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|---|---|---|---|---|---|---|---|---|---|---|---|
https://leetcode.com/problems/power-of-four/discuss/2667207/Python-solution-based-on-recursion
|
class Solution:
def isPowerOfFour(self, n: int) -> bool:
if n == 0:
return False
if n == 1 or n == 4:
return True
return self.isPowerOfFour(n/4)
|
power-of-four
|
Python solution based on recursion
|
MPoinelli
| 0 | 4 |
power of four
| 342 | 0.458 |
Easy
| 5,900 |
https://leetcode.com/problems/power-of-four/discuss/2664980/Python-Easy-Solution
|
class Solution:
def isPowerOfFour(self, n: int) -> bool:
if n < 1:
return False
if n == 1:
return True
return self.isPowerOfFour(n / 4)
|
power-of-four
|
Python Easy Solution
|
kruzhilkin
| 0 | 2 |
power of four
| 342 | 0.458 |
Easy
| 5,901 |
https://leetcode.com/problems/power-of-four/discuss/2662054/Python-solution-without-loopsrecursion
|
class Solution:
def isPowerOfFour(self, n: int) -> bool:
if n == 1:
return True
bit = "{0:b}".format(n)
if bit[-1] == "1":
return False
if bit[0] == "1" and "1" not in bit[1:] and len(bit) % 2 == 1:
return True
|
power-of-four
|
Python solution without loops/recursion
|
maomao1010
| 0 | 7 |
power of four
| 342 | 0.458 |
Easy
| 5,902 |
https://leetcode.com/problems/power-of-four/discuss/2614089/Python-Solution-Interview-Approach
|
class Solution:
def isPowerOfFour(self, n: int) -> bool:
if n<=0:
return False
while n>1:
if n%4!=0: return False
n//=4
return True
|
power-of-four
|
[Python Solution] Interview Approach
|
utsa_gupta
| 0 | 9 |
power of four
| 342 | 0.458 |
Easy
| 5,903 |
https://leetcode.com/problems/power-of-four/discuss/2559569/Python-(Simple-Solution-and-Beginner-Friendly)
|
class Solution:
def isPowerOfFour(self, n: int) -> bool:
while n>1:
n = n/4
return n == 1
|
power-of-four
|
Python (Simple Solution and Beginner-Friendly)
|
vishvavariya
| 0 | 39 |
power of four
| 342 | 0.458 |
Easy
| 5,904 |
https://leetcode.com/problems/power-of-four/discuss/2470113/Python3-bitwise-operator
|
class Solution(object):
def isPowerOfFour(self, num):
"""
:type n: int
:rtype: bool
"""
result = 0
if num <= 0:
return False
if num & (num - 1) != 0:
return False
for i in range(0, 31):
if i % 2 == 0:
result += 1 << i
return (num & result) == num
sol = Solution()
print(sol.isPowerOfFour(15))
|
power-of-four
|
Python3 - bitwise operator
|
user2354hl
| 0 | 4 |
power of four
| 342 | 0.458 |
Easy
| 5,905 |
https://leetcode.com/problems/power-of-four/discuss/2466684/Unique-Python-One-Liners
|
class Solution:
def isPowerOfFour(self, n: int) -> bool:
return n > 0 \
and f'{n:032b}'.count('1') == 1 \
and f'{n:032b}'.find('1') & 1
|
power-of-four
|
Unique Python One-Liners
|
timberg
| 0 | 16 |
power of four
| 342 | 0.458 |
Easy
| 5,906 |
https://leetcode.com/problems/power-of-four/discuss/2466530/Python-Simple-Python-Solution
|
class Solution:
def isPowerOfFour(self, n: int) -> bool:
power = 0
number = 4
while True:
num = number**power
if num == n:
return True
if num > n:
return False
else:
power = power + 1
return False
|
power-of-four
|
[ Python ] ✅✅ Simple Python Solution 🥳✌👍
|
ASHOK_KUMAR_MEGHVANSHI
| 0 | 22 |
power of four
| 342 | 0.458 |
Easy
| 5,907 |
https://leetcode.com/problems/power-of-four/discuss/2464578/Python-Short-Faster-Solution-log-base-4-oror-Documented
|
class Solution:
def isPowerOfFour(self, n: int) -> bool:
if n > 0:
power = int(math.log(n,4)) # find power using log base 4
return 4 ** power == n # if 4^power is same as n, return True, else False
return False
|
power-of-four
|
[Python] Short, Faster Solution - log base 4 || Documented
|
Buntynara
| 0 | 5 |
power of four
| 342 | 0.458 |
Easy
| 5,908 |
https://leetcode.com/problems/power-of-four/discuss/2463692/Python-Solution
|
class Solution:
def isPowerOfFour(self, n: int) -> bool:
result = 1
while n > result:
result *= 4
return result == n
|
power-of-four
|
Python Solution
|
hgalytoby
| 0 | 14 |
power of four
| 342 | 0.458 |
Easy
| 5,909 |
https://leetcode.com/problems/power-of-four/discuss/2462341/Python-Fast-and-trivial-solution
|
class Solution:
def isPowerOfFour(self, n: int) -> bool:
if n <= 0:
return False
while n > 1:
n, m = divmod(n, 4)
if m > 0:
return False
return True
|
power-of-four
|
[Python] Fast and trivial solution
|
RegInt
| 0 | 15 |
power of four
| 342 | 0.458 |
Easy
| 5,910 |
https://leetcode.com/problems/power-of-four/discuss/2462292/Three-Solution-using-recursion-iteration-and-Bit-Manipulation
|
class Solution:
def isPowerOfFour(self, n: int) -> bool:
# Solution 1 using recursion
while n % 4 == 0 and n > 0:
return self.isPowerOfFour(n/4)
return n == 1
# Solution 2 iteration
if n == 1:
return True
if n % 4:
return False
while n > 1:
if n % 4:
return False
n //= 4
return n == 1
# Solution 3 using bit manipulation
'''
Once we write numbers in it's binary representation, from there we can observe:=>
i. 000001 , power of 2 and 4
ii. 000010, power of only 2
iii. 000100 , power of 2 and 4
iv. 001000, power of only 2
v. 010000 , power of 2 and 4
vi. 100000, power of only 2
We can see if the set bit is at an odd position and is a power of 2, it's also power of 4.
