question_slug
stringlengths 3
77
| title
stringlengths 1
183
| slug
stringlengths 12
45
| summary
stringlengths 1
160
⌀ | author
stringlengths 2
30
| certification
stringclasses 2
values | created_at
stringdate 2013-10-25 17:32:12
2025-04-12 09:38:24
| updated_at
stringdate 2013-10-25 17:32:12
2025-04-12 09:38:24
| hit_count
int64 0
10.6M
| has_video
bool 2
classes | content
stringlengths 4
576k
| upvotes
int64 0
11.5k
| downvotes
int64 0
358
| tags
stringlengths 2
193
| comments
int64 0
2.56k
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
minimize-the-maximum-edge-weight-of-graph
|
Clear java solution easy to understand
|
clear-java-solution-easy-to-understand-b-ormm
|
Intuition
Use binary search with dfs.
Approach
First invert the edges for ease of dfs traversal.
Use binary search to check if the answer satisfied the given co
|
Sanjay0001
|
NORMAL
|
2025-01-14T16:16:27.979331+00:00
|
2025-01-14T16:16:27.979331+00:00
| 9 | false |
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
* Use binary search with dfs.
# Approach
<!-- Describe your approach to solving the problem. -->
* First invert the edges for ease of dfs traversal.
* Use binary search to check if the answer satisfied the given conditions are not, if satisfies then try lower values else try higher values.
* Conditions are simple, just check if we can traverse all nodes from 0th node using dfs graph traversal.
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
* O(n*nlong)
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
* O(n)
# Code
```java []
class Solution {
public void dfs(Map<Integer,List<Pair<Integer,Integer>>> graph, boolean[] visited, int i,int maxi) {
List<Pair<Integer,Integer>> nodes = graph.get(i);
visited[i] = true;
if (nodes == null) return;
for(Pair<Integer,Integer> node : nodes) {
if (!visited[node.getKey()] && node.getValue() <= maxi) {
dfs(graph, visited, node.getKey(), maxi);
}
}
}
public boolean isValid(int n, Map<Integer,List<Pair<Integer,Integer>>> graph, int maxi) {
boolean[] visited = new boolean[n];
dfs(graph, visited, 0, maxi);
for(int i =0;i<n;i++) {
if (!visited[i]) return false;
}
return true;
}
public int binarySearch(int n, Map<Integer,List<Pair<Integer,Integer>>> graph) {
int l = 1, r = 10000000, mid;
while(l<=r) {
mid = (l+r)/2;
if (isValid(n, graph, mid)) r = mid - 1;
else l = mid + 1;
}
return l;
}
public int minMaxWeight(int n, int[][] edges, int threshold) {
Map<Integer,List<Pair<Integer,Integer>>> graph = new HashMap<>();
for(int i=0;i<edges.length;i++) {
if (!graph.containsKey(edges[i][1])) {
graph.put(edges[i][1], new ArrayList<>());
}
graph.get(edges[i][1]).add(new Pair<Integer,Integer>(edges[i][0],edges[i][2]));
}
int ans = binarySearch(n,graph);
return ans == 10000001 ? -1 : ans;
}
}
```
| 0 | 0 |
['Java']
| 0 |
minimize-the-maximum-edge-weight-of-graph
|
Solution with BFS+Binary Search1 | Easily understandable |
|
solution-with-bfsbinary-search1-easily-u-tnnx
|
Complexity
Time complexity:
O(log(maxi) * (V+E))
Code
|
ap2112
|
NORMAL
|
2025-01-14T13:49:20.510857+00:00
|
2025-01-14T13:49:20.510857+00:00
| 10 | false |
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- **O(log(maxi) * (V+E))**
---
# Code
```cpp []
class Solution {
public:
bool isPossibleToZero(int n,vector<vector<pair<int,int>>>&adj,int mid){
queue<int> q;
vector<bool> vis(n) ;
q.push(0) ;
vis[0]=true ;
while(!q.empty()){
int u=q.front() ;
q.pop() ;
for(pair p : adj[u]){
int v=p.first ;
int w=p.second ;
if(w<=mid && !vis[v]){
vis[v]=true ;
q.push(v);
}
}
}
for(int i=0;i<n;i++){
if(!vis[i]){
return false;
}
}
return true;
}
int minMaxWeight(int n, vector<vector<int>>& edges, int threshold) {
vector<vector<pair<int,int>>> adj(n) ;
int maxWt=0 ;
for(int i=0;i<edges.size();i++){
int u=edges[i][0] ;
int v=edges[i][1] ;
int w=edges[i][2] ;
adj[v].push_back({u,w}) ;
maxWt=max(w,maxWt) ;
}
int result=INT_MAX;
int l=0,r=maxWt ;
while(l<=r){
int mid=l+(r-l)/2 ;
if(isPossibleToZero(n,adj,mid)){
result=mid;
r=mid-1 ;
}else{
l=mid+1 ;
}
}
return result==INT_MAX?-1:result ;
}
};
```
| 0 | 0 |
['C++']
| 0 |
minimize-the-maximum-edge-weight-of-graph
|
C++ | Prim | O(V+E)
|
c-prim-ove-by-aryonbe-hysp
|
Code
|
aryonbe
|
NORMAL
|
2025-01-14T13:36:08.672952+00:00
|
2025-01-14T13:36:08.672952+00:00
| 22 | false |
# Code
```cpp []
class Solution{
public:
int minMaxWeight(int n, vector<vector<int>>& edges, int threshold){
vector<unordered_map<int, int>> G(n);
for(auto& e: edges){
int u = e[0], v = e[1], w = e[2];
if(G[v].contains(u)) G[v][u] = min(G[v][u], w);
else G[v][u] = w;
}
int res = 0;
priority_queue<pair<int,int>, vector<pair<int,int>>, greater<>> pq;
pq.emplace(0,0);
unordered_set<int> visited;
while(pq.size()){
auto [w, u] = pq.top();
pq.pop();
if(visited.contains(u)) continue;
visited.insert(u);
res = max(res, w);
for(auto& [v,w]: G[u]){
pq.emplace(w,v);
}
}
return visited.size() == n ? res : -1;
}
};
```
| 0 | 0 |
['C++']
| 0 |
minimize-the-maximum-edge-weight-of-graph
|
Python | Prim's Algorithm | O(V+E)
|
python-prims-algorithm-ove-by-aryonbe-hrw8
|
Code
|
aryonbe
|
NORMAL
|
2025-01-14T12:54:20.694372+00:00
|
2025-01-14T12:54:20.694372+00:00
| 6 | false |
# Code
```python3 []
from heapq import heappush, heappop
class Solution:
def minMaxWeight(self, n: int, edges: List[List[int]], threshold: int) -> int:
G = defaultdict(list)
for u,v,w in edges:
G[v].append((u,w))
heap = [(0,0)]
visited = set()
res = 0
while heap:
w,u = heappop(heap)
if u in visited: continue
visited.add(u)
res = max(res, w)
for (v,w) in G[u]:
heappush(heap, (w,v))
return res if len(visited) == n else -1
```
| 0 | 0 |
['Python3']
| 0 |
minimize-the-maximum-edge-weight-of-graph
|
Binary search + dfs O(N*logn) || Beats 100%
|
binary-search-dfs-onlogn-beats-100-by-be-b2m0
|
Intuition
To find minimum possible value of the maximum edge weight ,we should do binary search on answer and check whether with the obtained weight using binar
|
be_fighter
|
NORMAL
|
2025-01-14T11:49:49.051589+00:00
|
2025-01-14T11:49:49.051589+00:00
| 49 | false |
