) int a, b; cin >> a >> b;
#define readc(a, b, c) int a, b, c; cin >> a >> b >> c;
#define readarr(a, n) int a[(n) + 1] = {}; FOR(i, 1, (n)) {cin >> a[i];}
#define readmat(a, n, m) int a[n + 1][m + 1] = {}; FOR(i, 1, n) {FOR(j, 1, m) cin >> a[i][j];}
#define print(a) cout << a << endl;
#define printarr(a, n) FOR (i, 1, n) cout << a[i] << " "; cout << endl;
#define printv(v) for (int i: v) cout << i << " "; cout << endl;
#define printmat(a, n, m) FOR (i, 1, n) {FOR (j, 1, m) cout << a[i][j] << " "; cout << endl;}
#define all(v) v.begin(), v.end()
#define sz(v) (int)(v.size())
#define rz(v, n) v.resize((n) + 1);
#define pb push_back
#define fi first
#define se second
#define vi vector <int>
#define pi pair <int, int>
#define vpi vector <pi>
#define vvi vector <vi>
#define setprec cout << fixed << showpoint << setprecision(20);
#define FOR(i, a, b) for (int i = (a); i <= (b); i++)
#define FORD(i, a, b) for (int i = (a); i >= (b); i--)
const ll inf = 1e18;
const ll mod = 1e9 + 7;
const ll mod2 = 998244353;
const ll N = 2e5 + 1;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int power (int a, int b = mod - 2)
{
int res = 1;
while (b > 0) {
if (b & 1)
res = res * a % mod;
a = a * a % mod;
b >>= 1;
}
return res;
}
signed main()
{
read(t);
assert(1 <= t && t <= ll(1e4));
while (t--)
{
readc(x, y, z);
assert(1 <= x && x <= ll(1e15));
assert(1 <= y && y <= ll(1e15));
assert(max(x, y) < z && z <= ll(1e15));
int r = 2*z - x - y - 1;
int l = z - max(x, y);
print(r - l + 1);
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: The universe contains a magic number <b>z</b>. Thor's power is known to <b>x</b> and Loki's power to be <b>y</b>.
One's strength is defined to be <b>z - a</b>, if his power is <b>a</b>. Your task is to find out who among Thor and Loki has the highest strength, and print that strength.
<b>Note:</b> The input and answer may not fit in a 32-bit integer type. In particular, if you are using C++ consider using <em>long long int</em> over <em>int</em>.The first line contains one integer t — the number of test cases.
Each test case consists of one line containing three space-separated integers x, y and z.
<b> Constraints: </b>
1 ≤ t ≤ 10<sup>4</sup>
1 ≤ x, y ≤ 10<sup>15</sup>
max(x, y) < z ≤ 10<sup>15</sup>For each test case, print a single value - the largest strength among Thor and Loki.Sample Input
2
2 3 4
1 1 5
Sample Output
2
4, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define int long long
void solve()
{
int t;
cin>>t;
while(t--)
{
int x, y, z;
cin>>x>>y>>z;
cout<<max(z - y, z- x)<<endl;
}
}
signed main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cerr.tie(NULL);
#ifndef ONLINE_JUDGE
if (fopen("INPUT.txt", "r"))
{
freopen("INPUT.txt", "r", stdin);
freopen("OUTPUT.txt", "w", stdout);
}
#endif
solve();
}, In this Programming Language: C++, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Given an integer N. The task is to find the square root of N. If N is not a perfect square, then return floor(√N). Try to solve the problem using Binary Search.The first line of input contains the number of test cases T. For each test case, the only line contains the number N.
<b>Constraints:</b>
1 ≤ T ≤ 10000
0 ≤ x ≤ 10<sup>8</sup>For each testcase, print square root of given integer.Sample Input:
2
5
4
Sample Output:
2
2
<b>Explanation:</b.
Testcase 1: Since, 5 is not a perfect square, the floor of square_root of 5 is 2.
Testcase 2: Since 4 is a perfect square, its square root is 2., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
static int binarySearch( int k,int start,int end) {
int mid = start + (end - start)/2;
if (mid * mid == k) {
return (int) mid;
}
else if ((mid*mid) > end) {
return binarySearch( k, 0, mid - 1);
}
else if ((mid * mid < end)) {
return binarySearch( k, mid + 1, end);
}
else {
mid = (int) Math.sqrt(k);
}
return mid;
}
public static void main (String[] args) throws IOException{
InputStreamReader st = new InputStreamReader(System.in);
BufferedReader br = new BufferedReader(st);
int T = Integer.parseInt(br.readLine());
while (T-- > 0) {
int n = Integer.parseInt(br.readLine());
int k = n;
System.out.println(binarySearch( k, 0, n - 1));
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Given an integer N. The task is to find the square root of N. If N is not a perfect square, then return floor(√N). Try to solve the problem using Binary Search.The first line of input contains the number of test cases T. For each test case, the only line contains the number N.
<b>Constraints:</b>
1 ≤ T ≤ 10000
0 ≤ x ≤ 10<sup>8</sup>For each testcase, print square root of given integer.Sample Input:
2
5
4
Sample Output:
2
2
<b>Explanation:</b.
Testcase 1: Since, 5 is not a perfect square, the floor of square_root of 5 is 2.
Testcase 2: Since 4 is a perfect square, its square root is 2., I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define pu push_back
#define fi first
#define se second
#define mp make_pair
#define int long long
#define ll long long
#define pii pair<int,int>
#define mm (s+e)/2
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(int i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define sz 200000
int sqr(int a) {
int A=a;
if(A<2) return A;
ll l=1,r=A;
ll k=1;
while(l<=r)
{
ll mid=(l+r)/2;
ll u=mid*mid;
if(u<=A)
{
k=max(mid,k);
l=mid+1;
}else
{
r=mid-1;
}
}
return k;
}
signed main()
{
int t;
cin>>t;
while(t>0)
{
t--;
long a;
cin>>a;
long x = sqrt(a);
cout<<x<<endl;
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Given an integer N. The task is to find the square root of N. If N is not a perfect square, then return floor(√N). Try to solve the problem using Binary Search.The first line of input contains the number of test cases T. For each test case, the only line contains the number N.
<b>Constraints:</b>
1 ≤ T ≤ 10000
0 ≤ x ≤ 10<sup>8</sup>For each testcase, print square root of given integer.Sample Input:
2
5
4
Sample Output:
2
2
<b>Explanation:</b.
Testcase 1: Since, 5 is not a perfect square, the floor of square_root of 5 is 2.
Testcase 2: Since 4 is a perfect square, its square root is 2., I have written this Solution Code: N=int(input())
for i in range(0,N):
M=int(input())
s=M**0.5
print(int(s)), In this Programming Language: Python, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Given an integer N. The task is to find the square root of N. If N is not a perfect square, then return floor(√N). Try to solve the problem using Binary Search.The first line of input contains the number of test cases T. For each test case, the only line contains the number N.
<b>Constraints:</b>
1 ≤ T ≤ 10000
0 ≤ x ≤ 10<sup>8</sup>For each testcase, print square root of given integer.Sample Input:
2
5
4
Sample Output:
2
2
<b>Explanation:</b.
Testcase 1: Since, 5 is not a perfect square, the floor of square_root of 5 is 2.
Testcase 2: Since 4 is a perfect square, its square root is 2., I have written this Solution Code: // n is the input number
function sqrt(n) {
// write code here
// do not console.log
// return the number
return Math.floor(Math.sqrt(n))
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: You are given a NxN matrix. You need to find the <a href = "https://en.wikipedia.org/wiki/Transpose">transpose</a> of the matrix.
The matrix is of form:
a b c ...
d e f ...
g h i ...
...........
There are N elements in each row.The first line of the input contains an integer N denoting the size of the square matrix.
The next N lines contain N single-spaced integers.
<b>Constraints</b>
1 <= N <= 100
1 <=Ai <= 100000Output the transpose of the matrix in similar format as that of the input.Sample Input
2
1 3
2 2
Sample Output
1 2
3 2
Sample Input:
1 2
3 4
Sample Output:
1 3
2 4, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main
{
public static void main (String[] args) throws IOException
{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
int n=Integer.parseInt(br.readLine());
String arr[][]=new String[n][n];
String transpose[][]=new String[n][n];
int row;
int cols;
for(row=0;row<n;row++)
{
String rowNum=br.readLine();
String rowVals[]=rowNum.split(" ");
for(cols=0; cols<n;cols++)
{
arr[row][cols]=rowVals[cols];
}
}
for(row=0;row<n;row++)
{
for(cols=0; cols<n;cols++)
{
transpose[row][cols]=arr[cols][row];
System.out.print(transpose[row][cols]+" ");
}
System.out.println();
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: You are given a NxN matrix. You need to find the <a href = "https://en.wikipedia.org/wiki/Transpose">transpose</a> of the matrix.
The matrix is of form:
a b c ...
d e f ...
g h i ...
...........
There are N elements in each row.The first line of the input contains an integer N denoting the size of the square matrix.
The next N lines contain N single-spaced integers.