'''
return n.bit_length() & 1 and not(n & (n-1))
|
power-of-four
|
Three Solution using recursion, iteration and Bit-Manipulation
|
__Asrar
| 0 | 25 |
power of four
| 342 | 0.458 |
Easy
| 5,911 |
https://leetcode.com/problems/power-of-four/discuss/2462065/30ms-or-fast-and-easy-approach
|
class Solution:
def isPowerOfFour(self, n: int) -> bool:
if n>0:
x=math.log(n,4)
return math.floor(x)==math.ceil(x)
return False
|
power-of-four
|
30ms | fast and easy approach
|
ayushigupta2409
| 0 | 16 |
power of four
| 342 | 0.458 |
Easy
| 5,912 |
https://leetcode.com/problems/power-of-four/discuss/2461942/python3-with-binary-bit-manipulation
|
class Solution:
def isPowerOfFour(self, n: int) -> bool:
if n<=0: return False
n_bin = bin(n)[2:]
return True if (len(n_bin) % 2 == 1) and (sum(map(int,n_bin)) == 1) else False
|
power-of-four
|
python3, with binary bit manipulation
|
pjy953
| 0 | 3 |
power of four
| 342 | 0.458 |
Easy
| 5,913 |
https://leetcode.com/problems/power-of-four/discuss/2461898/Python3-Solution
|
class Solution:
def isPowerOfFour(self, n: int) -> bool:
i = 0
curr = 0
while curr <= n:
curr = 4 ** i
if curr == n:
return True
i += 1
return False
|
power-of-four
|
Python3 Solution
|
creativerahuly
| 0 | 19 |
power of four
| 342 | 0.458 |
Easy
| 5,914 |
https://leetcode.com/problems/power-of-four/discuss/2461708/Python-Recursion-O(logn)-time-O(logn)-space.
|
class Solution:
def isPowerOfFour(self, n: int) -> bool:
if n == 1:
return True
if n < 1:
return False
return self.isPowerOfFour(n/4)
|
power-of-four
|
Python Recursion, O(logn) time, O(logn) space.
|
OsamaRakanAlMraikhat
| 0 | 17 |
power of four
| 342 | 0.458 |
Easy
| 5,915 |
https://leetcode.com/problems/power-of-four/discuss/2461452/Simple-Python-Solution
|
class Solution:
def isPowerOfFour(self, n: int) -> bool:
while n>1:
n = n/4
if n == 1:
return True
return False
|
power-of-four
|
Simple Python Solution
|
saiavunoori4187
| 0 | 15 |
power of four
| 342 | 0.458 |
Easy
| 5,916 |
https://leetcode.com/problems/power-of-four/discuss/2461394/Python-Simple-and-Easy-Solution
|
class Solution:
def isPowerOfFour(self, n: int) -> bool:
if n == 1: return True
while n > 1:
if n%4 == 0:
n = n // 4
else:
return False
if n == 1:
return True
return False
|
power-of-four
|
Python - Simple and Easy Solution
|
dayaniravi123
| 0 | 7 |
power of four
| 342 | 0.458 |
Easy
| 5,917 |
https://leetcode.com/problems/power-of-four/discuss/2461372/Easy-Python-Solution
|
class Solution:
def isPowerOfFour(self, n: int) -> bool:
if n<1:
return False
elif n==1:
return True
else:
while n!=1:
if n%4!=0:
return False
else:
n=n//4
return True
|
power-of-four
|
Easy Python Solution
|
a_dityamishra
| 0 | 13 |
power of four
| 342 | 0.458 |
Easy
| 5,918 |
https://leetcode.com/problems/power-of-four/discuss/2461309/Python-O(1)-oneliner-easy-to-understand-(no-loops-or-bitwise-math)
|
class Solution:
def isPowerOfFour(self, num):
return num == 4**(num.bit_length()//2)
|
power-of-four
|
[Python] O(1) oneliner - easy to understand (no loops or bitwise math)
|
user5285r
| 0 | 7 |
power of four
| 342 | 0.458 |
Easy
| 5,919 |
https://leetcode.com/problems/power-of-four/discuss/2461196/Python-bit-check
|
class Solution:
def isPowerOfFour(self, n: int) -> bool:
return n > 0 and n & (n - 1) == 0 and n & 0x55555555
|
power-of-four
|
Python, bit check
|
blue_sky5
| 0 | 20 |
power of four
| 342 | 0.458 |
Easy
| 5,920 |
https://leetcode.com/problems/power-of-four/discuss/2446455/python3-using-a-while-loop-faster-than-86
|
class Solution:
def isPowerOfFour(self, n: int) -> bool:
while n%4==0<n:
n/=4
return n==1
|
power-of-four
|
[python3] using a while loop, faster than 86%
|
hhlinwork
| 0 | 8 |
power of four
| 342 | 0.458 |
Easy
| 5,921 |
https://leetcode.com/problems/power-of-four/discuss/2366691/Simple-fast-Python-solution-or-Easy-to-Understand-or-73.70-faster
|
class Solution:
def isPowerOfFour(self, n: int) -> bool:
x = 0
y = pow(4, x)
while y<n:
x+=1
y = pow(4, x)
if (n - y) == 0:
return True
else:
return False
|
power-of-four
|
Simple fast Python solution | Easy to Understand | 73.70% faster
|
harishmanjunatheswaran
| 0 | 40 |
power of four
| 342 | 0.458 |
Easy
| 5,922 |
https://leetcode.com/problems/power-of-four/discuss/2120598/Python3-easy-understanding-solution
|
class Solution:
def isPowerOfFour(self, n: int) -> bool:
if n == 0:
return False
pow_value = 1
while pow_value < n:
pow_value *=4
return pow_value == n
|
power-of-four
|
Python3 - easy understanding solution
|
thanhinterpol
| 0 | 23 |
power of four
| 342 | 0.458 |
Easy
| 5,923 |
https://leetcode.com/problems/power-of-four/discuss/1936644/Python-One-Liner-or-Log-or-Math
|
class Solution:
def isPowerOfFour(self, n):
return n>0 and log(n,4).is_integer()
|
power-of-four
|
Python - One-Liner | Log | Math
|
domthedeveloper
| 0 | 44 |
power of four
| 342 | 0.458 |
Easy
| 5,924 |
https://leetcode.com/problems/power-of-four/discuss/1815199/Easy-to-understand-trick
|
class Solution:
def isPowerOfFour(self, n: int) -> bool:
arr=[]
for i in range(100):
arr.append(4**i)
if n in arr:
return True
else:
return False
|
power-of-four
|
Easy to understand trick
|
prashantdahiya711
| 0 | 76 |
power of four
| 342 | 0.458 |
Easy
| 5,925 |
https://leetcode.com/problems/integer-break/discuss/2830343/O(1)-oror-TC1-10-line-code
|
class Solution:
def integerBreak(self, n: int) -> int:
if(n<=3):
return n-1
n3=n//3
r3=n%3
if(r3==0):
return 3**n3
if(r3==1):
r3=4
n3-=1
return r3*(3**n3)
|
integer-break
|
O(1) || TC=1 10 line code
|
droj
| 7 | 71 |
integer break
| 343 | 0.554 |
Medium
| 5,926 |
https://leetcode.com/problems/integer-break/discuss/2180665/Python3-DP-top-down-approach-faster-than-96
|
class Solution:
result = {1:1,2:1,3:2,4:4,5:6,6:9,7:12,8:18,9:27,10:36}
def integerBreak(self, n: int) -> int:
try:
return self.result[n]
except:
x = float("-inf")
for i in range(1,n):
j = n-1
while j>0:
if i+j==n:
k = self.integerBreak(i)*self.integerBreak(j)
x = max(x,k)
j-=1
self.result[n] = x
return self.result[n]
|
integer-break
|
📌 Python3 DP top down approach faster than 96%
|
Dark_wolf_jss
| 4 | 76 |
integer break
| 343 | 0.554 |
Medium
| 5,927 |
https://leetcode.com/problems/integer-break/discuss/1699371/Python-solution-O(1)-method-Easy-to-understand-with-explanation
|
class Solution:
def integerBreak(self, n: int) -> int:
if n == 2 or n == 3:
return n - 1 # must break into 1 & (n-1)
num_3 = n // 3 # break the number into (3+3+3+...) + remainder
if n%3 == 0:
return 3**num_3 # (3+3+3+...)