# Intuition
1. To find minimum possible value of the maximum edge weight ,we should do binary search on answer and check whether with the obtained weight using binary search can reach to 0 from all nodes
2. If yes, then make answer as mid and high=mid-1 and again apply binary search
3. If No, the low=mid+1
4. For the checking whether it is possible to reach 0 from all nodes to make it linear
5. Reverse the edges
6. Using dfs and maintaing visited array check whether it is possible to vist all nodes from 0
7. If yes return true
8. Else false
# Complexity Analysis
- Time complexity:
O(N*logN) : O(logn) for binary search on weight and o(n) for dfs and check function
- Space complexity:
O(N): For visited vector
# Code
```cpp []
class Solution {
public:
// possible to reach from all nodes to 0
void dfs(int src,vector<pair<int,int>>adj[],vector<int>&vis,int mid){
vis[src]=1;
for(auto it:adj[src]){
if(!vis[it.first] && it.second<=mid){
dfs(it.first,adj,vis,mid);
}
}
}
// check function to find possiblity
bool check(int src,vector<pair<int,int>>adj[],int mid,int n){
vector<int>vis(n,0);
dfs(src,adj,vis,mid);
for(int i=0;i<vis.size();i++){
if(vis[i]==0){
return false;
}
}
return true;
}
int minMaxWeight(int n, vector<vector<int>>& edges, int threshold) {
vector<pair<int,int>>adj[n];
int mini=INT_MAX,maxi=INT_MIN;
//Reversing the edges
for(auto it:edges){
adj[it[1]].push_back({it[0],it[2]});
mini=min(mini,it[2]);
maxi=max(maxi,it[2]);
}
int ans=-1;
int low=mini,high=maxi;
//Binary search on answer
while(low<=high){
int mid=low+(high-low)/2;
if(check(0,adj,mid,n)){
ans=mid;
high=mid-1;
}
else{
low=mid+1;
}
}
return ans;
}
};
```
```java []
import java.util.*;
class Solution {
private void dfs(int src, List<List<int[]>> adj, boolean[] vis, int mid) {
vis[src] = true;
for (int[] neighbor : adj.get(src)) {
if (!vis[neighbor[0]] && neighbor[1] <= mid) {
dfs(neighbor[0], adj, vis, mid);
}
}
}
private boolean check(int src, List<List<int[]>> adj, int mid, int n) {
boolean[] vis = new boolean[n];
dfs(src, adj, vis, mid);
for (boolean v : vis) {
if (!v) return false;
}
return true;
}
public int minMaxWeight(int n, int[][] edges, int threshold) {
List<List<int[]>> adj = new ArrayList<>();
for (int i = 0; i < n; i++) {
adj.add(new ArrayList<>());
}
int mini = Integer.MAX_VALUE, maxi = Integer.MIN_VALUE;
for (int[] edge : edges) {
adj.get(edge[1]).add(new int[] { edge[0], edge[2] });
mini = Math.min(mini, edge[2]);
maxi = Math.max(maxi, edge[2]);
}
int ans = -1, low = mini, high = maxi;
while (low <= high) {
int mid = low + (high - low) / 2;
if (check(0, adj, mid, n)) {
ans = mid;
high = mid - 1;
} else {
low = mid + 1;
}
}
return ans;
}
}
```
```python []
class Solution:
def dfs(self, src, adj, vis, mid):
vis[src] = True
for neighbor, weight in adj[src]:
if not vis[neighbor] and weight <= mid:
self.dfs(neighbor, adj, vis, mid)
def check(self, src, adj, mid, n):
vis = [False] * n
self.dfs(src, adj, vis, mid)
return all(vis)
def minMaxWeight(self, n, edges, threshold):
adj = [[] for _ in range(n)]
mini, maxi = float('inf'), float('-inf')
for x, y, w in edges:
adj[y].append((x, w))
mini = min(mini, w)
maxi = max(maxi, w)
ans = -1
low, high = mini, maxi
while low <= high:
mid = low + (high - low) // 2
if self.check(0, adj, mid, n):
ans = mid
high = mid - 1
else:
low = mid + 1
return ans
```
```javascript []
class Solution {
dfs(src, adj, vis, mid) {
vis[src] = true;
for (const [neighbor, weight] of adj[src]) {
if (!vis[neighbor] && weight <= mid) {
this.dfs(neighbor, adj, vis, mid);
}
}
}
check(src, adj, mid, n) {
const vis = new Array(n).fill(false);
this.dfs(src, adj, vis, mid);
return vis.every(v => v);
}
minMaxWeight(n, edges, threshold) {
const adj = Array.from({ length: n }, () => []);
let mini = Infinity, maxi = -Infinity;
for (const [x, y, w] of edges) {
adj[y].push([x, w]);
mini = Math.min(mini, w);
maxi = Math.max(maxi, w);
}
let ans = -1, low = mini, high = maxi;
while (low <= high) {
const mid = Math.floor(low + (high - low) / 2);
if (this.check(0, adj, mid, n)) {
ans = mid;
high = mid - 1;
} else {
low = mid + 1;
}
}
return ans;
}
}
```
# If like the solution upvote
| 0 | 0 |
['Hash Table', 'Binary Search', 'Greedy', 'Depth-First Search', 'Breadth-First Search', 'Graph', 'Python', 'C++', 'Java', 'JavaScript']
| 1 |
minimize-the-maximum-edge-weight-of-graph
|
C# | O(nlog(n)) | Heap
|
c-onlogn-heap-by-alghi-8f7k
|
Complexity
Time complexity: O(nlogn)
Space complexity: O(n)
Code
|
Alghi
|
NORMAL
|
2025-01-14T11:00:48.761948+00:00
|
2025-01-14T11:03:08.125381+00:00
| 13 | false |
# Complexity
- Time complexity: O(nlogn)
- Space complexity: O(n)
# Code
```csharp []
public class Solution {
public int MinMaxWeight(int n, int[][] edges, int threshold) {
int max = int.MinValue;
PriorityQueue<(int, int), int> pq = new();
Dictionary<int, List<(int, int)>> adjMap = new();
HashSet<int> notVisited = new HashSet<int>();
for(int i = 0; i < n; i++)
{
notVisited.Add(i);
adjMap.Add(i, new List<(int, int)>());
}
for(int i = 0; i < edges.Length; i++)
{
adjMap[edges[i][1]].Add((edges[i][0], edges[i][2]));
}
pq.Enqueue((0, 0), 0);
while (pq.Count > 0 && notVisited.Count > 0)
{
(int currNode, int currWeight) = pq.Dequeue();
if (!notVisited.Contains(currNode)) continue;
notVisited.Remove(currNode);
max = Math.Max(max, currWeight);
foreach((int node, int weight) in adjMap[currNode])
{
pq.Enqueue((node, weight), weight);
}
}
return notVisited.Count > 0 ? -1 : max;
}
}
```
| 0 | 0 |
['Heap (Priority Queue)', 'Shortest Path', 'C#']
| 0 |
minimize-the-maximum-edge-weight-of-graph
|
Simple and Easy Java Code
|
simple-and-easy-java-code-by-sankeerth29-kt5z
|
IntuitionDo binary search as it is mentioned to find the The maximum edge weight in the resulting graph is minimized.
-> reverse the edges so that we know wheth
|
sankeerth2903
|
NORMAL
|
2025-01-14T09:26:41.467444+00:00
|
2025-01-14T09:26:41.467444+00:00
| 12 | false |
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
Do binary search as it is mentioned to find the The maximum edge weight in the resulting graph is minimized.