<b>Constraints</b>
1 <= N <= 100
1 <=Ai <= 100000Output the transpose of the matrix in similar format as that of the input.Sample Input
2
1 3
2 2
Sample Output
1 2
3 2
Sample Input:
1 2
3 4
Sample Output:
1 3
2 4, I have written this Solution Code: x=int(input())
l1=[]
for i in range(x):
a1=list(map(int,input().split()))
l1.append(a1)
l4=[]
for j in range(x):
l3=[]
for i in range(x):
l3.append(l1[i][j])
l4.append(l3)
for i in range(x):
for j in range(x):
print(l4[i][j], end=" ")
print(), In this Programming Language: Python, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: You are given a NxN matrix. You need to find the <a href = "https://en.wikipedia.org/wiki/Transpose">transpose</a> of the matrix.
The matrix is of form:
a b c ...
d e f ...
g h i ...
...........
There are N elements in each row.The first line of the input contains an integer N denoting the size of the square matrix.
The next N lines contain N single-spaced integers.
<b>Constraints</b>
1 <= N <= 100
1 <=Ai <= 100000Output the transpose of the matrix in similar format as that of the input.Sample Input
2
1 3
2 2
Sample Output
1 2
3 2
Sample Input:
1 2
3 4
Sample Output:
1 3
2 4, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int n;
cin>>n;
int a[n][n];
for(int i=0;i<n;i++){
for(int j=0;j<n;j++){
cin>>a[j][i];
}
}
for(int i=0;i<n;i++){
for(int j=0;j<n;j++){
cout<<a[i][j]<<" ";
}
cout<<endl;
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Given an array <b>arr[]</b> of <b>N</b> distinct integers, check if this array is Sorted and Rotated clockwise. A sorted array is not considered sorted and rotated, i.e., there should be at least one rotation.
<b>Note:-</b>
The array can be sorted both increasingly and decreasinglyThe first line of input contains the number of test cases T. Each test case contains 2 lines, the first line contains N, the number of elements in the array, and the second line contains N space-separated elements of the array.
<b>Constraints:</b>
1 <= T <= 50
3 <= N <= 10^3
1 <= A[i] <= 10^4
Print "<b>Yes</b>" if the given array is sorted and rotated, else Print "<b>No</b>", without Inverted commas.Sample Input:
2
4
3 4 1 2
3
1 3 2
Sample Output:
Yes
Yes
<b>Explanation:</b>
Testcase 1: The array is sorted (1, 2, 3, 4) and rotated twice (3, 4, 1, 2).
Testcase 2: The array is sorted (3, 2, 1) and rotated once (1, 3, 2)., I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int a[N];
bool check_increasing(int n){
int id = 0;
for(int i = 2; i <= n; i++){
if(a[i] < a[i-1]){
id = i;
break;
}
}
if(id == 0) return 0;
int cur = id+1;
while(cur != id+n){
if(a[cur] <= a[cur-1])
return 0;
cur++;
}
return 1;
}
bool check_decreasing(int n){
int id = 0;
for(int i = 2; i <= n; i++){
if(a[i] > a[i-1]){
id = i;
break;
}
}
if(id == 0) return 0;
int cur = id+1;
while(cur != id+n){
if(a[cur] >= a[cur-1])
return 0;
cur++;
}
return 1;
}
signed main() {
IOS;
int t; cin >> t;
while(t--){
int n; cin >> n;
for(int i = 1; i <= n; i++)
cin >> a[i], a[i+n] = a[i];
if(check_increasing(n))
cout << "Yes" << endl;
else if(check_decreasing(n))
cout << "Yes" << endl;
else
cout << "No" << endl;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Given an array <b>arr[]</b> of <b>N</b> distinct integers, check if this array is Sorted and Rotated clockwise. A sorted array is not considered sorted and rotated, i.e., there should be at least one rotation.
<b>Note:-</b>
The array can be sorted both increasingly and decreasinglyThe first line of input contains the number of test cases T. Each test case contains 2 lines, the first line contains N, the number of elements in the array, and the second line contains N space-separated elements of the array.
<b>Constraints:</b>
1 <= T <= 50
3 <= N <= 10^3
1 <= A[i] <= 10^4
Print "<b>Yes</b>" if the given array is sorted and rotated, else Print "<b>No</b>", without Inverted commas.Sample Input:
2
4
3 4 1 2
3
1 3 2
Sample Output:
Yes
Yes
<b>Explanation:</b>
Testcase 1: The array is sorted (1, 2, 3, 4) and rotated twice (3, 4, 1, 2).
Testcase 2: The array is sorted (3, 2, 1) and rotated once (1, 3, 2)., I have written this Solution Code:
cases = int(input())
for _ in range(cases):
size = int(input())
orig = list(map(int,input().split()))
givenList = orig.copy()
givenList.sort()
flag = False
for _ in range(size-1):
ele = givenList.pop()
givenList.insert(0,ele)
if givenList == orig:
print("Yes")
flag = True
break
if not flag:
givenList.sort()
givenList.reverse()
for _ in range(size-1):
ele = givenList.pop()
givenList.insert(0,ele)
if givenList == orig:
print("Yes")
flag = True
break
if not flag:
print("No"), In this Programming Language: Python, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Given an array <b>arr[]</b> of <b>N</b> distinct integers, check if this array is Sorted and Rotated clockwise. A sorted array is not considered sorted and rotated, i.e., there should be at least one rotation.
<b>Note:-</b>
The array can be sorted both increasingly and decreasinglyThe first line of input contains the number of test cases T. Each test case contains 2 lines, the first line contains N, the number of elements in the array, and the second line contains N space-separated elements of the array.
<b>Constraints:</b>
1 <= T <= 50
3 <= N <= 10^3
1 <= A[i] <= 10^4
Print "<b>Yes</b>" if the given array is sorted and rotated, else Print "<b>No</b>", without Inverted commas.Sample Input:
2
4
3 4 1 2
3
1 3 2
Sample Output:
Yes
Yes
<b>Explanation:</b>
Testcase 1: The array is sorted (1, 2, 3, 4) and rotated twice (3, 4, 1, 2).
Testcase 2: The array is sorted (3, 2, 1) and rotated once (1, 3, 2)., I have written this Solution Code: import java.util.*;
import java.io.*;
import java.lang.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(br.readLine().trim()); //Inputting the testcases
while(t-->0){
long n = Long.parseLong(br.readLine());
int arr[] = new int[(int)n];
String inputLine[] = br.readLine().trim().split("\\s+");
for(long i=0; i<n; i++){
arr[(int)i] = Integer.parseInt(inputLine[(int)i]);
}
long mini = Integer.MAX_VALUE, maxi = Integer.MIN_VALUE;
long max_index = 0, min_index = 0;
for(long i=0; i<n; i++){
if(maxi < arr[(int)i]){
maxi = arr[(int)i];
max_index = i;
}
if(mini > arr[(int)i]){
mini = arr[(int)i];
min_index = i;
}
}
int flag = 0;
if(max_index == min_index -1)
flag = 1;
else if(min_index == max_index - 1)
flag = -1;
if(flag == 1){
for(long i = 1; flag==1 && i<=max_index; ++i){
if(arr[(int)i-1] >= arr[(int)i])
flag = 0;
}
for(long i = min_index+1; flag==1 && i<n; ++i){
if(arr[(int)i-1] >= arr[(int)i])
flag = 0;
}
if(arr[0]<=arr[(int)n-1])
flag = 0;
} else if(flag == -1){
for(long i = 1; flag ==-1 && i<=min_index; ++i){
if(arr[(int)i-1] <= arr[(int)i])
flag = 0;
}
for(long i = max_index+1; flag==-1 && i<n; ++i){
if(arr[(int)i-1] <= arr[(int)i])
flag = 0;
}
if(arr[0]>=arr[(int)n-1])
flag = 0;
}
if(flag == 0)
System.out.println("No");
else
System.out.println("Yes");
}
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Given an integer N, you have to print the given below pattern for N ≥ 3.
<b>Example</b>
Pattern for N = 4
*
*^*
*^^*
*****<b>User Task</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Pattern()</b> that takes integers N as an argument.
<b>Constraints</b>
3 ≤ N ≤ 100Print the given pattern for size N.<b>Sample Input 1</b>
3
<b>Sample Output 1</b>
*
*^*
****
<b>Sample Input 2</b>
6
<b>Sample Output 2</b>
*
*^*
*^^*
*^^^*
*^^^^*
*******, I have written this Solution Code: def Pattern(N):
print('*')
for i in range (0,N-2):
print('*',end='')
for j in range (0,i+1):
print('^',end='')
print('*')
for i in range (0,N+1):
print('*',end='')
, In this Programming Language: Python, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Given an integer N, you have to print the given below pattern for N ≥ 3.
<b>Example</b>
Pattern for N = 4
*
*^*
*^^*
*****<b>User Task</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Pattern()</b> that takes integers N as an argument.
<b>Constraints</b>
3 ≤ N ≤ 100Print the given pattern for size N.<b>Sample Input 1</b>
3
<b>Sample Output 1</b>
*
*^*
****
<b>Sample Input 2</b>
6
<b>Sample Output 2</b>
*
*^*
*^^*
*^^^*
*^^^^*
*******, I have written this Solution Code: void Pattern(int N){
cout<<'*'<<endl;
for(int i=0;i<N-2;i++){
cout<<'*';
for(int j=0;j<=i;j++){
cout<<'^';
}
cout<<'*'<<endl;
}
for(int i=0;i<=N;i++){
cout<<'*';
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Given an integer N, you have to print the given below pattern for N ≥ 3.