if n%3 == 2:
return (3**num_3) * 2 # (3+3+3+...) + 2
if n%3 == 1:
return 3**(num_3-1) * 4 # (3+3+3+...) + 1 --> (3+3+...) + 3 + 1 -- > (3+3+...) + 4
|
integer-break
|
Python solution -- O(1) method Easy to understand with explanation
|
byroncharly3
| 4 | 130 |
integer break
| 343 | 0.554 |
Medium
| 5,928 |
https://leetcode.com/problems/integer-break/discuss/479923/Python-O(n)-Dynamic-Programming-with-Memoization
|
class Solution:
def integerBreak(self, n: int) -> int:
memo = {}
def intBreak(n):
if n in memo:
return memo[n]
if n<2:
memo[n] = 0
return 0
if n==2:
memo[2] = 2
return 1
maxval = 0
for i in range(1,n//2+1):
maxval = max(maxval,i*intBreak(n-i),i*(n-i))
memo[n] = maxval
return maxval
ans = intBreak(n)
return ans
|
integer-break
|
Python O(n) Dynamic Programming with Memoization
|
user8307e
| 3 | 692 |
integer break
| 343 | 0.554 |
Medium
| 5,929 |
https://leetcode.com/problems/integer-break/discuss/2407006/Python3-or-Solved-using-Bottom-Up-DP-%2B-Tabulation
|
class Solution:
#Time-Complexity: O(n^2)
#Space-Complexity: O(n)
def integerBreak(self, n: int) -> int:
#we know we can reduce n as
# n
# / \
# 1 n-1
# / \
# 1 n-2
# ...
#Basically, we can keep reducing n like this in this tree structure above!
#This is the pattern I recognized! I recognized for given n, there are
#potential sums of (1, n-1), (2, n-2), (3, n-3), ..., (n//2, n//2)!
#For each pair, I can compare the direct number with the max product decomposition
#and take the max of two!
#Reason for comparison: for each of the sum factor of given n, either leave it
#undecomposed or decompose it into further sum factors if the product of sum
#factors produce ultimately a number that exceeds the orignal sum factor! This way
#I am maximing product contribution for each and every sum factor!
#For example, for 5, we decompose it into 2 and 3, since 2*3 > 5, so it will
#maximize our product further!
#However, for 3, we don't decompose since we can maximally decompose to
#1 and 2 but 1*2 < 3!
#Do that for both numbers of each pair and take the product!
#Whatever is largest across the pairs will be answer for given input n!
dp = [-1] * (n+1)
#add dp-base!
dp[1] = 1
#this problem has only one state parameter: the given number to start decomposing #from!
#iterate through each subproblem or state!
#Bottom-Up
for i in range(2, n+1, 1):
upper_bound = (i // 2) + 1
#iterate through all possible pairs!
for j in range(1, upper_bound, 1):
#current pair (j, i-j), which we probably already solved its subproblems!
first = max(j, dp[j])
second = max(i-j, dp[i-j])
#get product for current pair!
sub_ans = first * second
#compare current pair's product against built up answer maximum!
dp[i] = max(dp[i], sub_ans)
#then, once we are done, we can return dp[n]!
return dp[n]
|
integer-break
|
Python3 | Solved using Bottom-Up DP + Tabulation
|
JOON1234
| 2 | 29 |
integer break
| 343 | 0.554 |
Medium
| 5,930 |
https://leetcode.com/problems/integer-break/discuss/1136369/O(n)-DP-Python3-bites-97
|
class Solution:
def integerBreak(self, n: int) -> int:
dp = [None, None, 1, 2, 4, 6, 9]
for i in range(7, n+1):
dp.append(dp[i-3] * 3)
return dp[n]
|
integer-break
|
O(n) - DP - Python3 bites 97%
|
ShSaktagan
| 2 | 167 |
integer break
| 343 | 0.554 |
Medium
| 5,931 |
https://leetcode.com/problems/integer-break/discuss/1778956/Python-3-Solution-Using-power-of-3
|
class Solution:
def integerBreak(self, n: int) -> int:
if n==2:
return 1
if n==3:
return 2
if n==4:
return 4
x=n%3
a=n//3
if x==0:
return 3**a
if x==1:
return 3**(a-1)*4
return 3**a*2
|
integer-break
|
[Python 3 Solution Using power of 3]
|
jiteshbhansali
| 1 | 81 |
integer break
| 343 | 0.554 |
Medium
| 5,932 |
https://leetcode.com/problems/integer-break/discuss/1148577/Python-math-solutiion-O(1)
|
class Solution:
def integerBreak(self, n: int) -> int:
q, r = divmod(n, 3)
if r == 0:
return 2 if n == 3 else 3**q
elif r == 1:
return 1 if n == 1 else 4*3**(q-1)
else: # r == 2
return 1 if n == 2 else 2*3**q
|
integer-break
|
Python, math solutiion O(1)
|
blue_sky5
| 1 | 136 |
integer break
| 343 | 0.554 |
Medium
| 5,933 |
https://leetcode.com/problems/integer-break/discuss/1007487/Python3Golang-Solution-with-using-dp
|
class Solution:
def integerBreak(self, n: int) -> int:
dp = [0] * (n + 1)
dp[1] = 1
dp[2] = 1
for i in range(3, n + 1):
for j in range(1, i // 2 + 1):
dp[i] = max(dp[i], max((i - j), dp[i - j]) * j)
return dp[-1]
|
integer-break
|
[Python3/Golang] Solution with using dp
|
maosipov11
| 1 | 55 |
integer break
| 343 | 0.554 |
Medium
| 5,934 |
https://leetcode.com/problems/integer-break/discuss/2842149/Python-Solution-with-O(1)-time-and-space-complexity
|
class Solution:
def integerBreak(self, n: int) -> int:
if n == 2:
return 1
if n == 3:
return 2
remainder = n%3
numberOfThree = n//3
if remainder == 0:
return 3 ** numberOfThree
if remainder == 1:
return (3 ** (numberOfThree-1)) * 4
else:
return (3 ** numberOfThree) * 2
|
integer-break
|
Python Solution with O(1) time and space complexity
|
jxlau
| 0 | 3 |
integer break
| 343 | 0.554 |
Medium
| 5,935 |
https://leetcode.com/problems/integer-break/discuss/2817001/intuitive-sub-optimal-solution-from-someone-who-is-terrible-at-math