-> reverse the edges so that we know whether we reach 0 node or not, haha yes I have done this cleverly, I have used dijkshtras to know it is possible or not
# Code
```java []
class Solution {
public int minMaxWeight(int n, int[][] edges, int threshold) {
ArrayList<ArrayList<Pair>> adj=new ArrayList<>();
for(int i=0;i<n;i++){
adj.add(new ArrayList<>());
}
int min=Integer.MAX_VALUE;
int max=Integer.MIN_VALUE;
for(int i=0;i<edges.length;i++){
int dist=edges[i][2];
adj.get(edges[i][1]).add(new Pair(edges[i][0], edges[i][2]));
min=Math.min(min, dist);
max=Math.max(max, dist);
}
int ans=-1;
while(min<=max){
int mid=(min+max)/2;
if(isPossible(adj, mid)){
ans=mid;
max=mid-1;
} else{
min=mid+1;
}
}
return ans;
}
private boolean isPossible(ArrayList<ArrayList<Pair>> adj, int maxEdge){
int n=adj.size();
int[] movement = new int[n];
Arrays.fill(movement, Integer.MAX_VALUE);
movement[0]=0;
PriorityQueue<Pair> pq = new PriorityQueue<>(new Comparator<Pair>(){
public int compare(Pair p, Pair q){
return p.dist-q.dist;
}
});
pq.add(new Pair(0, 0));
while(pq.size()>0){
Pair pair=pq.poll();
int ver=pair.ver;
int dist=pair.dist;
for(int j=0;j<adj.get(ver).size() ; j++){
int nextVer = adj.get(ver).get(j).ver;
int nextDist = adj.get(ver).get(j).dist;
if(movement[nextVer]> dist+nextDist && nextDist <= maxEdge){
movement[nextVer] = dist+nextDist;
pq.add(new Pair(nextVer, movement[nextVer]));
}
}
}
for(int i=0;i<n;i++){
if(movement[i]==Integer.MAX_VALUE){
return false;
}
}
return true;
}
class Pair{
int ver;
int dist;
Pair(int ver, int dist){
this.ver=ver;
this.dist=dist;
}
}
}
```
# DO UPVOTE IF YOU FOUND THIS HELPFUL 😀
| 0 | 0 |
['Java']
| 1 |
minimize-the-maximum-edge-weight-of-graph
|
Binary search
|
binary-search-by-siddharth96shukla-xjge
|
Code
|
siddharth96shukla
|
NORMAL
|
2025-01-14T07:56:24.508589+00:00
|
2025-01-14T07:56:24.508589+00:00
| 5 | false |
# Code
```cpp []
#define pb push_back
class Solution {
public:
vector<vector<pair<int, int>>>g;
int chk(int v, int m, vector<bool>&vis){
vis[v]=1;int cnt=1;
for(auto p:g[v]){
if(p.second<=m && !vis[p.first])cnt+=chk(p.first, m, vis);
}
return cnt;
}
int minMaxWeight(int n, vector<vector<int>>& edges, int threshold) {
g.resize(n);
for(auto e:edges)g[e[1]].pb({e[0], e[2]});
int l=0, r=1000001;
while((r-l) > 1){
int m=l+((r-l)>>1);
vector<bool>vis(n, 0);
if(chk(0, m, vis)==n)r=m;
else l=m;
}
return (r==1000001)?-1:r;
}
};
```
| 0 | 0 |
['Binary Search', 'C++']
| 0 |
minimize-the-maximum-edge-weight-of-graph
|
Brute Force
|
brute-force-by-fartalesuraj20-fsw0
|
IntuitionApproachComplexity
Time complexity:
Space complexity:
Code
|
fartalesuraj20
|
NORMAL
|
2025-01-14T05:08:18.271510+00:00
|
2025-01-14T05:08:18.271510+00:00
| 3 | false |
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```cpp []
class Solution {
public:
vector<vector<pair<int,int>>>adj;
int cnt;
bool flag;
bool isPossible(int node,int maxWeight,vector<int>&visited,vector<int>&outEdges,int threshold){
if(visited[node] == 0){
visited[node] = 1;
cnt--;
}
if(cnt == 0) return true;
bool res = false;
for(pair<int,int>&P : adj[node]){
if(P.second <= maxWeight && visited[P.first] == 0)
{
outEdges[P.first]++;
if(outEdges[P.first] > threshold) flag = false;
res |= isPossible(P.first,maxWeight,visited,outEdges,threshold);
}
}
return res;
}
int minMaxWeight(int n, vector<vector<int>>& edges, int threshold) {
// int n = edges.size();
adj.resize(n);
set<int>st;
int maxWeight = 0;
for(vector<int>&e : edges){
adj[e[1]].push_back({e[0],e[2]});
maxWeight = max(maxWeight,e[2]);
st.insert(e[2]);
}
vector<int>visited(n,0);
vector<int>outEdges(n,0);
vector<int>weights(begin(st),end(st));
cnt = n;
flag = true;
int l = 0 , r = weights.size()-1;
int res = -1;
while(l <= r){
int mid = l + (r-l)/2;
maxWeight = weights[mid];
vector<int>visited(n,0);
vector<int>outEdges(n,0);
cnt = n;
flag = true;
bool status = isPossible(0,maxWeight,visited,outEdges,threshold);
if(status && flag){
res = maxWeight;
r = mid - 1;
}
else l = mid + 1;
}
return res;
}
};
```
| 0 | 0 |
['C++']
| 0 |
minimize-the-maximum-edge-weight-of-graph
|
Binary search + bfs
|
binary-search-bfs-by-gwwsc25-bfff
|
IntuitionApproach
reverse the map, threshold is useless
use bfs to search the min max problem
Complexity
Time complexity:
Space complexity:
Code
|
gwwsc25
|
NORMAL
|
2025-01-14T04:43:17.745921+00:00
|
2025-01-14T04:43:17.745921+00:00
| 4 | false |
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
1. reverse the map, threshold is useless
2. use bfs to search the min max problem
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```cpp []
class Solution {
public:
int minMaxWeight(int n, vector<vector<int>>& edges, int threshold) {
vector<vector<pair<int,int>>> graph(n);
int l = INT_MAX;
int r = INT_MIN;
for(auto & edge : edges){
graph[edge[1]].push_back({edge[0],edge[2]});
l = min(l, edge[2]);
r = max(r, edge[2]);
}
while(l < r){
int m = l + (r-l)/2;
if(bfs(graph, m, n)) // m is good
r = m;
else // m is not enough
l = m+1;
}
if(bfs(graph,r,n))
return r;
return -1;
}
bool bfs(vector<vector<pair<int,int>>> &graph, int k, int n){
int ans = 0;
vector<bool> visited(n, false);
queue<int> que;
que.push(0);
ans++;
visited[0] = true;
while(!que.empty()){
int curr = que.front();
que.pop();
for(auto & edge : graph[curr]){
if(!visited[edge.first]){
if(edge.second <=k){
que.push(edge.first);
ans++;
visited[edge.first] = true;
}
}
}
}
return ans == n;
}
};
```
| 0 | 0 |
['C++']
| 0 |
minimize-the-maximum-edge-weight-of-graph
|
Using mst
|
using-mst-by-risabhuchiha-q5z6
| null |
risabhuchiha
|
NORMAL
|
2025-01-13T17:55:12.974593+00:00
|
2025-01-13T17:55:12.974593+00:00
| 9 | false |
```java []
class Solution {
public int minMaxWeight(int n, int[][] edges, int threshold) {
List<List<int[]>> adj = new ArrayList<>();
int[] visited = new int[n];
for(int i=0;i<n;i++) adj.add(new ArrayList<>());
for(int[] x:edges){
adj.get(x[1]).add(new int[]{x[0],x[2]});
}
// visited[0] = 1;
int count = 1;
int ans=0;
PriorityQueue<int[]> q = new PriorityQueue<>((a,b)->a[1]-b[1]);
q.offer(new int[]{0,0});
while(!q.isEmpty()){
int[] cur = q.poll();
if(visited[cur[0]]==1)continue;
ans=Math.max(ans,cur[1]);
visited[cur[0]]=1;
for(int[] x:adj.get(cur[0])){
if(visited[x[0]] == 0){
q.offer(x);
}
}
}
for(int i=0;i<n;i++){
if(visited[i]==0)return -1;
}
return ans;
}
}
```
| 0 | 0 |
['Java']
| 0 |
minimize-the-maximum-edge-weight-of-graph
|
3419. Minimize the Maximum Edge Weight of Graph | Binary Search - Ignore Threshold Value ✨✅
|
3419-minimize-the-maximum-edge-weight-of-6rqc
|
Optimizing Graph Edge Weights Under Constraints 🛠️IntuitionThe problem asks us to minimize the maximum edge weight in a graph while meeting three constraints:
E
|
ayushkumar08032003
|
NORMAL
|
2025-01-13T16:46:42.737594+00:00
|
2025-01-13T16:46:42.737594+00:00
| 5 | false |
# **Optimizing Graph Edge Weights Under Constraints 🛠️**
## **Intuition**
The problem asks us to **minimize the maximum edge weight** in a graph while meeting three constraints:
1. Every node must be able to **reach node 0**.
2. Each node must have at most **`threshold` outgoing edges**.
3. The weight of the maximum edge in the graph should be minimized.
The **key insight** is that the maximum weight allowed (`limit`) directly determines the edges included in the graph. If we repeatedly adjust this `limit` using **binary search**, we can find the smallest `limit` that meets all conditions.