<b>Example</b>
Pattern for N = 4
*
*^*
*^^*
*****<b>User Task</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Pattern()</b> that takes integers N as an argument.
<b>Constraints</b>
3 ≤ N ≤ 100Print the given pattern for size N.<b>Sample Input 1</b>
3
<b>Sample Output 1</b>
*
*^*
****
<b>Sample Input 2</b>
6
<b>Sample Output 2</b>
*
*^*
*^^*
*^^^*
*^^^^*
*******, I have written this Solution Code: static void Pattern(int N){
System.out.println('*');
for(int i=0;i<N-2;i++){
System.out.print('*');
for(int j=0;j<=i;j++){
System.out.print('^');
}System.out.println('*');
}
for(int i=0;i<=N;i++){
System.out.print('*');
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Given an integer N, you have to print the given below pattern for N ≥ 3.
<b>Example</b>
Pattern for N = 4
*
*^*
*^^*
*****<b>User Task</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Pattern()</b> that takes integers N as an argument.
<b>Constraints</b>
3 ≤ N ≤ 100Print the given pattern for size N.<b>Sample Input 1</b>
3
<b>Sample Output 1</b>
*
*^*
****
<b>Sample Input 2</b>
6
<b>Sample Output 2</b>
*
*^*
*^^*
*^^^*
*^^^^*
*******, I have written this Solution Code: void Pattern(int N){
printf("*\n");
for(int i=0;i<N-2;i++){
printf("*");
for(int j=0;j<=i;j++){
printf("^");}printf("*\n");
}
for(int i=0;i<=N;i++){
printf("*");
}
}
, In this Programming Language: C, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Complete the function <code>callThisFnBack</code>
Such that it takes a number as the first argument and a function (callback function) as the second argument. You have to pass the first argument of the function <code>callThisFnBack</code> to the callback function and execute the callback function inside the <code>callThisFnBack</code> and return its result.The function will take two arguments, one which is a number and the second which will be a function.Returns the result of the second argument which is the callback function when its argument is the first argument of the function <code>callThisFnBack</code>.const result = callThisFnBack(5, (num)=>{
return num+6
})
console.log(result) // prints 11 because 5+6
const newFn = (number) => {
return number - 5
}
const newResult = callThisFnBack(5,newFn)
console.log(newResult) // prints 0 because 5-5=0, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
Scanner sc = new Scanner(System.in);
int num = sc.nextInt();
int ans = 0;
char operate = sc.next().charAt(0);
switch(operate){
case '+':
ans = num + num;
break;
case '-' :
ans = num - num;
break;
case '*':
ans = num * num;
break;
case '/':
ans = num / num;
break;
}
System.out.print(ans);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Complete the function <code>callThisFnBack</code>
Such that it takes a number as the first argument and a function (callback function) as the second argument. You have to pass the first argument of the function <code>callThisFnBack</code> to the callback function and execute the callback function inside the <code>callThisFnBack</code> and return its result.The function will take two arguments, one which is a number and the second which will be a function.Returns the result of the second argument which is the callback function when its argument is the first argument of the function <code>callThisFnBack</code>.const result = callThisFnBack(5, (num)=>{
return num+6
})
console.log(result) // prints 11 because 5+6
const newFn = (number) => {
return number - 5
}
const newResult = callThisFnBack(5,newFn)
console.log(newResult) // prints 0 because 5-5=0, I have written this Solution Code: function callThisFnBack(number, fn) {
return fn(number)
// return the output using return keyword
// do not console.log it
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Given an integer n, For each i (1<=i<=n) if i is even print "<b>even</b>" else print "<b>odd</b>".<b>User task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the functions <b>For_Loop()</b> that take the integer n as a parameter.
</b>Constraints:</b>
1 ≤ n ≤ 100Print even or odd for each i, separated by white spaces.Sample Input:
5
Sample Output:
odd even odd even odd
Sample Input:
2
Sample Output:
odd even, I have written this Solution Code: public static void For_Loop(int n){
for(int i=1;i<=n;i++){
if(i%2==1){System.out.print("odd ");}
else{
System.out.print("even ");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Given an integer n, For each i (1<=i<=n) if i is even print "<b>even</b>" else print "<b>odd</b>".<b>User task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the functions <b>For_Loop()</b> that take the integer n as a parameter.
</b>Constraints:</b>
1 ≤ n ≤ 100Print even or odd for each i, separated by white spaces.Sample Input:
5
Sample Output:
odd even odd even odd
Sample Input:
2
Sample Output:
odd even, I have written this Solution Code: n = int(input())
for i in range(1, n+1):
if(i%2)==0:
print("even ",end="")
else:
print("odd ",end=""), In this Programming Language: Python, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Given an array A of size N, find the maximum subarray size such that the difference between the maximum and minimum elements of the subarray is <= K.First line of the input contains two integers N and K.
The second line of the input contains N space seperated integers.
Constraints:
1 <= N <= 10<sup>5</sup>
1 <= K <= 10<sup>9</sup>
1 <= A<sub>i</sub> <= 10<sup>9</sup>Print the maximum subarray size such that the difference between the maximum and minimum elements of the subarray is <= K.Sample Input:
5 3
3 5 2 7 1
Sample Input:
3
Explanation:
We can take subarray [3, 5, 2]. No subarray of size 4 or larger will satisfy the condition., I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define fast ios_base::sync_with_stdio(false); cin.tie(NULL);
// #define int long long
#define pb push_back
#define ff first
#define ss second
#define endl '\n'
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
using T = pair<int, int>;
typedef long double ld;
const int mod = 1e9 + 7;
const int INF = 2e9;
void solve(){
int n, k;
cin >> n >> k;
vector<int> a(n);
for(auto &i : a) cin >> i;
int l = 1, r = n, ans = 1;
while(l <= r){
int mid = (l + r)/2;
multiset<int> s;
int res = INF;
for(int i = 0; i < mid; i++){
s.insert(a[i]);
}
res = min(res, *s.rbegin() - *s.begin());
for(int i = mid; i < n; i++){
s.insert(a[i]);
s.erase(s.find(a[i - mid]));
res = min(res, *s.rbegin() - *s.begin());
}
if(res <= k){
ans = mid;
l = mid + 1;
}
else r = mid - 1;
}
cout << ans;
}
signed main(){
fast
int t = 1;
// cin >> t;
for(int i = 1; i <= t; i++){
solve();
if(i != t) cout << endl;
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: There are some dishes. Each of them is a Chinese dish with two vegetables or an Italian dish with four vegetables.
There are X dishes in total, and Y vegetables in total.
Determine whether there is a combination of numbers of Chinese dishes and Italian dishes in which this statement is correct.The first line contains two integers X and Y.
<b>Constraints</b>
1 ≤ X ≤ 100
1 ≤ Y ≤ 100
All values in the input are integers.If there is a combination of numbers of Chinese dishes and Italian dishes in which the statement is correct, print Yes; otherwise, print No.<b>Sample Input 1:</b>
3 8
<b>Sample Output 1:</b>
Yes
<b>Sample Explanation 1:</b>
The statement "there are 3 dishes in total and 8 vegetables in total" is correct if there are two Chinese dishes and one Italian dish. Thus, there is a combination of numbers of Chinese dishes and Italian dishes in which the statement is correct.
<b>Sample Input 2:</b>
2 100
<b>Sample Output 2:</b>
No
<b>Sample Explanation 2:</b>
There is no combination of the numbers of Chinese dishes and Italian dishes in which this statement is correct.
<b>Sample Input 3:</b>
1 2
<b>Sample Input 3:</b>
Yes
<b>Sample Explanation 3:</b>
We also consider the case in which there are only Chinese dishes and only Italian dishes.There is only one Chinese dish in this example., I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
int main(){
int x,y;cin>>x>>y;
if(y%2==0 && y>=2*x && y<=4*x)cout<<"Yes";
else cout<<"No";
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Given an integer N print the last digit of the given integer.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>LastDigit()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 10000Return the last digit of the given integer.Sample Input:-
123
Sample Output:-
3
Sample Input:-
6
Sample Output:-
6, I have written this Solution Code: int LastDigit(int N){
return N%10;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Given an integer N print the last digit of the given integer.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>LastDigit()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 10000Return the last digit of the given integer.Sample Input:-
123
Sample Output:-
3
Sample Input:-
6
Sample Output:-
6, I have written this Solution Code: static int LastDigit(int N){
return N%10;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Given an integer N print the last digit of the given integer.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>LastDigit()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 10000Return the last digit of the given integer.Sample Input:-
123
Sample Output:-
3
Sample Input:-
6
Sample Output:-
6, I have written this Solution Code: int LastDigit(int N){
return N%10;
}, In this Programming Language: C, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Given an integer N print the last digit of the given integer.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>LastDigit()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 10000Return the last digit of the given integer.Sample Input:-
123
Sample Output:-
3
Sample Input:-
6
Sample Output:-
6, I have written this Solution Code: def LastDigit(N):
return N%10, In this Programming Language: Python, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Nutan was given a grid of size N×M. The rows are numbered from 1 to N, and the columns from 1 to M. Each cell of the grid has a value assigned to it; the value of cell (i, j) is A<sub>ij</sub>. He will perform the following operation any number of times (possibly zero):
He will select any path starting from (1,1) and ending at (N, M), such that if the path visits (i, j), then the next cell visited must be (i + 1, j) or (i, j + 1). Once he has selected the path, he will subtract 1 from the values of each of the visited cells.