|
class Solution:
def integerBreak(self, n: int) -> int:
# to max product, try to break into k similar-sized pieces
def build_pieces( k ):
if k < 2:
return None
# k >= 2
div, mod = divmod(n,k)
pieces = [div] * k
while mod >0:
pieces[mod] += 1 # spread the residule evenly
mod -= 1
return pieces
def product( nums ):
ans = 1
for num in nums:
ans *= num
return ans
ans = 0
for k in range(2, n+1):
pieces = build_pieces( k ) # for debug purpose
ans = max(ans, product( pieces ) )
return ans
|
integer-break
|
intuitive, sub-optimal solution from someone who is terrible at math
|
SkinheadBob
| 0 | 2 |
integer break
| 343 | 0.554 |
Medium
| 5,936 |
https://leetcode.com/problems/integer-break/discuss/2776999/Simple-Python-Solution-or-Math
|
class Solution:
def integerBreak(self, n: int) -> int:
three = 0
if n<4: return n-1
while n>4:
n-=3
three+=1
return 3**three*n
|
integer-break
|
Simple Python Solution | Math
|
RajatGanguly
| 0 | 6 |
integer break
| 343 | 0.554 |
Medium
| 5,937 |
https://leetcode.com/problems/integer-break/discuss/2740476/Python-solution
|
class Solution:
def integerBreak(self, n: int) -> int:
if n == 2:
return 1
if n == 3:
return 2
if n % 3 == 0:
return pow(3, n//3)
else:
r3 = 0
while n > 4:
r3 += 1
n -= 3
return pow(3, r3) * n
|
integer-break
|
Python solution
|
maomao1010
| 0 | 5 |
integer break
| 343 | 0.554 |
Medium
| 5,938 |
https://leetcode.com/problems/integer-break/discuss/2733624/Very-short-dynamic-programming-solution-beat-99
|
class Solution:
def integerBreak(self, n: int) -> int:
dp = [0] + [1] * n
for i in range(3,n+1):
dp[i] = max(max(i-2,dp[i-2]) * 2, max(i-3,dp[i-3]) * 3)
return dp[n]
|
integer-break
|
Very short dynamic programming solution, beat 99%
|
zhaoqiang
| 0 | 7 |
integer break
| 343 | 0.554 |
Medium
| 5,939 |
https://leetcode.com/problems/integer-break/discuss/2691100/python-working-solution-or-dp
|
class Solution:
def integerBreak(self, n: int) -> int:
dp = {}
def dfs(n):
if n == 1:
return 1
if n in dp:
return dp[n]
ans = float('-inf')
for l in range(1,n):
r = n-l
left,right = dfs(l),dfs(r)
ans = max(ans ,left * r , l * right ,r*l)
dp[n] = ans
return ans
return dfs(n)
|
integer-break
|
python working solution | dp
|
Sayyad-Abdul-Latif
| 0 | 10 |
integer break
| 343 | 0.554 |
Medium
| 5,940 |
https://leetcode.com/problems/integer-break/discuss/2682681/Easy-to-Understand.-Not-the-best-solution-I-feel.-Faster-than-50
|
class Solution:
def integerBreak(self, n: int) -> int:
def dfs(a):
if a == 1:
return 1
max_res = 1
for right in range(1, a):
left = a - right
max_res = max(max_res, left * dfs(right), left * right)
return max_res
return dfs(n)
|
integer-break
|
Easy to Understand. Not the best solution I feel. Faster than 50%
|
devanshsolani30
| 0 | 18 |
integer break
| 343 | 0.554 |
Medium
| 5,941 |
https://leetcode.com/problems/integer-break/discuss/2662453/python-easy-dp-solution
|
class Solution:
def integerBreak(self, n: int) -> int:
dp = [0]*(n+1)
dp[1] = 1
for i in range(2,n+1):
for j in range(1,i):
dp[i] =max(dp[i],j*(i-j),j*dp[i-j])
return dp[n]
|
integer-break
|
python easy dp solution
|
Akash_chavan
| 0 | 13 |
integer break
| 343 | 0.554 |
Medium
| 5,942 |
https://leetcode.com/problems/integer-break/discuss/2657315/Simple-1D-Bottom-Up-DP
|
class Solution:
def integerBreak(self, n: int) -> int:
dp = [0] * (n+1)
for i in range(1, n+1):
for j in range(1, i):
dp[i] = max(dp[i], j * (i-j), j * dp[i-j])
return dp[-1]
|
integer-break
|
Simple 1D Bottom-Up DP
|
jonathanbrophy47
| 0 | 6 |
integer break
| 343 | 0.554 |
Medium
| 5,943 |
https://leetcode.com/problems/integer-break/discuss/2296772/Python3-or-Memoization-or-Recursion
|
class Solution:
def integerBreak(self, n: int) -> int:
@lru_cache(None)
def recursive(n):
if n==1:
return 1
max_prod=0
for index in range(1,n):
max_prod=max(max_prod,index*recursive(n-index))
max_prod=max(max_prod,index*(n-index))
return max_prod
return recursive(n)
|
integer-break
|
Python3 | Memoization | Recursion
|
Sefinehtesfa34
| 0 | 19 |
integer break
| 343 | 0.554 |
Medium
| 5,944 |
https://leetcode.com/problems/integer-break/discuss/1642913/Liner-solution-90-speed
|
class Solution:
def integerBreak(self, n: int) -> int:
max_prod = 1
for k in range(2, n // 2 + 2):
num, rem = divmod(n, k)
max_prod = max(max_prod, pow(num, k - rem) * pow(num + 1, rem))
return max_prod
|
integer-break
|
Liner solution, 90% speed
|
EvgenySH
| 0 | 92 |
integer break
| 343 | 0.554 |
Medium
| 5,945 |
https://leetcode.com/problems/integer-break/discuss/1593305/Python-Straightforward-Solution-or-Pure-Easy-Mathematics
|
class Solution:
def integerBreak(self, n: int) -> int:
if n <= 3:
return n-1
three = n//3
rem = n % 3
if rem == 0:
return 3**three
elif rem == 1:
return (3**(three-1))*4
else:
return (3**three)*2
# 6 = 3x3 = 9
# 7 = 3x4 = 12
# 8 = 3x3x2 = 18
# 9 = 3x3x3 = 27
# 10 = 3x3x4 = 36
# 54 = 3 ^ 18 = 387420489
# 55 = (3 ^ 17)*4 = 516560652
# 56 = (3 ^ 18)*2 = 774840978
|
integer-break
|
Python Straightforward Solution | Pure Easy Mathematics
|
leet_satyam
| 0 | 94 |
integer break
| 343 | 0.554 |
Medium
| 5,946 |
https://leetcode.com/problems/integer-break/discuss/1498409/we-only-need-2-and-3.