---
## **Approach**
### **Key Idea**
- Use **binary search** over the range of possible edge weights.
- For each candidate `limit`, check whether all nodes can reach node 0 using only edges with weights ≤ `limit`.
### **Steps**
1. **Reverse the edges:**
To simplify checking if every node can reach node 0, we **reverse all edges** in the graph. This transforms the problem into checking if node 0 can reach every other node.
2. **Graph Construction:**
Build an adjacency list for the reversed graph, where each edge stores its weight.
3. **Feasibility Check (`help` function):**
Using **BFS**, traverse the graph, keeping track of visited nodes. Only use edges with weights ≤ `limit`.
If all nodes are reachable, the current `limit` is valid.
4. **Binary Search on Edge Weights:**
- Start with the range of edge weights (`[min_edge_weight, max_edge_weight]`).
- Use the `help` function to check if a given mid-point (`limit`) is feasible.
- Adjust the search range based on the feasibility result.
5. **Threshold Constraint:**
The `threshold` constraint can be ignored here because the BFS traversal ensures at most one outgoing edge per node is processed.
---
## **Code with Explanation**
```cpp
#define vi vector<int>
#define vvi vector<vi>
#define pi pair<int, int>
#define vpi vector<pi>
#define vvpi vector<vpi>
class Solution {
public:
// Helper function to check feasibility with a given weight limit.
bool help(int n, vvi& edges, int limit) {
// Create a reversed graph adjacency list
vvpi graph(n);
for (vi& edge : edges) {
graph[edge[1]].push_back(make_pair(edge[0], edge[2])); // Reverse the edge
}
// BFS to check reachability to node 0 with weight ≤ limit
queue<int> q;
q.push(0);
vector<bool> vis(n, false);
vis[0] = true; // Mark node 0 as visited
while (!q.empty()) {
int node = q.front();
q.pop();
for (pi it : graph[node]) { // Check neighbors
if (!vis[it.first] && it.second <= limit) { // Only visit if weight ≤ limit
vis[it.first] = true;
q.push(it.first);
}
}
}
// Check if all nodes are visited
for (bool x : vis) {
if (!x)
return false; // If any node is unreachable, return false
}
return true; // All nodes are reachable
}
// Function to minimize the maximum edge weight
int minMaxWeight(int n, vector<vector<int>>& edges, int threshold) {
// Find the range of edge weights
int maxe = INT_MIN, mine = INT_MAX;
for (const vi& edge : edges) {
maxe = max(maxe, edge[2]); // Maximum weight
mine = min(mine, edge[2]); // Minimum weight
}
// Binary search to find the minimum maximum weight
int low = mine, high = maxe, ans = -1;
while (low <= high) {
int mid = low + (high - low) / 2; // Mid-point
if (help(n, edges, mid)) { // Check feasibility with current weight limit
ans = mid; // Update answer
high = mid - 1; // Try for a smaller maximum weight
} else {
low = mid + 1; // Increase the limit
}
}
return ans; // Return the minimum maximum weight
}
};
```
---
## **Time Complexity**
1. **Graph Construction:** \(O(E)\), where \(E\) is the number of edges.
2. **BFS Traversal:** \(O(E + V)\) for each `help` call, where \(V\) is the number of nodes.
3. **Binary Search:** Runs \(O(log(max\_weight - min\_weight))\) iterations.
#### **Total Complexity: O(E+V)⋅O(log(max_weight−min_weight))**
## **Space Complexity**
- **Graph Storage:** \(O(E + V)\) for adjacency list.
- **Visited Array:** \(O(V)\).
**Total Space: \(O(E + V)\).**
---
# **PLEASE UPVOTE IF YOU LIKED IT ✅**
| 0 | 0 |
['C++']
| 0 |
minimize-the-maximum-edge-weight-of-graph
|
Simple Binary Search
|
simple-binary-search-by-catchme999-jg62
|
IntuitionApproachComplexity
Time complexity:
Space complexity:
Code
|
catchme999
|
NORMAL
|
2025-01-13T16:11:58.962621+00:00
|
2025-01-13T16:11:58.962621+00:00
| 5 | false |
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```python3 []
from collections import deque, defaultdict
class Solution:
def minMaxWeight(self, n, edges, threshold):
# Create the adjacency list
graph = defaultdict(list)
max_weight = 0
# Populate the graph
for u, v, w in edges:
graph[v].append((u, w)) # Create incoming edges
max_weight = max(max_weight, w)
def bfs(mid):
visited = [0] * n
queue = deque([0])
visited[0] = 1
count = 1
while queue:
node = queue.popleft()
if count == n:
return True
for neighbor, weight in graph[node]:
if weight <= mid and not visited[neighbor]:
visited[neighbor] = 1
count += 1
queue.append(neighbor)
# Check if each node has at most `threshold` incoming edges
for node in range(n):
incoming_count = sum(1 for _, weight in graph[node] if weight <= mid)
if incoming_count > threshold:
return False
return count == n
# If the max_weight BFS traversal fails, return -1
if not bfs(max_weight):
return -1
# Binary search to find the minimum max weight
lo, hi = 0, max_weight
answer = max_weight
while lo <= hi:
mid = (lo + hi) // 2
if bfs(mid):
answer = mid
hi = mid - 1
else:
lo = mid + 1
return answer
```
| 0 | 0 |
['Python3']
| 0 |
minimize-the-maximum-edge-weight-of-graph
|
Simple BFS
|
simple-bfs-by-abinashgupta712-uvkf
|
IntuitionStart with node 0 and find minimize the weight in the path to reach other nodes .Approachreversed the edges for smooth traversal
Maintained a queue of
|
abinashgupta712
|
NORMAL
|
2025-01-13T15:06:45.435608+00:00
|
2025-01-13T15:06:45.435608+00:00
| 5 | false |
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
Start with node 0 and find minimize the weight in the path to reach other nodes .
# Approach
<!-- Describe your approach to solving the problem. -->
reversed the edges for smooth traversal
Maintained a queue of (node , max weight traversed to reach that node)
then traverse all the neighbour nodes
update weight of the neighbour node min(neighbour node weight, max(edgeWeight, srcNodeWeight));
if neighbour weight decreases in this step then we need to consider this node and its neighbours again
hence push it into the queue
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
O(E) we will traverse each node at most once
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
O(N) all nodes will be stored in the worst case scenario
# Code
```cpp []
class Solution {
public:
int minMaxWeight(int n, vector<vector<int>>& edges, int threshold) {
// maintain queue with node and curr min of max edge traversed in this path
// update weight of the dst node min(dst node weight, max(edgeWeight, srcNodeWeight));
// if dst node weight value changes then push
// else ignore
// return the max of the all the weight of the nodes
vector<vector<pair<int,int>>>g(n);
for(auto it:edges){
g[it[1]].push_back({it[0],it[2]});
}
queue<pair<int,int>>q;
vector<int>nodeWeight(n,1e9);
nodeWeight[0]=0;
q.push({0,0});
while(q.size()){
int node= q.front().first;
int srcWt=q.front().second;
q.pop();
for(auto it:g[node]){
int dstNode=it.first;
int edgeWt=it.second;
if(nodeWeight[dstNode]>max(edgeWt,srcWt)){
nodeWeight[dstNode]=max(edgeWt,srcWt);
q.push({dstNode,nodeWeight[dstNode]});
// cout<<"source Node :"<<node<<" source Weight :"<<srcWt<<" dstNode :"<<dstNode<<" dstEdgeWt :"<<nodeWeight[dstNode]<<" edge Weight"<<edgeWt<<endl;
}
}
}
int ans=-1;
for(auto it:nodeWeight){
ans=max(ans,it);
}
return ans==1e9?-1:ans;
}
};
```
| 0 | 0 |
['C++']
| 0 |
minimize-the-maximum-edge-weight-of-graph
|
Prim's Algorithm
|
prims-algorithm-by-dr0nser-l7je
|
I got this solution by reading through some of the posted solutions. I found this approach the most intuitive.The idea here is simply to check if you are able t
|
souvikdas
|
NORMAL
|
2025-01-13T14:11:09.545636+00:00
|
2025-01-13T14:11:33.876428+00:00
| 7 | false |
I got this solution by reading through some of the posted solutions. I found this approach the most intuitive.