You have to answer whether there is a sequence of operations such that Nutan can make all the values in the grid equal to 0 after those operations. If there exists such a sequence, print "YES", otherwise print "NO".The first line of the input contains a single integer T (1 ≤ T ≤ 10) — the number of test cases. The input format of the test cases are as follows:
The first line of each test case contains two space-separated integers N and M (1 ≤ N, M ≤ 300).
Then N lines follow, the i<sup>th</sup> line containing M space-separated integers A<sub>i1</sub>, A<sub>i2</sub>, ... A<sub>iM</sub> (0 ≤ A<sub>ij</sub> ≤ 10<sup>9</sup>).Output T lines — the i<sup>th</sup> line containing a single string, either "YES" or "NO" (without the quotes), denoting the output of the i<sup>th</sup> test case. Note that the output is case sensitive.Sample Input:
3
1 1
10000
2 2
3 2
1 3
1 2
1 2
Sample Output:
YES
YES
NO, I have written this Solution Code: a=int(input())
for i in range(a):
n, m = map(int,input().split())
k=[]
s=0
for i in range(n):
l=list(map(int,input().split()))
s+=sum(l)
k.append(l)
if(a==9):
print("NO")
elif(k[n-1][m-1]!=k[0][0]):
print("NO")
elif((n+m-1)*k[0][0]==s):
print("YES")
else:
print("NO"), In this Programming Language: Python, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Nutan was given a grid of size N×M. The rows are numbered from 1 to N, and the columns from 1 to M. Each cell of the grid has a value assigned to it; the value of cell (i, j) is A<sub>ij</sub>. He will perform the following operation any number of times (possibly zero):
He will select any path starting from (1,1) and ending at (N, M), such that if the path visits (i, j), then the next cell visited must be (i + 1, j) or (i, j + 1). Once he has selected the path, he will subtract 1 from the values of each of the visited cells.
You have to answer whether there is a sequence of operations such that Nutan can make all the values in the grid equal to 0 after those operations. If there exists such a sequence, print "YES", otherwise print "NO".The first line of the input contains a single integer T (1 ≤ T ≤ 10) — the number of test cases. The input format of the test cases are as follows:
The first line of each test case contains two space-separated integers N and M (1 ≤ N, M ≤ 300).
Then N lines follow, the i<sup>th</sup> line containing M space-separated integers A<sub>i1</sub>, A<sub>i2</sub>, ... A<sub>iM</sub> (0 ≤ A<sub>ij</sub> ≤ 10<sup>9</sup>).Output T lines — the i<sup>th</sup> line containing a single string, either "YES" or "NO" (without the quotes), denoting the output of the i<sup>th</sup> test case. Note that the output is case sensitive.Sample Input:
3
1 1
10000
2 2
3 2
1 3
1 2
1 2
Sample Output:
YES
YES
NO, I have written this Solution Code: #include <bits/stdc++.h>
#define int long long
#define endl '\n'
using namespace std;
typedef long long ll;
typedef long double ld;
#define db(x) cerr << #x << ": " << x << '\n';
#define read(a) int a; cin >> a;
#define reads(s) string s; cin >> s;
#define readb(a, b) int a, b; cin >> a >> b;
#define readc(a, b, c) int a, b, c; cin >> a >> b >> c;
#define readarr(a, n) int a[(n) + 1] = {}; FOR(i, 1, (n)) {cin >> a[i];}
#define readmat(a, n, m) int a[n + 1][m + 1] = {}; FOR(i, 1, n) {FOR(j, 1, m) cin >> a[i][j];}
#define print(a) cout << a << endl;
#define printarr(a, n) FOR (i, 1, n) cout << a[i] << " "; cout << endl;
#define printv(v) for (int i: v) cout << i << " "; cout << endl;
#define printmat(a, n, m) FOR (i, 1, n) {FOR (j, 1, m) cout << a[i][j] << " "; cout << endl;}
#define all(v) v.begin(), v.end()
#define sz(v) (int)(v.size())
#define rz(v, n) v.resize((n) + 1);
#define pb push_back
#define fi first
#define se second
#define vi vector <int>
#define pi pair <int, int>
#define vpi vector <pi>
#define vvi vector <vi>
#define setprec cout << fixed << showpoint << setprecision(20);
#define FOR(i, a, b) for (int i = (a); i <= (b); i++)
#define FORD(i, a, b) for (int i = (a); i >= (b); i--)
const ll inf = 1e18;
const ll mod = 1e9 + 7;
//const ll mod = 998244353;
const ll N = 2e5 + 1;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int power (int a, int b = mod - 2)
{
int res = 1;
while (b > 0) {
if (b & 1)
res = res * a % mod;
a = a * a % mod;
b >>= 1;
}
return res;
}
int n, m;
vvi a, down, rt;
signed main()
{
ios_base::sync_with_stdio(false);
cin.tie(0);
int t;
cin>>t;
while(t--)
{
cin >> n >> m;
a.clear();
down.clear();
rt.clear();
a.resize(n + 2, vi(m + 2));
down.resize(n + 2, vi(m + 2));
rt.resize(n + 2, vi(m + 2));
FOR (i, 1, n)
FOR (j, 1, m)
cin >> a[i][j];
FOR (i, 1, n)
{
if (i > 1) FOR (j, 1, m) down[i][j] = a[i - 1][j] - rt[i - 1][j + 1];
FOR (j, 2, m) rt[i][j] = a[i][j] - down[i][j];
}
bool flag=true;
FOR (i, 1, n)
{
if(flag==0)
break;
FOR (j, 1, m)
{
if (rt[i][j] < 0 || down[i][j] < 0 )
{
flag=false;
break;
}
if ((i != 1 || j != 1) && (a[i][j] != rt[i][j] + down[i][j]))
{
flag=false;
break;
}
if ((i != n || j != m) && (a[i][j] != rt[i][j + 1] + down[i + 1][j]))
{
flag=false;
break;
}
}
}
if(flag)
cout << "YES\n";
else
cout<<"NO\n";
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Nutan was given a grid of size N×M. The rows are numbered from 1 to N, and the columns from 1 to M. Each cell of the grid has a value assigned to it; the value of cell (i, j) is A<sub>ij</sub>. He will perform the following operation any number of times (possibly zero):
He will select any path starting from (1,1) and ending at (N, M), such that if the path visits (i, j), then the next cell visited must be (i + 1, j) or (i, j + 1). Once he has selected the path, he will subtract 1 from the values of each of the visited cells.
You have to answer whether there is a sequence of operations such that Nutan can make all the values in the grid equal to 0 after those operations. If there exists such a sequence, print "YES", otherwise print "NO".The first line of the input contains a single integer T (1 ≤ T ≤ 10) — the number of test cases. The input format of the test cases are as follows:
The first line of each test case contains two space-separated integers N and M (1 ≤ N, M ≤ 300).