|
class Solution:
def integerBreak(self, n: int) -> int:
if n == 2: return 1
if n == 3: return 2
num3 = n//3
maxv = 1
for i in range(num3+1):
res = n - 3*i
if res %2 == 0:
num2 = res//2
maxv = max(maxv, 2**num2*3**i)
return maxv
|
integer-break
|
we only need 2 and 3.
|
byuns9334
| 0 | 103 |
integer break
| 343 | 0.554 |
Medium
| 5,947 |
https://leetcode.com/problems/integer-break/discuss/1492277/Python-solution-using-DP
|
class Solution:
def integerBreak(self, n: int) -> int:
if n == 2:
return 1
if n == 3:
return 2
# 1
dp = [sys.maxsize] * (n+1)
# 2
dp[0] = 0
dp[1] = 1
dp[2] = 2
# 3
for i in range(3,n+1):
# 4
# Generate all combos that add up to i
# Figure out what is bigger: i, or max((a x dp[b]) for (a,b) in combos)
max_combo = max([(j * dp[i-j]) for j in range(1,i)])
# If i is bigger than the max_combo at i, just use i (this only applies to 3 and 4)
dp[i] = max(i, max_combo)
return dp[n]
|
integer-break
|
Python solution using DP
|
mikewillard87
| 0 | 46 |
integer break
| 343 | 0.554 |
Medium
| 5,948 |
https://leetcode.com/problems/integer-break/discuss/792197/Python3-dp-and-math
|
class Solution:
def integerBreak(self, n: int) -> int:
@lru_cache(None)
def fn(n):
"""Return the max product by splitting n."""
if n == 1: return 1
return max(max(i, fn(i))*max(n-i, fn(n-i)) for i in range(1, n//2+1))
return fn(n)
|
integer-break
|
[Python3] dp & math
|
ye15
| 0 | 91 |
integer break
| 343 | 0.554 |
Medium
| 5,949 |
https://leetcode.com/problems/integer-break/discuss/792197/Python3-dp-and-math
|
class Solution:
def integerBreak(self, n: int) -> int:
ans = [None, 1] + [0]*(n-1)
for k in range(2, n+1):
for i in range(1, k//2+1):
ans[k] = max(ans[k], max(i, ans[i])*max(k-i, ans[k-i]))
return ans[-1]
|
integer-break
|
[Python3] dp & math
|
ye15
| 0 | 91 |
integer break
| 343 | 0.554 |
Medium
| 5,950 |
https://leetcode.com/problems/integer-break/discuss/792197/Python3-dp-and-math
|
class Solution:
def integerBreak(self, n: int) -> int:
if n == 2: return 1
if n == 3: return 2
ans = 1
while n > 4:
ans *= 3
n -= 3
return ans*n
|
integer-break
|
[Python3] dp & math
|
ye15
| 0 | 91 |
integer break
| 343 | 0.554 |
Medium
| 5,951 |
https://leetcode.com/problems/integer-break/discuss/457138/python-DP-solution-with-explanation-80-faster
|
class Solution:
def integerBreak(self, n: int) -> int:
dp = [0] * (n+3) # extra buffer
dp[0] = 0
dp[1] = 1
dp[2] = 1
if n <= 2: return dp[n]
for i in range(3, n+1):
for j in range(1, i // 2 + 1):
dp[i] = max(
dp[i],
j * dp[i-j],
j * (i-j)
)
return dp[n]
|
integer-break
|
python DP solution with explanation 80% faster
|
sokketsu
| 0 | 141 |
integer break
| 343 | 0.554 |
Medium
| 5,952 |
https://leetcode.com/problems/integer-break/discuss/362181/Solution-in-Python-3-(beats-100)-(one-line)-(direct-solution-using-Calculus)
|
class Solution:
def integerBreak(self, n: int) -> int:
return 3**(n//3-((n-3*(n//3))<=1))*((n-3*(n//3))+3*((n-3*(n//3))<=1)) if n>3 else n-1
|
integer-break
|
Solution in Python 3 (beats 100%) (one line) (direct solution using Calculus)
|
junaidmansuri
| 0 | 241 |
integer break
| 343 | 0.554 |
Medium
| 5,953 |
https://leetcode.com/problems/reverse-string/discuss/670137/Python-3-~actually~-easiest-solution
|
class Solution:
def reverseString(self, s: List[str]) -> None:
s[:] = s[::-1]
|
reverse-string
|
Python 3 ~actually~ easiest solution
|
drblessing
| 77 | 6,800 |
reverse string
| 344 | 0.762 |
Easy
| 5,954 |
https://leetcode.com/problems/reverse-string/discuss/1480872/Python-and-Java-One-Line-Iterative-Recursive-Solutions
|
class Solution:
def reverseString(self, s: List[str]) -> None:
s.reverse()
|
reverse-string
|
Python and Java - One Line, Iterative, Recursive Solutions
|
Pootato
| 8 | 554 |
reverse string
| 344 | 0.762 |
Easy
| 5,955 |
https://leetcode.com/problems/reverse-string/discuss/1480872/Python-and-Java-One-Line-Iterative-Recursive-Solutions
|
class Solution:
def reverseString(self, s: List[str]) -> None:
N = len(s)
for i in range(N // 2):
s[i], s[-i-1] == s[-i-1], s[i]
|
reverse-string
|
Python and Java - One Line, Iterative, Recursive Solutions
|
Pootato
| 8 | 554 |
reverse string
| 344 | 0.762 |
Easy
| 5,956 |
https://leetcode.com/problems/reverse-string/discuss/1480872/Python-and-Java-One-Line-Iterative-Recursive-Solutions
|
class Solution:
def reverse(self, l, r, s):
if l >= r:
return
s[l], s[r] = s[r], s[l]
self.reverse(l+1, r-1, s)
def reverseString(self, s: List[str]) -> None:
self.reverse(0, len(s)-1, s)
|
reverse-string
|
Python and Java - One Line, Iterative, Recursive Solutions
|
Pootato
| 8 | 554 |
reverse string
| 344 | 0.762 |
Easy
| 5,957 |
https://leetcode.com/problems/reverse-string/discuss/946287/Python-Easy-Solution
|
class Solution:
def reverseString(self, s: List[str]) -> None:
for i in range(len(s)//2):
s[i],s[-i-1]=s[-i-1],s[i]
return s
|
reverse-string
|
Python Easy Solution
|
lokeshsenthilkumar
| 8 | 1,400 |
reverse string
| 344 | 0.762 |
Easy
| 5,958 |
https://leetcode.com/problems/reverse-string/discuss/1902443/Python-Easy-Solutions-or-One-Liner
|
class Solution:
def reverseString(self, s: List[str]) -> None:
lo, hi = 0, len(s) - 1
while lo < hi:
s[lo], s[hi] = s[hi], s[lo]
lo += 1
hi -= 1
|
reverse-string
|
✅ Python Easy Solutions | One Liner
|
dhananjay79
| 6 | 952 |
reverse string
| 344 | 0.762 |
Easy
| 5,959 |
https://leetcode.com/problems/reverse-string/discuss/1902443/Python-Easy-Solutions-or-One-Liner
|
class Solution:
def reverseString(self, s: List[str]) -> None:
def reverse(lo,hi):
if lo > hi: return
s[lo], s[hi] = s[hi], s[lo]
reverse(lo+1, hi-1)
reverse(0, len(s)-1)
|
reverse-string
|
✅ Python Easy Solutions | One Liner
|
dhananjay79
| 6 | 952 |
reverse string
| 344 | 0.762 |
Easy
| 5,960 |
https://leetcode.com/problems/reverse-string/discuss/1902443/Python-Easy-Solutions-or-One-Liner
|
class Solution:
def reverseString(self, s: List[str]) -> None:
for i in range(1,len(s) // 2 + 1): s[i-1], s[-i] = s[-i], s[i-1]
|
reverse-string
|
✅ Python Easy Solutions | One Liner
|
dhananjay79
| 6 | 952 |
reverse string
| 344 | 0.762 |
Easy
| 5,961 |
https://leetcode.com/problems/reverse-string/discuss/2403620/Pythonoror4-methodoror-beginner-friendlyororeasyto-understandororvery-simple
|
class Solution:
def reverseString(self, s: List[str]) -> None:
i=0
j=len(s)-1
def rev(s,i,j):
if i>=j:
return
s[i],s[j]=s[j],s[i]
rev(s,i+1,j-1)
rev(s,i,j)
|
reverse-string
|
Python||4 method|| beginner-friendly||easyto understand||very simple
|
shikhar_srivastava391
| 4 | 184 |
reverse string
| 344 | 0.762 |
Easy
| 5,962 |
https://leetcode.com/problems/reverse-string/discuss/2403620/Pythonoror4-methodoror-beginner-friendlyororeasyto-understandororvery-simple
|
class Solution:
def reverseString(self, s: List[str]) -> None:
size = len(s)
for i in range(size//2):
s[i], s[-i-1] = s[-i-1], s[i]
|
reverse-string
|
Python||4 method|| beginner-friendly||easyto understand||very simple
|
shikhar_srivastava391
| 4 | 184 |
reverse string
| 344 | 0.762 |
Easy
| 5,963 |
https://leetcode.com/problems/reverse-string/discuss/1764369/Python-3-(200ms)-or-Two-Pointers-Approach-or-2-Solutions
|
class Solution:
def reverseString(self, s: List[str]) -> None:
st,e=0,len(s)-1
while st<=e:
s[st],s[e]=s[e],s[st]
st+=1
e-=1
|
reverse-string
|
Python 3 (200ms) | Two Pointers Approach | 2 Solutions
|
MrShobhit
| 4 | 338 |
reverse string
| 344 | 0.762 |
Easy
| 5,964 |
https://leetcode.com/problems/reverse-string/discuss/1764369/Python-3-(200ms)-or-Two-Pointers-Approach-or-2-Solutions
|
class Solution:
def reverseString(self, s: List[str]) -> None:
for i in range(len(s)//2): s[i], s[-i-1] = s[-i-1], s[i]
|
reverse-string
|
Python 3 (200ms) | Two Pointers Approach | 2 Solutions
|
MrShobhit
| 4 | 338 |
reverse string
| 344 | 0.762 |
Easy
| 5,965 |
https://leetcode.com/problems/reverse-string/discuss/1264496/Easy-Python-Solution
|
class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
for i in range(int(len(s)/2)):
s[i],s[len(s)-i-1]=s[len(s)-i-1],s[i]
|
reverse-string
|
Easy Python Solution
|
Sneh17029
| 3 | 1,000 |
reverse string
| 344 | 0.762 |
Easy
| 5,966 |
https://leetcode.com/problems/reverse-string/discuss/2095725/Simple-python-solutions-(3-different-methods)
|
class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
s.reverse()
|
reverse-string
|
Simple python solutions (3 different methods)
|
tylerpruitt
| 2 | 143 |
reverse string
| 344 | 0.762 |
Easy
| 5,967 |
https://leetcode.com/problems/reverse-string/discuss/2095725/Simple-python-solutions-(3-different-methods)
|
class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
s[:] = s[-1::-1]
|
reverse-string
|
Simple python solutions (3 different methods)
|
tylerpruitt
| 2 | 143 |
reverse string
| 344 | 0.762 |
Easy
| 5,968 |
https://leetcode.com/problems/reverse-string/discuss/2095725/Simple-python-solutions-(3-different-methods)
|
class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
# First pointer (start of array)
i = 0
# Second pointer (end of array)
j = len(s) - 1
while i < j:
s[i], s[j] = s[j], s[i]
i += 1
j -= 1
|
reverse-string
|
Simple python solutions (3 different methods)
|
tylerpruitt
| 2 | 143 |
reverse string
| 344 | 0.762 |
Easy
| 5,969 |
https://leetcode.com/problems/reverse-string/discuss/2346952/3-SOLUTIONS-oror-SINGLE-LINE-PYTHON-oror-oror-Beginner-friendly
|
class Solution:
def reverseString(self, s: List[str]) -> None:
s[:] = s[::-1]
|
reverse-string
|
3 SOLUTIONS || SINGLE LINE PYTHON || ✔✔|| Beginner friendly
|
adithya_s_k
| 1 | 100 |
reverse string
| 344 | 0.762 |
Easy
| 5,970 |
https://leetcode.com/problems/reverse-string/discuss/2346952/3-SOLUTIONS-oror-SINGLE-LINE-PYTHON-oror-oror-Beginner-friendly
|
class Solution(object):
def reverseString(self, s):
l = len(s)
if l < 2:
return s
return self.reverseString(s[l/2:]) + self.reverseString(s[:l/2])
**other two general solutions**
class SolutionClassic(object):
def reverseString(self, s):
r = list(s)
i, j = 0, len(r) - 1
while i < j:
r[i], r[j] = r[j], r[i]
i += 1
j -= 1
return "".join(r)
|
reverse-string
|
3 SOLUTIONS || SINGLE LINE PYTHON || ✔✔|| Beginner friendly
|
adithya_s_k
| 1 | 100 |
reverse string
| 344 | 0.762 |
Easy
| 5,971 |
https://leetcode.com/problems/reverse-string/discuss/2308497/Python-2-Easy-Solutions
|
class Solution:
def reverseString(self, s: List[str]) -> None:
left = 0
right = len(s) - 1
while left <= right:
s[left], s[right] = s[right], s[left]
left += 1
right -= 1
return s
|
reverse-string
|
✅Python 2 Easy Solutions
|
Skiper228
| 1 | 121 |
reverse string
| 344 | 0.762 |
Easy
| 5,972 |
https://leetcode.com/problems/reverse-string/discuss/2308497/Python-2-Easy-Solutions
|
class Solution:
def reverseString(self, s: List[str]) -> None:
s[:] = s[::-1]
|
reverse-string
|
✅Python 2 Easy Solutions
|
Skiper228
| 1 | 121 |
reverse string
| 344 | 0.762 |
Easy
| 5,973 |
https://leetcode.com/problems/reverse-string/discuss/2253500/O(n)-Runtime%3A-185-ms-faster-than-99.86-of-Python3-online-submissions-for-Reverse-String.