The idea here is simply to check if you are able to reach all nodes from `0`. While traversing take the path with minimum weight. Since, we will take the minimum required path (i.e. minimum edges and weights) during traversing so the threshold checking is not required at all.
# Code
```java []
class Solution {
public int minMaxWeight(int n, int[][] edges, int threshold) {
List<List<Pair<Integer, Integer>>> adjList = new ArrayList<>();
for (int i = 0; i < n; i++)
adjList.add(new ArrayList<>());
// store the edges in reverse
// as we need to traverse from 0 to other nodes
for (int[] edge : edges) {
int u = edge[0], v = edge[1], w = edge[2];
adjList.get(v).add(new Pair(u, w));
}
// min heap
// arranges nodes in ascending order by edge weight
PriorityQueue<Pair<Integer, Integer>> pq = new PriorityQueue<>((a, b) -> a.getValue() - b.getValue());
int maxWeight = 0;
boolean[] vis = new boolean[n];
pq.add(new Pair(0, 0));
// run prims algo
while (!pq.isEmpty()) {
int node = pq.peek().getKey(), weight = pq.peek().getValue();
pq.remove();
if (vis[node])
continue;
vis[node] = true;
// keeping track of the maxWeight on the path
maxWeight = Math.max(weight, maxWeight);
for (Pair<Integer, Integer> adj : adjList.get(node)) {
int adjNode = adj.getKey();
if (!vis[adjNode])
pq.add(adj);
}
}
// check if all nodes are visited
for (int i = 0; i < n; i++)
if (!vis[i])
return -1;
return maxWeight;
}
}
```
| 0 | 0 |
['Java']
| 0 |
minimize-the-maximum-edge-weight-of-graph
|
Reverse Graph | Prism MST Algorithm | Intuitive Approach
|
reverse-graph-prism-mst-algorithm-intuit-25ag
|
IntuitionApproachComplexity
Time complexity:
Space complexity:
Code
|
itz_hemant_071
|
NORMAL
|
2025-01-13T09:47:03.629091+00:00
|
2025-01-13T10:07:52.409282+00:00
| 20 | false |
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
O((V+E)logV)
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
O(V+E)
# Code
```cpp []
class Solution {
public:
int MST(unordered_map<int,vector<pair<int,int>> > &adj,int n,vector<int> &visited){
priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> pq;
pq.push({0,0}); // {wt,node}
int maxWt = -1;
while(!pq.empty()){
int node = pq.top().second;
int wt = pq.top().first;
pq.pop();
if(visited[node]) continue;
visited[node]=1;
maxWt=max(maxWt,wt);
for(auto it : adj[node]){
int v = it.first;
int ed_Wt = it.second;
if(!visited[v]){
pq.push({ed_Wt,v});
}
}
}
for(int i=0;i<n;i++){
if(!visited[i]) return -1;
}
return maxWt;
}
int minMaxWeight(int n, vector<vector<int>>& edges, int threshold) {
unordered_map<int,vector<pair<int,int>> > adj;
for(auto it : edges){
int u = it[0];
int v = it[1];
int wt = it[2];
adj[v].push_back({u,wt});
}
vector<int> visited(n,0);
return MST(adj,n,visited);
}
};
```
| 0 | 0 |
['Breadth-First Search', 'Heap (Priority Queue)', 'Minimum Spanning Tree', 'C++']
| 0 |
minimize-the-maximum-edge-weight-of-graph
|
MBST solution
|
mbst-solution-by-blackzlq-3b6s
|
IntuitionThe problem requires us to construct a graph such that:
Node 0 is reachable from all other nodes.
Each node has at most threshold outgoing edges.
The m
|
blackzlq
|
NORMAL
|
2025-01-13T08:55:33.004279+00:00
|
2025-01-13T08:55:33.004279+00:00
| 6 | false |
# Intuition
The problem requires us to construct a graph such that:
1. Node 0 is reachable from all other nodes.
2. Each node has at most `threshold` outgoing edges.
3. The maximum edge weight in the resulting graph is minimized.
This boils down to:
- Reducing the graph to satisfy the outgoing edge constraint per node.
- Using a priority-based search (e.g., Dijkstra's algorithm) to find the optimal configuration of edges that minimizes the maximum edge weight.
The idea is to use a modified shortest path algorithm to explore all possible configurations while respecting the conditions.
# Approach
1. **Graph Construction**:
- Parse the input edges into an adjacency list for easier traversal.
2. **Priority Queue**:
- Use a priority queue (min-heap) to keep track of nodes and the current maximum edge weight. The priority queue ensures that we process paths with smaller weights first.
3. **Relaxation and Edge Constraint**:
- While traversing, maintain the `dis` array to track the smallest maximum weight to reach each node.
- For each node, enforce the threshold constraint by ensuring the number of outgoing edges considered does not exceed `threshold`. This can be done by sorting or selectively picking the smallest weights.
4. **Early Exit**:
- If all nodes are reachable with the given constraints, return the minimized maximum edge weight. Otherwise, return `-1`.
5. **Edge Validation**:
- After constructing the path, verify that all conditions are met (e.g., `threshold` outgoing edges, connectivity).
# Complexity
- **Time Complexity**:
- Constructing the adjacency list: $$O(E)$$, where \(E\) is the number of edges.
- Dijkstra's traversal with \(n\) nodes and up to \(E\) edges: $$O((E + n) \log n)$$.
- Enforcing the outgoing edge threshold: $$O(n \cdot \log n)$$ for sorting edges.
- Total: $$O(E \log n + n \log n)$$.
- **Space Complexity**:
- Adjacency list: $$O(E)$$.
- Priority queue and distance array: $$O(n)$$.
- Total: $$O(E + n)$$.