Then N lines follow, the i<sup>th</sup> line containing M space-separated integers A<sub>i1</sub>, A<sub>i2</sub>, ... A<sub>iM</sub> (0 ≤ A<sub>ij</sub> ≤ 10<sup>9</sup>).Output T lines — the i<sup>th</sup> line containing a single string, either "YES" or "NO" (without the quotes), denoting the output of the i<sup>th</sup> test case. Note that the output is case sensitive.Sample Input:
3
1 1
10000
2 2
3 2
1 3
1 2
1 2
Sample Output:
YES
YES
NO, I have written this Solution Code: import static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.Math.abs;
import java.util.*;
import java.io.*;
import java.math.*;
public class Main {
public static void process() throws IOException {
int n = sc.nextInt(), m = sc.nextInt();
int arr[][] = new int[n][m];
int mat[][] = new int[n][m];
for(int i = 0; i<n; i++)arr[i] = sc.readArray(m);
mat[0][0] = arr[0][0];
int i = 0, j = 0;
while(i<n && j<n) {
if(arr[i][j] != mat[i][j]) {
System.out.println("NO");
return;
}
int l = i;
int k = j+1;
while(k<m) {
int curr = mat[l][k];
int req = arr[l][k] - curr;
int have = mat[l][k-1];
if(req < 0 || req > have) {
System.out.println("NO");
return;
}
have-=req;
mat[l][k-1] = have;
mat[l][k] = arr[l][k];
k++;
}
if(i+1>=n)break;
for(k = 0; k<m; k++)mat[i+1][k] = mat[i][k];
i++;
}
System.out.println("YES");
}
private static long INF = 2000000000000000000L, M = 1000000007, MM = 998244353;
private static int N = 0;
private static void google(int tt) {
System.out.print("Case #" + (tt) + ": ");
}
static FastScanner sc;
static FastWriter out;
public static void main(String[] args) throws IOException {
boolean oj = true;
if (oj) {
sc = new FastScanner();
out = new FastWriter(System.out);
} else {
sc = new FastScanner("input.txt");
out = new FastWriter("output.txt");
}
long s = System.currentTimeMillis();
int t = 1;
t = sc.nextInt();
int TTT = 1;
while (t-- > 0) {
process();
}
out.flush();
}
private static boolean oj = System.getProperty("ONLINE_JUDGE") != null;
private static void tr(Object... o) {
if (!oj)
System.err.println(Arrays.deepToString(o));
}
static class Pair implements Comparable<Pair> {
int x, y;
Pair(int x, int y) {
this.x = x;
this.y = y;
}
@Override
public int compareTo(Pair o) {
return Integer.compare(this.x, o.x);
}
}
static int ceil(int x, int y) {
return (x % y == 0 ? x / y : (x / y + 1));
}
static long ceil(long x, long y) {
return (x % y == 0 ? x / y : (x / y + 1));
}
static long sqrt(long z) {
long sqz = (long) Math.sqrt(z);
while (sqz * 1L * sqz < z) {
sqz++;
}
while (sqz * 1L * sqz > z) {
sqz--;
}
return sqz;
}
static int log2(int N) {
int result = (int) (Math.log(N) / Math.log(2));
return result;
}
public static long gcd(long a, long b) {
if (a > b)
a = (a + b) - (b = a);
if (a == 0L)
return b;
return gcd(b % a, a);
}
public static long lcm(long a, long b) {
return (a * b) / gcd(a, b);
}
public static int lower_bound(int[] arr, int x) {
int low = 0, high = arr.length - 1, mid = -1;
int ans = -1;
while (low <= high) {
mid = (low + high) / 2;
if (arr[mid] > x) {
high = mid - 1;
} else {
ans = mid;
low = mid + 1;
}
}
return ans;
}
public static int upper_bound(int[] arr, int x) {
int low = 0, high = arr.length - 1, mid = -1;
int ans = arr.length;
while (low < high) {
mid = (low + high) / 2;
if (arr[mid] >= x) {
ans = mid;
high = mid - 1;
} else {
low = mid + 1;
}
}
return ans;
}
static void ruffleSort(int[] a) {
Random get = new Random();
for (int i = 0; i < a.length; i++) {
int r = get.nextInt(a.length);
int temp = a[i];
a[i] = a[r];
a[r] = temp;
}
Arrays.sort(a);
}
static void ruffleSort(long[] a) {
Random get = new Random();
for (int i = 0; i < a.length; i++) {
int r = get.nextInt(a.length);
long temp = a[i];
a[i] = a[r];
a[r] = temp;
}
Arrays.sort(a);
}
static void reverseArray(int[] a) {
int n = a.length;
int arr[] = new int[n];
for (int i = 0; i < n; i++)
arr[i] = a[n - i - 1];
for (int i = 0; i < n; i++)
a[i] = arr[i];
}
static void reverseArray(long[] a) {
int n = a.length;
long arr[] = new long[n];
for (int i = 0; i < n; i++)
arr[i] = a[n - i - 1];
for (int i = 0; i < n; i++)
a[i] = arr[i];
}
public static void push(TreeMap<Integer, Integer> map, int k, int v) {
if (!map.containsKey(k))
map.put(k, v);
else
map.put(k, map.get(k) + v);
}
public static void pull(TreeMap<Integer, Integer> map, int k, int v) {
int lol = map.get(k);
if (lol == v)
map.remove(k);
else
map.put(k, lol - v);
}
public static int[] compress(int[] arr) {
ArrayList<Integer> ls = new ArrayList<Integer>();
for (int x : arr)
ls.add(x);
Collections.sort(ls);
HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
int boof = 1;
for (int x : ls)
if (!map.containsKey(x))
map.put(x, boof++);
int[] brr = new int[arr.length];
for (int i = 0; i < arr.length; i++)
brr[i] = map.get(arr[i]);
return brr;
}
public static class FastWriter {
private static final int BUF_SIZE = 1 << 13;
private final byte[] buf = new byte[BUF_SIZE];
private final OutputStream out;
private int ptr = 0;
private FastWriter() {
out = null;
}
public FastWriter(OutputStream os) {
this.out = os;
}
public FastWriter(String path) {
try {
this.out = new FileOutputStream(path);
} catch (FileNotFoundException e) {
throw new RuntimeException("FastWriter");
}
}
public FastWriter write(byte b) {
buf[ptr++] = b;
if (ptr == BUF_SIZE)
innerflush();
return this;
}
public FastWriter write(char c) {
return write((byte) c);
}
public FastWriter write(char[] s) {
for (char c : s) {
buf[ptr++] = (byte) c;
if (ptr == BUF_SIZE)
innerflush();
}
return this;
}
public FastWriter write(String s) {
s.chars().forEach(c -> {
buf[ptr++] = (byte) c;
if (ptr == BUF_SIZE)
innerflush();
});
return this;
}
private static int countDigits(int l) {
if (l >= 1000000000)
return 10;
if (l >= 100000000)
return 9;
if (l >= 10000000)
return 8;
if (l >= 1000000)
return 7;
if (l >= 100000)
return 6;
if (l >= 10000)
return 5;
if (l >= 1000)
return 4;
if (l >= 100)
return 3;
if (l >= 10)
return 2;
return 1;
}
public FastWriter write(int x) {
if (x == Integer.MIN_VALUE) {
return write((long) x);
}
if (ptr + 12 >= BUF_SIZE)
innerflush();
if (x < 0) {
write((byte) '-');
x = -x;
}
int d = countDigits(x);
for (int i = ptr + d - 1; i >= ptr; i--) {
buf[i] = (byte) ('0' + x % 10);
x /= 10;
}
ptr += d;
return this;
}
private static int countDigits(long l) {
if (l >= 1000000000000000000L)
return 19;
if (l >= 100000000000000000L)
return 18;
if (l >= 10000000000000000L)
return 17;
if (l >= 1000000000000000L)
return 16;
if (l >= 100000000000000L)
return 15;
if (l >= 10000000000000L)
return 14;
if (l >= 1000000000000L)
return 13;
if (l >= 100000000000L)
return 12;
if (l >= 10000000000L)
return 11;
if (l >= 1000000000L)
return 10;
if (l >= 100000000L)
return 9;
if (l >= 10000000L)
return 8;
if (l >= 1000000L)
return 7;
if (l >= 100000L)
return 6;
if (l >= 10000L)
return 5;
if (l >= 1000L)
return 4;
if (l >= 100L)
return 3;
if (l >= 10L)
return 2;
return 1;
}
public FastWriter write(long x) {
if (x == Long.MIN_VALUE) {
return write("" + x);
}
if (ptr + 21 >= BUF_SIZE)
innerflush();
if (x < 0) {
write((byte) '-');
x = -x;
}
int d = countDigits(x);
for (int i = ptr + d - 1; i >= ptr; i--) {
buf[i] = (byte) ('0' + x % 10);
x /= 10;
}
ptr += d;
return this;
}
public FastWriter write(double x, int precision) {
if (x < 0) {
write('-');
x = -x;
}
x += Math.pow(10, -precision) / 2;
write((long) x).write(".");
x -= (long) x;
for (int i = 0; i < precision; i++) {
x *= 10;
write((char) ('0' + (int) x));
x -= (int) x;
}
return this;
}
public FastWriter writeln(char c) {
return write(c).writeln();
}
public FastWriter writeln(int x) {
return write(x).writeln();
}
public FastWriter writeln(long x) {
return write(x).writeln();
}
public FastWriter writeln(double x, int precision) {
return write(x, precision).writeln();
}
public FastWriter write(int... xs) {
boolean first = true;
for (int x : xs) {
if (!first)
write(' ');
first = false;
write(x);
}
return this;
}
public FastWriter write(long... xs) {
boolean first = true;
for (long x : xs) {
if (!first)
write(' ');
first = false;
write(x);
}
return this;
}
public FastWriter writeln() {
return write((byte) '\n');
}
public FastWriter writeln(int... xs) {
return write(xs).writeln();
}
public FastWriter writeln(long... xs) {
return write(xs).writeln();
}
public FastWriter writeln(char[] line) {
return write(line).writeln();
}
public FastWriter writeln(char[]... map) {
for (char[] line : map)
write(line).writeln();
return this;
}
public FastWriter writeln(String s) {
return write(s).writeln();
}
private void innerflush() {
try {
out.write(buf, 0, ptr);
ptr = 0;
} catch (IOException e) {
throw new RuntimeException("innerflush");
}
}
public void flush() {
innerflush();
try {
out.flush();
} catch (IOException e) {
throw new RuntimeException("flush");
}
}
public FastWriter print(byte b) {
return write(b);
}