|
class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
start=0
end = len(s)-1
while start<end:
s[start],s[end] = s[end], s[start]
start = start+1
end = end-1
|
reverse-string
|
O(n) Runtime: 185 ms, faster than 99.86% of Python3 online submissions for Reverse String.
|
grishptl
| 1 | 105 |
reverse string
| 344 | 0.762 |
Easy
| 5,974 |
https://leetcode.com/problems/reverse-string/discuss/2134039/2-Approaches.-Easy-to-Understand-in-Python.
|
class Solution:
def reverseString(self, s: List[str]) -> None:
s[:]=s[::-1]
|
reverse-string
|
2 Approaches. Easy to Understand in Python.
|
AY_
| 1 | 103 |
reverse string
| 344 | 0.762 |
Easy
| 5,975 |
https://leetcode.com/problems/reverse-string/discuss/2134039/2-Approaches.-Easy-to-Understand-in-Python.
|
class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
sLen = len(s)
j = sLen-1
for i in range(sLen//2):
s[i],s[j] = s[j],s[i]
j -= 1
|
reverse-string
|
2 Approaches. Easy to Understand in Python.
|
AY_
| 1 | 103 |
reverse string
| 344 | 0.762 |
Easy
| 5,976 |
https://leetcode.com/problems/reverse-string/discuss/2131067/Python3-Two-pointer-Solution-faster-than-90.31-easily-explained.
|
class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
# so we use two pointers; left at 0 index and right at last index (len(numbers)-1)
# then in a while loop we swap our left index value to the right index value
# in one liner. That's it.
left, right = 0, len(s)-1
while left < right:
s[left], s[right] = s[right], s[left]
left += 1
right -= 1
# return s
|
reverse-string
|
Python3 Two-pointer Solution; faster than 90.31%, easily explained.
|
shubhamdraj
| 1 | 56 |
reverse string
| 344 | 0.762 |
Easy
| 5,977 |
https://leetcode.com/problems/reverse-string/discuss/2098031/Python3-1-optimal-solution-1-brute-force-2-builtIn
|
class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
return self.reverseStringOptimal(s)
# O(n) || O(1)
# runtime: 342 19.09%
def reverseStringOptimal(self, string):
if not string: return string
left, right = 0, len(string) - 1
while left < right:
string[left], string[right] = string[right], string[left]
left += 1
right -= 1
return string
# O(n) || O(n)
# brute force
def reverseStringWithNewList(self, string):
if not string:
return string
newList = [0] * len(string)
j = 0
for i in reversed(range(len(string))):
newList[i] = string[j]
j += 1
return newList
# below are just 'some' python built in
def reverseStringWithListCompression(self, string):
if not string: return string
return [string[i] for i in reversed(range(len(string)))]
def reversedStringWithReverse(self, string):
string.reverse()
return string or string[::-1]
|
reverse-string
|
Python3 1 optimal solution, 1 brute force, 2 builtIn
|
arshergon
| 1 | 69 |
reverse string
| 344 | 0.762 |
Easy
| 5,978 |
https://leetcode.com/problems/reverse-string/discuss/1787196/Dumb-python-solution-oror-faster-than-93-with-space-less-than-99
|
class Solution:
def reverseString(self, s: List[str]) -> None:
for i, v in enumerate(s[::-1]):
s[i] = v
|
reverse-string
|
Dumb python solution || faster than 93% with space less than 99%
|
shafzgamer
| 1 | 239 |
reverse string
| 344 | 0.762 |
Easy
| 5,979 |
https://leetcode.com/problems/reverse-string/discuss/1751782/Python-Simple-and-Clean-Python-Solution-By-Swapping-First-and-Last-Number
|
class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
length = len(s)
for i in range(length//2):
s[i],s[-(i+1)]=s[-(i+1)],s[i]
|
reverse-string
|
[ Python ] ✔✔ Simple and Clean Python Solution By Swapping First and Last Number
|
ASHOK_KUMAR_MEGHVANSHI
| 1 | 103 |
reverse string
| 344 | 0.762 |
Easy
| 5,980 |
https://leetcode.com/problems/reverse-string/discuss/1736515/Python-93.24-faster-ororSimplest-solution-with-explanationoror-Beg-to-Advoror-Two-Pointer
|
class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
s.reverse()
return s
|
reverse-string
|
Python 93.24% faster ||Simplest solution with explanation|| Beg to Adv|| Two Pointer
|
rlakshay14
| 1 | 152 |
reverse string
| 344 | 0.762 |
Easy
| 5,981 |
https://leetcode.com/problems/reverse-string/discuss/1736515/Python-93.24-faster-ororSimplest-solution-with-explanationoror-Beg-to-Advoror-Two-Pointer
|
class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
left, right = 0, len(s) - 1
while left < right:
#swaping the positions of the elements.
s[left], s[right] = s[right], s[left]
# incrementing both the pointers.
right -= 1
left += 1
|
reverse-string
|
Python 93.24% faster ||Simplest solution with explanation|| Beg to Adv|| Two Pointer
|
rlakshay14
| 1 | 152 |
reverse string
| 344 | 0.762 |
Easy
| 5,982 |
https://leetcode.com/problems/reverse-string/discuss/1621867/faster-than-82.25-of-Python3-online-submissions-for-Reverse-String.
|
class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
i, j = 0, len(s) - 1
while i != j and i < j:
s[i], s[j] = s[j], s[i]
i += 1
j -= 1
|
reverse-string
|
faster than 82.25% of Python3 online submissions for Reverse String.