# Code
```java []
class Solution {
public int minMaxWeight(int n, int[][] edges, int threshold) {
Map<Integer, List<int[]>> map = new HashMap<>();
for (int[] edge : edges) {
List<int[]> list = map.getOrDefault(edge[1], new ArrayList<>());
list.add(edge);
map.put(edge[1], list);
}
int[] dis = new int[n];
Arrays.fill(dis, Integer.MAX_VALUE);
PriorityQueue<int[]> queue = new PriorityQueue<>((int[] x, int[] y) -> {
return x[2] < y[2] ? -1 : 1;
});
int result = -1;
queue.add(new int[]{0,0,0});
while (!queue.isEmpty()) {
int[] node = queue.poll();
if (node[1] < dis[node[0]]) {
result = Math.max(result, node[2]);
dis[node[0]] = node[2];
for (int[] next : map.getOrDefault(node[0], new ArrayList<>())) {
int to = next[0];
if (dis[node[0]] + next[2] < dis[to]) {
queue.add(new int[]{to, dis[node[0]] + next[2], next[2]});
}
}
}
}
for (int d : dis) if (d == Integer.MAX_VALUE) return -1;
return result;
}
}
```
| 0 | 0 |
['Java']
| 0 |
minimize-the-maximum-edge-weight-of-graph
|
Very Easy Java code
|
very-easy-java-code-by-sdflaskjdfj435345-oy21
|
Code
|
sdflaskjdfj435345
|
NORMAL
|
2025-01-13T08:35:30.164872+00:00
|
2025-01-13T08:35:30.164872+00:00
| 6 | false |
# Code
```java []
class Solution {
public int minMaxWeight(int n, int[][] edges, int threshold) {
ArrayList<ArrayList<int[]>> adj = new ArrayList<>();
for(int i=0;i<n;i++) adj.add(new ArrayList<>());
for(int j=0;j<edges.length;j++){
//dest,wt
adj.get(edges[j][1]).add(new int[]{edges[j][0],edges[j][2]});
}
PriorityQueue<int[]> pq = new PriorityQueue<>((a,b)->{
return a[2]-b[2];
});
// src,par,wt
pq.add(new int[]{0,-1,0});
int ans=0;
boolean vis[]=new boolean[n];
while(!pq.isEmpty()){
int arr[]=pq.remove();
int src=arr[0],par=arr[1],wt=arr[2];
if(par!=-1&&!vis[src]) ans=Math.max(ans,wt);
vis[src]=true;
for(int i[]:adj.get(src)){
if(!vis[i[0]]){
pq.add(new int[]{i[0],src,i[1]});
}
}
}
for(int i=0;i<n;i++){
if(!vis[i]) return -1;
}
return ans;
}
}
```
| 0 | 0 |
['Java']
| 0 |
minimize-the-maximum-edge-weight-of-graph
|
Easy BFS Solution | CPP | O(N)
|
easy-bfs-solution-cpp-on-by-sparkking29-yzmk
|
IntuitionWe need to traverse the graph while maintaining the maximum weight encountered so far, updating the relevant values as we go. Finally, we need to deter
|
SparkKing29
|
NORMAL
|
2025-01-13T06:01:28.676990+00:00
|
2025-01-13T06:01:28.676990+00:00
| 7 | false |
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
We need to traverse the graph while maintaining the maximum weight encountered so far, updating the relevant values as we go. Finally, we need to determine the maximum value from the resulting array.
# Complexity
- Time complexity:$$O(n)$$
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:$$O(n)$$
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```cpp []
class Solution {
public:
int minMaxWeight(int n, vector<vector<int>>& edges, int threshold) {
vector<vector<pair<int,int>>> adj(n,vector<pair<int,int>>());
for(vector<int> i: edges){
adj[i[1]].push_back({i[0],i[2]});
}
vector<int> a(n,INT_MAX);
queue<pair<int,int>> q;
q.push({0,INT_MIN});
while(!q.empty()){
int s = q.size();
for(int i = 0; i < s;i++){
pair<int,int> p = q.front();
a[p.first] = min(p.second,a[p.first]);
q.pop();
for(pair<int,int> t : adj[p.first]){
if( t.first != 0 && a[t.first] > max(t.second,p.second)){q.push({t.first,max(t.second,p.second)});}
}
}
}
int mx = INT_MIN;
for(int i = 1; i < a.size(); i++){
mx = max(mx,a[i]);
}
if(mx == INT_MAX) return -1;
return mx;
}
};
```
| 0 | 0 |
['Breadth-First Search', 'Graph', 'Queue', 'C++']
| 0 |
minimize-the-maximum-edge-weight-of-graph
|
🧠 Think Simply: Modified Dijkstra's Beats 100%
|
think-simply-modified-dijkstras-beats-10-ywrh
|
Link to Faster Than 100% Runhttps://leetcode.com/problems/minimize-the-maximum-edge-weight-of-graph/submissions/1506924000/Core ConceptsA solid understanding of
|
0p3r4t0r
|
NORMAL
|
2025-01-13T05:58:13.829834+00:00
|
2025-01-13T06:02:39.133325+00:00
| 13 | false |
# Link to Faster Than 100% Run
https://leetcode.com/problems/minimize-the-maximum-edge-weight-of-graph/submissions/1506924000/
# Core Concepts
A solid understanding of the following is needed to understand the solution.
- Dijkstra's: https://en.wikipedia.org/wiki/Dijkstra%27s_algorithm
- minHeap: https://en.wikipedia.org/wiki/Min-max_heap
# Key Insight
- We should reverse all edges in the graph. This allows us to use `0` as the source node for Dijkstra's.
- Standard Dijksta's prioritizes the shortest path,
but we need to prioritize the path with the lowest edge-weight.
- To do this the entries in our minHeap must take the form
`(max_edge_weight, vertex)`
This ensures that the entry with the minimum `max_edge_weight` always gets prioritized.
- Finally, because we are only interested in the global max_edge_weight,
there is no need to store an array of the minimum weights to all verticies
as in the standard Dijkstra's.
# Complexity
- Time complexity:
$O(E + V\log(V))$
- Space complexity:
$O(E + V)$
# Code
```python3 []
class Solution:
def minMaxWeight(self, n: int, edges: List[List[int]], threshold: int) -> int:
"""
The Algorithm ----------------------------------------------
1. Use modified version of Dijkstra's. Reverse all edges.
2. Track the max_edge_weight along the path as we go (ans).
3. In modified Dijkstra's, **prioritize paths with a low max_edge_weight**.
NOT THE SHORTEST PATHS.
4. In the end, validate that all nodes are reachable.
5. If all nodes are reachable return the max_edge_weight.
"""
adj = { i: list() for i in range(n) }
for A, B, W in edges:
# 1: Reverse edges
adj[B].append((W, A))
visited = set()
# 2: track global max_edge weight
ans = 0
# 3: heap of (max_edge_weight, vertex)
heap = [ (0, 0) ]
while heap:
max_edge_weight, v1 = heappop(heap)
if v1 in visited:
continue
# 2: track global max_edge_weight
ans = max(ans, max_edge_weight)
visited.add(v1)
for w2, v2 in adj[v1]:
if v2 not in visited:
heappush(heap, (max(max_edge_weight, w2), v2))
# 4: verify all nodes visited
if len(visited) < n:
return -1
return ans
```
| 0 | 0 |
['Python3']
| 0 |
minimize-the-maximum-edge-weight-of-graph
|
Easy Sol
|
easy-sol-by-singhmohit9868nsitco_re_ring-4h3a
|
IntuitionApproachOnce a node is visited this means a edge is reaching the node and so if some other edge also reaches the same node it is of no use just we have
|
Singhmohit9868NSITCO_Re_Ring06
|
NORMAL
|
2025-01-13T05:10:36.628276+00:00
|
2025-01-13T05:10:36.628276+00:00
| 8 | false |
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
Once a node is visited this means a edge is reaching the node and so if some other edge also reaches the same node it is of no use just we have to ignore this as we have already visited the node and so we are doing during the dfs part
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```cpp []
class Solution {
public:
void dfs(int i,vector<vector<pair<int,int>>>&graph,vector<int>&st,int wght){
if(st[i]){
return;
}
st[i]=1;
for(auto [v,w]:graph[i]){
if(w>wght)continue;
dfs(v,graph,st,wght);
}
}
int minMaxWeight(int n, vector<vector<int>>& edges, int threshold) {
int ans=-1;
int l=0;
int h=1e6+10;
vector<vector<pair<int,int>>>graph(n);
for(auto i:edges){
int u=i[0];
int v=i[1];
int w=i[2];
graph[v].push_back({u,w});
}
auto check = [&](int mid){
vector<int>edge(n);
vector<int>st(n);
dfs(0,graph,st,mid);
if(all_of(st.begin(),st.end(),[](int x){
return x==1;
}))return 1;
return 0;
};
while(l<=h){
int mid=(l+h)/2;
if(check(mid)){
ans=mid;
h=mid-1;
}
else{
l=mid+1;
}
}
return ans;
}
};
```
| 0 | 0 |
['Binary Search', 'Greedy', 'Depth-First Search', 'Graph', 'Sorting', 'C++']
| 0 |
minimize-the-maximum-edge-weight-of-graph
|
Minimize the Maximum Edge Weight of Graph
|
minimize-the-maximum-edge-weight-of-grap-sh3r
|
IntuitionApproachBinary Search and BFSComplexity
Time complexity:O(V+E)log(w)
Space complexity:O(N)
Code
|
Subham_Pujari23
|
NORMAL
|
2025-01-13T04:24:16.626955+00:00
|
2025-01-13T04:24:16.626955+00:00
| 5 | false |
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
Binary Search and BFS
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:O(V+E)log(w)
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:O(N)
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```java []
import java.util.*;
class Solution {
public int minMaxWeight(int n, int[][] edges, int threshold) {
int min = Integer.MAX_VALUE;
int max = Integer.MIN_VALUE;
// Find the minimum and maximum edge weights
for (int[] edge : edges) {
min = Math.min(edge[2], min);
max = Math.max(edge[2], max);
}
int ans = -1;
int l = min;
int h = max;
// Binary search on the edge weight to find the minimal weight that satisfies the condition
while (l <= h) {
int mid = l + (h - l) / 2;
if (isPossible(n, edges, mid)) {
ans = mid;
h = mid - 1; // Search in the lower half
} else {
l = mid + 1; // Search in the upper half
}
}
return ans;
}
public boolean isPossible(int n, int[][] edges, int target) {
// Adjacency list to store the graph
List<List<Integer>> list = new ArrayList<>();
for (int i = 0; i < n; i++) {
list.add(new ArrayList<>());
}
// Build the graph by adding edges with weight <= target
for (int[] edge : edges) {
if (edge[2] <= target) {
list.get(edge[1]).add(edge[0]); // Since it's an undirected graph
}
}
// Perform BFS to check if all nodes can be visited
Queue<Integer> q = new LinkedList<>();
boolean[] visited = new boolean[n];
q.offer(0);
visited[0] = true;
int count = 1; // Start with 1 since node 0 is already visited
while (!q.isEmpty()) {
int curr = q.poll();
for (int next : list.get(curr)) {
if (!visited[next]) {
q.add(next);
visited[next] = true;
count++;
}
}
}
// If all nodes are visited, return true
return count == n;
}
}
```
| 0 | 0 |
['Java']
| 0 |
minimize-the-maximum-edge-weight-of-graph
|
Reverse Graph | BFS | Binary search | Easy to understand
|
reverse-graph-bfs-binary-search-easy-to-rd6f2
|
IntuitionThe goal is to minimize the maximum edge weight in a directed graph while ensuring:
All nodes can reach node 0.