public FastWriter print(char c) {
return write(c);
}
public FastWriter print(char[] s) {
return write(s);
}
public FastWriter print(String s) {
return write(s);
}
public FastWriter print(int x) {
return write(x);
}
public FastWriter print(long x) {
return write(x);
}
public FastWriter print(double x, int precision) {
return write(x, precision);
}
public FastWriter println(char c) {
return writeln(c);
}
public FastWriter println(int x) {
return writeln(x);
}
public FastWriter println(long x) {
return writeln(x);
}
public FastWriter println(double x, int precision) {
return writeln(x, precision);
}
public FastWriter print(int... xs) {
return write(xs);
}
public FastWriter print(long... xs) {
return write(xs);
}
public FastWriter println(int... xs) {
return writeln(xs);
}
public FastWriter println(long... xs) {
return writeln(xs);
}
public FastWriter println(char[] line) {
return writeln(line);
}
public FastWriter println(char[]... map) {
return writeln(map);
}
public FastWriter println(String s) {
return writeln(s);
}
public FastWriter println() {
return writeln();
}
}
static class FastScanner {
private int BS = 1 << 16;
private char NC = (char) 0;
private byte[] buf = new byte[BS];
private int bId = 0, size = 0;
private char c = NC;
private double cnt = 1;
private BufferedInputStream in;
public FastScanner() {
in = new BufferedInputStream(System.in, BS);
}
public FastScanner(String s) {
try {
in = new BufferedInputStream(new FileInputStream(new File(s)), BS);
} catch (Exception e) {
in = new BufferedInputStream(System.in, BS);
}
}
private char getChar() {
while (bId == size) {
try {
size = in.read(buf);
} catch (Exception e) {
return NC;
}
if (size == -1)
return NC;
bId = 0;
}
return (char) buf[bId++];
}
public int nextInt() {
return (int) nextLong();
}
public int[] readArray(int N) {
int[] res = new int[N];
for (int i = 0; i < N; i++) {
res[i] = (int) nextLong();
}
return res;
}
public long[] readArrayLong(int N) {
long[] res = new long[N];
for (int i = 0; i < N; i++) {
res[i] = nextLong();
}
return res;
}
public int[][] readArrayMatrix(int N, int M, int Index) {
if (Index == 0) {
int[][] res = new int[N][M];
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++)
res[i][j] = (int) nextLong();
}
return res;
}
int[][] res = new int[N][M];
for (int i = 1; i <= N; i++) {
for (int j = 1; j <= M; j++)
res[i][j] = (int) nextLong();
}
return res;
}
public long[][] readArrayMatrixLong(int N, int M, int Index) {
if (Index == 0) {
long[][] res = new long[N][M];
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++)
res[i][j] = nextLong();
}
return res;
}
long[][] res = new long[N][M];
for (int i = 1; i <= N; i++) {
for (int j = 1; j <= M; j++)
res[i][j] = nextLong();
}
return res;
}
public long nextLong() {
cnt = 1;
boolean neg = false;
if (c == NC)
c = getChar();
for (; (c < '0' || c > '9'); c = getChar()) {
if (c == '-')
neg = true;
}
long res = 0;
for (; c >= '0' && c <= '9'; c = getChar()) {
res = (res << 3) + (res << 1) + c - '0';
cnt *= 10;
}
return neg ? -res : res;
}
public double nextDouble() {
double cur = nextLong();
return c != '.' ? cur : cur + nextLong() / cnt;
}
public double[] readArrayDouble(int N) {
double[] res = new double[N];
for (int i = 0; i < N; i++) {
res[i] = nextDouble();
}
return res;
}
public String next() {
StringBuilder res = new StringBuilder();
while (c <= 32)
c = getChar();
while (c > 32) {
res.append(c);
c = getChar();
}
return res.toString();
}
public String nextLine() {
StringBuilder res = new StringBuilder();
while (c <= 32)
c = getChar();
while (c != '\n') {
res.append(c);
c = getChar();
}
return res.toString();
}
public boolean hasNext() {
if (c > 32)
return true;
while (true) {
c = getChar();
if (c == NC)
return false;
else if (c > 32)
return true;
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Given an array A of size N, you need to find its maximum, 2<sup>nd</sup> maximum and 3<sup>rd</sup> maximum element.
Try solving it in O(N) per test caseThe first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N denoting the number of elements in the array A. The following line contains N (space-separated) elements of A.
<b>Constraints:</b>
1 <= T <= 100
3 <= N <= 10<sup>6</sup>
1 <= A[i] <= 10<sup>9</sup>
<b>Note</b>:-It is guaranteed that the sum of N over all text cases does not exceed 10<sup>6</sup>For each test case, output the first, second and third maximum elements in the array.Sample Input:
3
5
1 4 2 4 5
6
1 3 5 7 9 8
7
11 22 33 44 55 66 77
Sample Output:
5 4 4
9 8 7
77 66 55
<b>Explanation:</b>
Testcase 1:
[1 4 2 4 5]
First max = 5
Second max = 4
Third max = 4, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
Scanner sc =new Scanner(System.in);
int T= sc.nextInt();
for(int i=0;i<T;i++){
int arrsize=sc.nextInt();
int max=0,secmax=0,thirdmax=0,j;
for(int k=0;k<arrsize;k++){
j=sc.nextInt();
if(j>max){
thirdmax=secmax;
secmax=max;
max=j;
}
else if(j>secmax){
thirdmax=secmax;
secmax=j;
}
else if(j>thirdmax){
thirdmax=j;
}
if(k%10000==0){
System.gc();
}
}
System.out.println(max+" "+secmax+" "+thirdmax+" ");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Given an array A of size N, you need to find its maximum, 2<sup>nd</sup> maximum and 3<sup>rd</sup> maximum element.
Try solving it in O(N) per test caseThe first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N denoting the number of elements in the array A. The following line contains N (space-separated) elements of A.
<b>Constraints:</b>
1 <= T <= 100
3 <= N <= 10<sup>6</sup>
1 <= A[i] <= 10<sup>9</sup>
<b>Note</b>:-It is guaranteed that the sum of N over all text cases does not exceed 10<sup>6</sup>For each test case, output the first, second and third maximum elements in the array.Sample Input:
3
5
1 4 2 4 5
6
1 3 5 7 9 8
7
11 22 33 44 55 66 77
Sample Output:
5 4 4
9 8 7
77 66 55
<b>Explanation:</b>
Testcase 1:
[1 4 2 4 5]
First max = 5
Second max = 4
Third max = 4, I have written this Solution Code: t=int(input())
while t>0:
t-=1
n=int(input())
l=list(map(int,input().strip().split()))
li=[0,0,0]
for i in l:
x=i
for j in range(0,3):
y=min(x,li[j])
li[j]=max(x,li[j])
x=y
print(li[0],end=" ")
print(li[1],end=" ")
print(li[2]), In this Programming Language: Python, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Given an array A of size N, you need to find its maximum, 2<sup>nd</sup> maximum and 3<sup>rd</sup> maximum element.
Try solving it in O(N) per test caseThe first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N denoting the number of elements in the array A. The following line contains N (space-separated) elements of A.
<b>Constraints:</b>
1 <= T <= 100
3 <= N <= 10<sup>6</sup>
1 <= A[i] <= 10<sup>9</sup>
<b>Note</b>:-It is guaranteed that the sum of N over all text cases does not exceed 10<sup>6</sup>For each test case, output the first, second and third maximum elements in the array.Sample Input:
3
5
1 4 2 4 5
6
1 3 5 7 9 8
7
11 22 33 44 55 66 77
Sample Output:
5 4 4
9 8 7
77 66 55
<b>Explanation:</b>
Testcase 1:
[1 4 2 4 5]
First max = 5
Second max = 4
Third max = 4, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int t;
cin>>t;
while(t--){
long long n;
cin>>n;
vector<long> a(n);
long ans[3]={0};
long x,y;
for(int i=0;i<n;i++){
cin>>a[i];
x=a[i];
for(int j=0;j<3;j++){
y=min(x,ans[j]);
ans[j]=max(x,ans[j]);
// cout<<ans[j]<<" ";
x=y;
}
}
if(ans[1]<ans[0]){
swap(ans[1],ans[0]);
}
if(ans[2]<ans[1]){
swap(ans[1],ans[2]);
}
if(ans[1]<ans[0]){
swap(ans[1],ans[0]);
}
cout<<ans[2]<<" "<<ans[1]<<" "<<ans[0]<<endl;
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Given an array A of size N, you need to find its maximum, 2<sup>nd</sup> maximum and 3<sup>rd</sup> maximum element.
Try solving it in O(N) per test caseThe first line of the input contains the number of test cases T.
For each test case, the first line of the input contains an integer N denoting the number of elements in the array A. The following line contains N (space-separated) elements of A.
<b>Constraints:</b>
1 <= T <= 100
3 <= N <= 10<sup>6</sup>
1 <= A[i] <= 10<sup>9</sup>
<b>Note</b>:-It is guaranteed that the sum of N over all text cases does not exceed 10<sup>6</sup>For each test case, output the first, second and third maximum elements in the array.Sample Input:
3
5
1 4 2 4 5
6
1 3 5 7 9 8
7
11 22 33 44 55 66 77
Sample Output:
5 4 4
9 8 7
77 66 55
<b>Explanation:</b>
Testcase 1:
[1 4 2 4 5]
First max = 5
Second max = 4
Third max = 4, I have written this Solution Code: function maxNumbers(arr,n) {
// write code here
// do not console.log the answer
// return the answer as an array of 3 numbers
return arr.sort((a,b)=>b-a).slice(0,3)
};
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Being told that an unsorted array contains (n - 1) of n consecutive numbers (where the bounds are defined), find the missing number in O(n) time.