|
sagarhasan273
| 1 | 46 |
reverse string
| 344 | 0.762 |
Easy
| 5,983 |
https://leetcode.com/problems/reverse-string/discuss/1559714/Python-two-pointers-O(n)-time-O(1)-space-beats-85-explained
|
class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
# two pointers, one at the beginning, the other at the end
# swap, unti you meet at center
# O(N) time, O(1) space
left_pointer, right_pointer = 0, len(s) - 1
while left_pointer < right_pointer:
s[left_pointer], s[right_pointer] = s[right_pointer], s[left_pointer]
left_pointer += 1
right_pointer -= 1
class Solution2:
def reverseString(self, s:List[str]) -> None:
# for loop + bitwise inversion might be faser, but while loop makes it more readable
for l in range(len(s) // 2):
s[l], s[~l] = s[~l], s[l]
|
reverse-string
|
[Python] two-pointers O(n) time, O(1) space, beats 85%, explained
|
mateoruiz5171
| 1 | 135 |
reverse string
| 344 | 0.762 |
Easy
| 5,984 |
https://leetcode.com/problems/reverse-string/discuss/1388002/Python3-faster-than-95.34
|
class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
l = len(s)
for i in range(l//2):
s[i], s[l-i-1] = s[l-i-1],s[i]
return
|
reverse-string
|
Python3 faster than 95.34%
|
saumyasuvarna
| 1 | 258 |
reverse string
| 344 | 0.762 |
Easy
| 5,985 |
https://leetcode.com/problems/reverse-string/discuss/1259231/Python3-dollarolution
|
class Solution:
def reverseString(self, s: List[str]) -> None:
i, j = 0, len(s)-1
while i < j:
temp = s[i]
s[i] = s[j]
s[j] = temp
i += 1
j -= 1
|
reverse-string
|
Python3 $olution
|
AakRay
| 1 | 296 |
reverse string
| 344 | 0.762 |
Easy
| 5,986 |
https://leetcode.com/problems/reverse-string/discuss/670009/Python-one-liner-solution
|
class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
s=s.reverse()
|
reverse-string
|
Python one liner solution
|
rajesh_26
| 1 | 80 |
reverse string
| 344 | 0.762 |
Easy
| 5,987 |
https://leetcode.com/problems/reverse-string/discuss/669559/simplest-solution-for-reverse-string-python3
|
class Solution:
def reverseString(self, s: List[str]) -> None:
s[:] = s[::-1]
|
reverse-string
|
simplest solution for reverse string [python3]
|
user4410vu
| 1 | 118 |
reverse string
| 344 | 0.762 |
Easy
| 5,988 |
https://leetcode.com/problems/reverse-string/discuss/291391/Python-basic-recursive-and-iterative-formula
|
class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
for i in range(len(s) // 2):
s[i], s[len(s) - i - 1] = s[len(s) - i - 1], s[i]
|
reverse-string
|
Python -- basic recursive and iterative formula
|
CorvusEtiam
| 1 | 293 |
reverse string
| 344 | 0.762 |
Easy
| 5,989 |
https://leetcode.com/problems/reverse-string/discuss/291391/Python-basic-recursive-and-iterative-formula
|
class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
return self.loop(s, 0, len(s) // 2)
def loop(self, s, i, final):
if i == final:
return
else:
s[i], s[len(s) - i - 1] = s[len(s) - i - 1], s[i]
return self.loop(s, i + 1, final)
|
reverse-string
|
Python -- basic recursive and iterative formula
|
CorvusEtiam
| 1 | 293 |
reverse string
| 344 | 0.762 |
Easy
| 5,990 |
https://leetcode.com/problems/reverse-string/discuss/2847979/Quick-Python-solution-to-reverse-string
|
class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
for i in range(0,len(s)):
s.insert(i,s.pop())
print(s)
sol=Solution()
sol.reverseString(["h","e","l","l","o"])
|
reverse-string
|
Quick Python solution to reverse string
|
dassdipanwita
| 0 | 1 |
reverse string
| 344 | 0.762 |
Easy
| 5,991 |
https://leetcode.com/problems/reverse-string/discuss/2846086/python3-oror-in-place-o(1)-memory
|
class Solution:
def reverseString(self, s: List[str]) -> None:
l, r = 0, len(s) - 1 # left and right pointer
while(l < r):
# switch chars at l and r
s[l], s[r] = s[r], s[l]
l += 1
r -= 1
|
reverse-string
|
python3 || in-place, o(1) memory
|
wduf
| 0 | 2 |
reverse string
| 344 | 0.762 |
Easy
| 5,992 |
https://leetcode.com/problems/reverse-string/discuss/2842646/Python-solution.
|
class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
l = 0
h = len(s) - 1
for i in range(len(s) // 2):
s[i], s[h] = s[h], s[l]
l += 1
h -= 1
|
reverse-string
|
Python solution.
|
ahti1405
| 0 | 2 |
reverse string
| 344 | 0.762 |
Easy
| 5,993 |
https://leetcode.com/problems/reverse-string/discuss/2840063/Reverse-String-or-Python-3.x-or-Two-pointers
|
class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
head = -1
tail = len(s) - 1
while head != tail:
tmp = s.pop(tail)
tail -= 1
s.append(tmp)
|
reverse-string
|
Reverse String | Python 3.x | Two pointers
|
SnLn
| 0 | 1 |
reverse string
| 344 | 0.762 |
Easy
| 5,994 |
https://leetcode.com/problems/reverse-string/discuss/2830609/Python-Stack-Two-Pointers
|
class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
stack = []
l, r = 0, len(s) - 1
stack.append((l, r))
while stack and l < r:
if stack:
l, r = stack.pop()
s[l], s[r] = s[r], s[l]
l += 1
r -= 1
stack.append((l , r))
|
reverse-string
|
Python Stack Two Pointers
|
user6168bh
| 0 | 1 |
reverse string
| 344 | 0.762 |
Easy
| 5,995 |
https://leetcode.com/problems/reverse-string/discuss/2819025/Simple-and-Fast-Python-Solution-One-Liner
|
class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
s.reverse()
|
reverse-string
|
Simple and Fast Python Solution - One-Liner
|
PranavBhatt
| 0 | 5 |
reverse string
| 344 | 0.762 |
Easy
| 5,996 |
https://leetcode.com/problems/reverse-string/discuss/2816647/list-reverse
|
class Solution:
def reverseString(self, s: List[str]) -> None:
s2 = s.copy()
lens = len(s)
for i in range(lens):
s[i] = s2.pop()
|
reverse-string
|
list-reverse
|
dexck7770
| 0 | 1 |
reverse string
| 344 | 0.762 |
Easy
| 5,997 |
https://leetcode.com/problems/reverse-string/discuss/2815904/Simple-Python-using-2-Pointers
|
class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
first, last = 0, len(s)-1
while first < last:
s[first],s[last] = s[last],s[first]
first +=1
last -= 1
return s
|
reverse-string
|
Simple Python using 2 Pointers
|
BhavyaBusireddy
| 0 | 2 |
reverse string
| 344 | 0.762 |
Easy
| 5,998 |
https://leetcode.com/problems/reverse-string/discuss/2813967/TC-%3A-95.40-Simple-solution
|
class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
s[:] = s[::-1]
|
reverse-string
|
😎 TC : 95.40 % Simple solution
|
Pragadeeshwaran_Pasupathi
| 0 | 3 |
reverse string
| 344 | 0.762 |
Easy
| 5,999 |
Subsets and Splits
Top 2 Solutions by Upvotes
Identifies the top 2 highest upvoted Python solutions for each problem, providing insight into popular approaches.