Each node has at most threshold outgoin
|
Abhidnya98
|
NORMAL
|
2025-01-13T03:35:34.990602+00:00
|
2025-01-13T03:35:34.990602+00:00
| 15 | false |
# Intuition
The goal is to minimize the maximum edge weight in a directed graph while ensuring:
1. All nodes can reach node `0`.
2. Each node has at most `threshold` outgoing edges.
This problem can be tackled by treating the edge weights as a binary search problem. We iteratively check if the graph can satisfy the conditions for a given maximum edge weight.
# Approach
1. **Binary Search for Edge Weight**:
- Identify the range of edge weights, from the minimum weight in `edges` to the maximum.
- Perform binary search over this range to find the smallest possible maximum weight for which the conditions are satisfied.
2. **Validation using BFS**:
- For a given maximum weight (`mid`), construct a reversed adjacency list of edges with weight ≤ `mid`.
- Use BFS to check if all nodes can reach node `0` in this reduced graph.
3. **Handling Edge Removal**:
- During graph construction, ignore edges with weight > `mid` and validate only those that satisfy the threshold condition.
4. **Result**:
- If the conditions are satisfied for a given `mid`, move the search range towards smaller weights.
- If not, adjust the range towards larger weights.
# Complexity
- **Time Complexity**:
- Let `m` be the number of edges and `n` be the number of nodes.
- Binary search runs in `O(log(maxW - minW))`.
- Each validation (BFS) step involves traversing all edges once, i.e., `O(m)`.
- Total: `O(mlog(maxW - minW))`.
- **Space Complexity**:
- Storing the adjacency list requires `O(m)`.
- BFS uses `O(n)` for the visited array and queue.
- Total: `O(m + n)`.
# Code
```java []
class Solution {
public int minMaxWeight(int n, int[][] edges, int threshold) {
int minW = Integer.MAX_VALUE;
int maxW = Integer.MIN_VALUE;
for (int[] edge : edges) {
minW = Math.min(minW, edge[2]);
maxW = Math.max(maxW, edge[2]);
}
int l = minW;
int r = maxW;
int ans = -1;
while (l <= r) {
int mid = l + (r - l) / 2;
if (possible(n, edges, mid)) {
ans = mid;
r = mid - 1;
} else {
l = mid + 1;
}
}
return ans;
}
private boolean possible(int n, int[][] edges, int max) {
HashMap<Integer, List<Integer>> adj = new HashMap<>();
for (int[] edge : edges) {
int from = edge[0];
int to = edge[1];
int wt = edge[2];
if (wt <= max) {
if (!adj.containsKey(to)) {
adj.put(to, new ArrayList<>());
}
adj.get(to).add(from);
}
}
boolean[] visited = new boolean[n];
Queue<Integer> q = new LinkedList<>();
q.add(0);
int count = 0;
while (!q.isEmpty()) {
int curr = q.remove();
if (visited[curr]) {
continue;
}
visited[curr] = true;
count++;
if (adj.get(curr) != null)
for (int next : adj.get(curr)) {
q.add(next);
}
}
return count == n;
}
}
```
| 0 | 0 |
['Binary Search', 'Breadth-First Search', 'Graph', 'Java']
| 0 |
minimize-the-maximum-edge-weight-of-graph
|
Priority Queue Java Solution
|
priority-queue-java-solution-by-haozhenb-ccdc
|
IntuitionWe first build a reverse graph of the original edges, so that we can try to reach all nodes from 0. We use a priority queue to store all the reversed e
|
Haozhenbo
|
NORMAL
|
2025-01-13T00:58:33.533994+00:00
|
2025-01-13T00:58:33.533994+00:00
| 6 | false |
# Intuition
We first build a reverse graph of the original edges, so that we can try to reach all nodes from 0. We use a priority queue to store all the reversed edges, so that we always using the edge with the minimum weight.
# Approach
From the edges, we build a reverse graph, where edge `a->b` is saved as an reverse edge `b->a` with same weights. We ignore the original `0->` outward paths, as we only want to make sure 0 is accessible for all nodes by the original edges.
We use a set to store the reachable nodes from 0 by the reversed edge.
We use a priority queue to store the reversed edges, starting from the reversed edges from 0, with weight as the key (from small to large).
While the number of reachable nodes are less than n and that the priority queue is not empty, we pop the reversed edge with smallest weight, we add the node to the reachable set, and update the min max weight with the edge's weight, as it is used by us to connect the node to 0. Then, for each neighbor node of the end node of the reversed edge, if it is not already in the reachable set, we add the reversed edge into the priority queue.
We return if the reachable size is n in the end.