The bounds tell that all the numbers between the lower bound and the upper bound are present in the array except one number which is missing. You have to find that missing number.The function takes three arguments , the first argument is the array of integers, the second argument is the upper bound and the third argument is the lower bound.
All the numbers between the lower and the upper bounds are present in the array (inclusive of both upper and lower bound) except one number which is missing.
Input is provided in the form of an array which would have 3 elements. The first element is the array of integers, the second element is the upper bound and the third element is the lower bound. All three elements are used internally to call the function.
Example: [[1, 4, 3] ,4, 1]
Here, [1,4,3] is the array of integers
4 is the upper bound
1 is the lower boundThe function should print the missing number in the console.const input = [[1, 4, 3] ,4, 1];
const arr = input[0];
const upper_bound = input[1];
const lower_bound = input[2];
findMissingNumber(arr, upper_bound, lower_bound); //prints 2
// Explanation: From numbers 1 to 4, only 2 is missing from the array, I have written this Solution Code:
function findMissingNumber(arrayOfIntegers, upperBound, lowerBound) {
// Iterate through array to find the sum of the numbers
let sumOfIntegers = 0;
for (let i = 0; i < arrayOfIntegers.length; i++) {
sumOfIntegers += arrayOfIntegers[i];
}
// Find theoretical sum of the consecutive numbers using a variation of Gauss Sum.
// Formula: [(N * (N + 1)) / 2] - [(M * (M - 1)) / 2];
// N is the upper bound and M is the lower bound
upperLimitSum = (upperBound * (upperBound + 1)) / 2;
lowerLimitSum = (lowerBound * (lowerBound - 1)) / 2;
theoreticalSum = upperLimitSum - lowerLimitSum;
console.log(theoreticalSum - sumOfIntegers);
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Given three integers your task is to calculate the maximum integer among the given integers.The input contains three integers a, b, and c
<b>Constraint:</b>
1<=integers<=10000Print the maximum integer among the given integers.Sample Input:-
2 6 3
Sample Output:-
6
Sample Input:-
48 100 100
Sample Output:
100, I have written this Solution Code: a,b,c=[int(a) for a in input().split()]
print(max(a,b,c)), In this Programming Language: Python, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Given three integers your task is to calculate the maximum integer among the given integers.The input contains three integers a, b, and c
<b>Constraint:</b>
1<=integers<=10000Print the maximum integer among the given integers.Sample Input:-
2 6 3
Sample Output:-
6
Sample Input:-
48 100 100
Sample Output:
100, I have written this Solution Code: import java.util.Scanner;
class Main {
public static void main (String[] args)
{
//Capture the user's input
Scanner scanner = new Scanner(System.in);
//Storing the captured value in a variable
int p = scanner.nextInt();
int tm = scanner.nextInt();
int r = scanner.nextInt();
int intrst = MaxInteger(p,tm,r);
System.out.println(intrst);
}
static int MaxInteger(int a ,int b, int c){
if(a>=b && a>=c){return a;}
if(b>=a && b>=c){return b;}
return c;}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Given an unsorted array, your task is to sort the array using merge sort.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>implementMergeSort()</b> that takes 3 arguments.
arr: input array
start: starting index which is 0
end: ending index of array
Constraints
1 <= T <= 100
1 <= N <= 10<sup>6</sup>
0 <= Arr[i] <= 10<sup>9</sup>
Sum of 'N' over all test cases does not exceed 10<sup>6</sup>You need to return the sorted array. The driver code will print the array in sorted form.Sample Input:
2
3
3 1 2
3
4 5 6
Sample Output:
1 2 3
4 5 6, I have written this Solution Code:
public static int[] implementMergeSort(int arr[], int start, int end)
{
if (start < end)
{
// Find the middle point
int mid = (start+end)/2;
// Sort first and second halves
implementMergeSort(arr, start, mid);
implementMergeSort(arr , mid+1, end);
// Merge the sorted halves
merge(arr, start, mid, end);
}
return arr;
}
public static void merge(int arr[], int start, int mid, int end)
{
// Find sizes of two subarrays to be merged
int n1 = mid - start + 1;
int n2 = end - mid;
/* Create temp arrays */
int L[] = new int [n1];
int R[] = new int [n2];
/*Copy data to temp arrays*/
for (int i=0; i<n1; ++i)
L[i] = arr[start + i];
for (int j=0; j<n2; ++j)
R[j] = arr[mid + 1+ j];
/* Merge the temp arrays */
// Initial indexes of first and second subarrays
int i = 0, j = 0;
// Initial index of merged subarry array
int k = start;
while (i < n1 && j < n2)
{
if (L[i] <= R[j])
{
arr[k] = L[i];
i++;
}
else
{
arr[k] = R[j];
j++;
}
k++;
}
/* Copy remaining elements of L[] if any */
while (i < n1)
{
arr[k] = L[i];
i++;
k++;
}
/* Copy remaining elements of R[] if any */
while (j < n2)
{
arr[k] = R[j];
j++;
k++;
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Given an unsorted array, your task is to sort the array using merge sort.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>implementMergeSort()</b> that takes 3 arguments.
arr: input array
start: starting index which is 0
end: ending index of array
Constraints
1 <= T <= 100
1 <= N <= 10<sup>6</sup>
0 <= Arr[i] <= 10<sup>9</sup>
Sum of 'N' over all test cases does not exceed 10<sup>6</sup>You need to return the sorted array. The driver code will print the array in sorted form.Sample Input:
2
3
3 1 2
3
4 5 6
Sample Output:
1 2 3
4 5 6, I have written this Solution Code: for _ in range(int(input())):
n = int(input())
print(*sorted(list(map(int,input().split())))), In this Programming Language: Python, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Given a string S find minimum number of changes required to make the string palindrome. In a change you can change any index of the string to any character.
A palindrome is a string that remains the same if reversed.First and only line of input contains a string S.
Constraints
1 <= |S| <= 100000
S contains only lowercase lettersOutput a single integer which is the minimum number of changes required to make string palindrome.sample input 1
abc
sample output 1
1
sample input 2
abcdef
sample output 2
3
, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String str = br.readLine();
int i=0,ans=0;
int j=str.length()-1;
while(i<=j){
if(str.charAt(i)!=str.charAt(j)){
ans++;
}
i++;
j--;
}
System.out.println(ans);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Given a string S find minimum number of changes required to make the string palindrome. In a change you can change any index of the string to any character.
A palindrome is a string that remains the same if reversed.First and only line of input contains a string S.
Constraints
1 <= |S| <= 100000
S contains only lowercase lettersOutput a single integer which is the minimum number of changes required to make string palindrome.sample input 1
abc
sample output 1
1
sample input 2
abcdef
sample output 2
3
, I have written this Solution Code: n = input()
count = 0
i = 0
j = len(n)-1
while i <= j:
if n[i] != n[j]:
count += 1
i += 1
j -= 1
print(count), In this Programming Language: Python, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Given a string S find minimum number of changes required to make the string palindrome. In a change you can change any index of the string to any character.
A palindrome is a string that remains the same if reversed.First and only line of input contains a string S.
Constraints
1 <= |S| <= 100000
S contains only lowercase lettersOutput a single integer which is the minimum number of changes required to make string palindrome.sample input 1
abc
sample output 1
1
sample input 2
abcdef
sample output 2
3
, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> oset;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define pii pair<int,int>
/////////////
signed main()
{
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
string s;
cin>>s;
int ans=0;
for(int i=0;i<s.length()/2;++i)
{
if(s[i]!=s[s.length()-i-1])
++ans;
}
cout<<ans;
#ifdef ANIKET_GOYAL
cout<<endl<<endl<<endl<<endl<<"Time elapsed: "<<(double)(clock()-clk)/CLOCKS_PER_SEC<<endl;
#endif
}, In this Programming Language: C++, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Given a string S find minimum number of changes required to make the string palindrome. In a change you can change any index of the string to any character.
A palindrome is a string that remains the same if reversed.First and only line of input contains a string S.
Constraints
1 <= |S| <= 100000
S contains only lowercase lettersOutput a single integer which is the minimum number of changes required to make string palindrome.sample input 1
abc
sample output 1
1
sample input 2
abcdef
sample output 2
3
, I have written this Solution Code: function palindrome(s) {
// write code here
// do not console.log
// return the number of changes
const n = s.length;
let cc = 0;
for (let i = 0; i < n / 2; i++) {
if (s[i] == s[n - i - 1])
continue;
cc += 1;
if (s[i] < s[n - i - 1])
s[n - i - 1] = s[i];
else
s[i] = s[n - i - 1];
}
return cc;
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: A number X (X>=0) is called strange if the sum of its digits is divisible by 9. Given an integer N, your task is to find the Nth strange number.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>StrangeNumber()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 1000Return the Nth strange number.Sample Input:-
3
Sample Output:-
18
Explanation:-
0, 9, and 18 are the first three strange numbers.
Sample Input:-
2
Sample Output:-
9, I have written this Solution Code: def StrangeNumber(N):
return 9*(N-1)
, In this Programming Language: Python, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: A number X (X>=0) is called strange if the sum of its digits is divisible by 9. Given an integer N, your task is to find the Nth strange number.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>StrangeNumber()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 1000Return the Nth strange number.Sample Input:-
3
Sample Output:-
18
Explanation:-
0, 9, and 18 are the first three strange numbers.
Sample Input:-
2
Sample Output:-
9, I have written this Solution Code: int StrangeNumber(int N){
return 9*(N-1);
}, In this Programming Language: C++, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: A number X (X>=0) is called strange if the sum of its digits is divisible by 9. Given an integer N, your task is to find the Nth strange number.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>StrangeNumber()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 1000Return the Nth strange number.Sample Input:-
3
Sample Output:-
18
Explanation:-
0, 9, and 18 are the first three strange numbers.
Sample Input:-
2
Sample Output:-
9, I have written this Solution Code: int StrangeNumber(int N){
return 9*(N-1);
}, In this Programming Language: C, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: A number X (X>=0) is called strange if the sum of its digits is divisible by 9. Given an integer N, your task is to find the Nth strange number.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>StrangeNumber()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 1000Return the Nth strange number.Sample Input:-
3
Sample Output:-
18
Explanation:-
0, 9, and 18 are the first three strange numbers.
Sample Input:-
2
Sample Output:-
9, I have written this Solution Code: static int StrangeNumber(int N){
return 9*(N-1);
}, In this Programming Language: Java, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Given a number n find the number of prime numbers less than equal to that number.There is only one integer containing value of n.
Constraints:-
1 <= n <= 10000000Return number of primes less than or equal to nSample Input
5
Sample Output
3
Explanation:-
2 3 and 5 are the required primes.
Sample Input
5000
Sample Output
669, I have written this Solution Code: #include <bits/stdc++.h>
// #define ll long long
using namespace std;
#define ma 10000001
bool a[ma];
int main()
{
int n;
cin>>n;
for(int i=0;i<=n;i++){
a[i]=false;
}
for(int i=2;i<=n;i++){
if(a[i]==false){
for(int j=i+i;j<=n;j+=i){
a[j]=true;
}
}
}
int cnt=0;
for(int i=2;i<=n;i++){
if(a[i]==false){cnt++;}
}
cout<<cnt;
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Given a number n find the number of prime numbers less than equal to that number.There is only one integer containing value of n.
Constraints:-
1 <= n <= 10000000Return number of primes less than or equal to nSample Input
5
Sample Output
3
Explanation:-
2 3 and 5 are the required primes.
Sample Input
5000
Sample Output
669, I have written this Solution Code: import java.io.*;
import java.util.*;
import java.lang.Math;
class Main {
public static void main (String[] args)
throws IOException
{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
long n = Integer.parseInt(br.readLine());
long i=2,j,count,noOfPrime=0;
if(n<=1)
System.out.println("0");
else{
while(i<=n)
{
count=0;
for(j=2; j<=Math.sqrt(i); j++)
{
if( i%j == 0 ){
count++;
break;
}
}
if(count==0){
noOfPrime++;
}
i++;
}
System.out.println(noOfPrime);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Given a number n find the number of prime numbers less than equal to that number.There is only one integer containing value of n.
Constraints:-
1 <= n <= 10000000Return number of primes less than or equal to nSample Input
5
Sample Output
3
Explanation:-
2 3 and 5 are the required primes.
Sample Input
5000
Sample Output
669, I have written this Solution Code: function numberOfPrimes(N)
{
let arr = new Array(N+1);
for(let i = 0; i <= N; i++)
arr[i] = 0;
for(let i=2; i<= N/2; i++)
{
if(arr[i] === -1)
{
continue;
}
let p = i;
for(let j=2; p*j<= N; j++)
{
arr[p*j] = -1;
}
}
//console.log(arr);
let count = 0;
for(let i=2; i<= N; i++)
{
if(arr[i] === 0)
{
count++;
}
}
//console.log(arr);
return count;
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Given a number n find the number of prime numbers less than equal to that number.There is only one integer containing value of n.
Constraints:-
1 <= n <= 10000000Return number of primes less than or equal to nSample Input
5
Sample Output
3
Explanation:-
2 3 and 5 are the required primes.
Sample Input
5000
Sample Output
669, I have written this Solution Code: import math
n = int(input())
n=n+1
if n<3:
print(0)
else:
primes=[1]*(n//2)
for i in range(3,int(math.sqrt(n))+1,2):
if primes[i//2]:primes[i*i//2::i]=[0]*((n-i*i-1)//(2*i)+1)
print(sum(primes)), In this Programming Language: Python, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Complete the object <code>sayHiMixin</code> and the function <code>setPrototype</code>. You should implement the following:
<b>setPrototype</b>, this should set the prototype of a pre- defined class (not visible to you)<code>User</code> to <code> sayHiMixin</code>.
<b>sayHiMixin</b>, this must override the method <b>say</b> of an already defined object (not visible to you) <code>sayMixin</code>. This overridden method <b>say</b> should call the a <say> function of the already defined object <code>sayMixin</code>
<b>Note: </b> The method "say" in <code>sayMixin</code> takes a single parameter as string (which in this case the <code>name</code> attribute of the User class) and returns a number based on it.Name of the person as <b>string</b> s.You should return the same value as the <code>say</code> function in <b>sayMixin</b> returns.<pre>
// This is what you have to implement.
let sayHiMixin = {
// this should inherit from sayMixin, and invoke it's "say" method somehow
// for example,
const say = (name) => sayMixin.say(name);
// name is a string, which will infact be the `name` attribute of a `User` object
};
// You must setPrototype of User class to that of sayHiMixin object.
function setPrototype(){
// some code here
}
// Finally, when
// const obj = new User("Newton School")
// console.log(obj.say()) // this should print whatever sayHiMixin.say returns, which in this case is 39
</pre>, I have written this Solution Code: function setPrototype() {
Object.assign(User.prototype, sayHiMixin);
}
let sayHiMixin = {
__proto__: sayMixin, // (or we could use Object.setPrototypeOf to set the prototype here)
say() {
// call parent method
return super.say(`${this.name}`); // (*)
},
};
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Given a directed graph of N vertices, Print 1 if you can perform a topological sort on the given graph. Otherwise, print 0.
There are no self-loops or multiple edges.The first line contains two integers N and E, denoting the number of nodes and number of edges in the graph respectively. Next E lines contain two integers u and v, denoting an edge from u to v
Constraints
1 <= N, E <= 1000
0 <= u, v <= N-1
u != vPrint 1 if topological sort can be done correctly. Otherwise, print 0.Sample Input 1:
4 5
0 1
1 2
2 3
3 0
0 2
Sample Output 1:
0
Sample Input 2:
4 3
0 1
1 2
2 3
Sample Output 2:
1
, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
static ArrayList<ArrayList<Integer>> adj;
static void graph(int v){
adj =new ArrayList<>();
for(int i=0;i<=v;i++) adj.add(new ArrayList<Integer>());
}
static void addedge(int u, int v){
adj.get(u).add(v);
}
static boolean dfs(int i, boolean [] vis, int parent[]){
vis[i] = true; parent[i] = 1;
for(Integer it: adj.get(i)){
if(vis[it]==false) {
if(dfs(it, vis, parent)==true) return true;
}
else if(parent[it]==1) return true;
}
parent[i] = 0;
return false;
}
static boolean helper(int n){
boolean vis[] = new boolean [n+1];
int parent[] = new int[n+1];
int i=1;
if(vis[i]==false){
if(dfs(i, vis, parent)) return true;
}
return false;
}
public static void main (String[] args) throws Exception{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String t[] = br.readLine().split(" ");
int n = Integer.parseInt(t[0]);
graph(n);
int e = Integer.parseInt(t[1]);
for(int i=0;i<e;i++) {
String num[] = br.readLine().split(" ");
int u = Integer.parseInt(num[0]);
int v = Integer.parseInt(num[1]);
addedge(u, v);
}
if(helper(n)) System.out.println("0");
else System.out.println("1");
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not?
|
Compilable
|
For this Question: Given a directed graph of N vertices, Print 1 if you can perform a topological sort on the given graph. Otherwise, print 0.
There are no self-loops or multiple edges.The first line contains two integers N and E, denoting the number of nodes and number of edges in the graph respectively. Next E lines contain two integers u and v, denoting an edge from u to v
Constraints
1 <= N, E <= 1000
0 <= u, v <= N-1
u != vPrint 1 if topological sort can be done correctly. Otherwise, print 0.Sample Input 1:
4 5
0 1
1 2
2 3
3 0
0 2
Sample Output 1:
0
Sample Input 2:
4 3
0 1
1 2
2 3
Sample Output 2:
1
, I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
vector<int> g[N];
int vis[N];
bool flag = 0;
void dfs(int u, int p){
vis[u] = 1;
for(auto i: g[u]){
if(i == p) continue;
if(vis[i] == 1)
flag = 1;
if(vis[i] == 0) dfs(i, u);
}
vis[u] = 2;
}
signed main() {
IOS;
int n, m;
cin >> n >> m;
for(int i = 1; i <= m; i++){
int u, v;
cin >> u >> v;
g[u].push_back(v);
}
for(int i = 0; i < n; i++){
if(vis[i]) continue;
dfs(i, n);
}
if(!flag)
cout << 1;
else
cout << 0;
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not?
|
Compilable
|
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