# Complexity
- Time complexity:
$$O(E \log E)$$
- Space complexity:
$$O(E)$$
# Code
```java []
class Solution {
public int minMaxWeight(int n, int[][] edges, int threshold) {
// reverseGraph[node] is the reversed path for node
List<Map<Integer,Integer>> reverseGraph = new ArrayList<>();
for (int i = 0; i < n; i++) {
reverseGraph.add(new HashMap<>()); // initialize the list
}
for (int[] edge: edges) {
if (edge[0] == 0) continue; // No need to add edge starting from 0
Map<Integer, Integer> edgeMap = reverseGraph.get(edge[1]);
// start: edge[1], end: edge[0], weight: edge[2]
// For the current start, we either:
// 1. add an edge to end with weight, if there isn't path start -> end
// 2. if there is such a path, we update its weight to be the smaller one
edgeMap.put(edge[0],
Math.min(
edgeMap.getOrDefault(edge[0], Integer.MAX_VALUE),
edge[2])
);
}
// reachableSet is the nodes that can reach 0
Set<Integer> reachableSet = new HashSet<>();
reachableSet.add(0);
int ans = 0; // the min max weight
// use a pq to store all edges, with weight as the key
PriorityQueue<int[]> pq = new PriorityQueue<>((a, b) -> a[0] - b[0]);
// add all paths from 0 to pq
// the int[] is {weight, from, to}
for (Map.Entry<Integer, Integer> edge: reverseGraph.get(0).entrySet()) {
pq.offer(new int[] {edge.getValue(), 0, edge.getKey()});
}
while (reachableSet.size() < n && !pq.isEmpty()) {
int[] edge = pq.poll();
int weight = edge[0], start = edge[1], end = edge[2];
if (reachableSet.contains(end)) continue;
// since we add nodes with path weights small to large,
// if the node is already reachable, no need to process again,
// as the weight must be larger
// add the current end to the reachable
reachableSet.add(end);
// udpate the min max weight
ans = Math.max(ans, weight);
Map<Integer, Integer> edgeMap = reverseGraph.get(end);
// for each outgoing paths of the end, add to the pq
for (Map.Entry<Integer, Integer> e: edgeMap.entrySet()) {
pq.offer(new int[] {e.getValue(), end, e.getKey()});
}
}
return reachableSet.size() == n ? ans : -1;
}
}
```
| 0 | 0 |
['Heap (Priority Queue)', 'Java']
| 0 |
minimize-the-maximum-edge-weight-of-graph
|
C++ BFS, beats > 90%
|
c-bfs-beats-90-by-jakonen-aul3
|
IntuitionConsider what the solution will look like: All of the nodes must be able to reach node 0, but otherwise we can remove as many edges as we want. So it b
|
jakonen
|
NORMAL
|
2025-01-12T23:49:53.179904+00:00
|
2025-01-12T23:49:53.179904+00:00
| 5 | false |
# Intuition
Consider what the solution will look like: All of the nodes must be able to reach node 0, but otherwise we can remove as many edges as we want. So it becomes a tree, with node 0 as the root and each edge [a,b] denoting a child-parent relationship. Thinking it this way, the threshold becomes irrelevant because each node only has one parent.
# Approach
Three steps:
1. First convert edges to a structure that can be traversed more easily - a vector of child nodes for each node. We can ignore all outgoing edges from 0.
2. Use BFS starting from node 0. Keep track of maximum weight in path to root node. For each child node, check if the current maximum is larger than max of the parent + weight of the edge from child to parent. If so, update the maximum (i.e. minimize it).
3. Lastly, go through all nodes to get min-max for all nodes. If any node was not updated during step 2, it means the graph was disconnected return -1, otherwise the answer is the max.
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```cpp []
class Solution {
public:
vector<vector<int>> childEdges;
vector<int> maxWeight;
int minMaxWeight(int n, vector<vector<int>>& edges, int threshold) {
childEdges.resize(n);
maxWeight.resize(n, INT_MAX);
maxWeight[0] = 0;
// Step 1
for (int i = 0; i < edges.size(); ++i) {
if (edges[i][0] != 0) {
childEdges[edges[i][1]].push_back(i);
}
}
// Step 2
unordered_set<int> nodes = {0};
while (!nodes.empty()) {
unordered_set<int> nextNodes;
for (int node : nodes) {
for (int edgeIndex : childEdges[node]) {
int child = edges[edgeIndex][0];
int weight = edges[edgeIndex][2];
if (maxWeight[child] > max(maxWeight[node], weight)) {
maxWeight[child] = max(maxWeight[node], weight);
nextNodes.insert(child);
}
}
}
nodes = std::move(nextNodes);
}
// Step 3
int ans = 0;
for (int weight : maxWeight) {
if (weight == INT_MAX)
return -1;
ans = max(ans, weight);
}
return ans;
}
};
```
| 0 | 0 |
['C++']
| 0 |
invalid-tweets
|
MySQL: LENGTH() is incorrect. Important difference between CHAR_LENGTH() vs LENGTH()
|
mysql-length-is-incorrect-important-diff-90ue
|
Using LENGTH() is incorrect. The question is asking for the number of characters used in the content. LENGTH() returns the length of the string measured in byte
|
vine9
|
NORMAL
|
2020-12-11T17:36:13.024510+00:00
|
2020-12-12T15:36:19.040235+00:00
| 35,295 | false |
Using ```LENGTH()``` is incorrect. The question is asking for ```the number of characters used in the content```. ```LENGTH()``` returns the length of the string measured in bytes. ```CHAR_LENGTH()``` returns the length of the string measured in characters.\n\n```sql\nSELECT tweet_id\nFROM Tweets\nWHERE CHAR_LENGTH(content) > 15\n```\n\nReference:\nhttps://stackoverflow.com/questions/1734334/mysql-length-vs-char-length?rq=1\n\n* LENGTH() returns the length of the string measured in bytes.\n* CHAR_LENGTH() returns the length of the string measured in characters.\n\nThis is especially relevant for Unicode, in which most characters are encoded in two bytes, or UTF-8, where the number of bytes varies. \n\nExample:\n```sql \nSELECT LENGTH(\'\u20AC\') # is equal to 3\n```\n```sql \nSELECT CHAR_LENGTH(\'\u20AC\') # is equal to 1\n```\n\n*Important Note:* Using ```LENGTH()``` will pass the testcases. If the testcases included characters such as ```\u20AC``` then it would fail as shown in the examples above.
| 606 | 2 |
[]
| 35 |
invalid-tweets
|
Solution With Extended Problem Statement
|
solution-with-extended-problem-statement-ri5t
|
Initial Problem Statement:\nThe task is to find the IDs of invalid tweets where the number of characters in the content exceeds 15.\n# Solution\n\nSELECT tweet_
|
Ahmed_Abdelmoneim
|
NORMAL
|
2024-06-22T07:31:55.034227+00:00
|
2024-06-22T07:31:55.034255+00:00
| 58,269 | false |
# Initial Problem Statement:\nThe task is to find the IDs of invalid tweets where the number of characters in the content exceeds 15.\n# Solution\n```\nSELECT tweet_id FROM Tweets\nWHERE LENGTH(content) > 15;\n```\n# Extended Problem Statement:\nIf the problem statement is modified such that a tweet is considered invalid if it contains more than 2 words, the following solution can be used:\n# Modified Solution:\n```\nSELECT tweet_id\nFROM Tweets\nWHERE LENGTH(content) - LENGTH(REPLACE(content, \' \', \'\')) + 1 > 2;\n```\n- LENGTH(content): This returns the total length of the content\n- REPLACE(content, \' \', \'\'): This removes all spaces from the content.\n- LENGTH(content) - LENGTH(REPLACE(content, \' \', \'\')): This calculates the number of spaces in the content\n- Adding 1 to the number of spaces gives the number of words.\n- The condition LENGTH(content) - LENGTH(REPLACE(content, \' \', \'\')) + 1 > 2 checks if the number of words is greater than 2.\n\n# Don\'t forget to upvote plz\n\n\n
| 413 | 0 |
['MySQL']
| 6 |
invalid-tweets
|
🔥💯 Pandas || MySQL: An Effortless Simple Approach! 🔥💯
|
pandas-mysql-an-effortless-simple-approa-hxy3
|
\n# Approach\n Describe your approach to solving the problem. \n- Filter the rows where the length of the content column is strictly greater than 15. This can b
|
sriganesh777
|
NORMAL
|
2023-08-02T12:02:51.504497+00:00
|
2023-08-02T12:02:51.504522+00:00
| 27,010 | false |
\n# Approach\n<!-- Describe your approach to solving the problem. -->\n- Filter the rows where the length of the content column is strictly greater than 15. This can be done using the str.len() method of the content column.\n- Select only the tweet_id column from the filtered DataFrame. We are interested in getting the IDs of invalid tweets.\n- Return the result_df, which will contain the IDs of the invalid tweets.\n# Pandas Code\n```\nimport pandas as pd\n\ndef invalid_tweets(tweets: pd.DataFrame) -> pd.DataFrame:\n # Filter rows where the length of \'content\' is strictly greater than 15\n invalid_tweets_df = tweets[tweets[\'content\'].str.len() > 15]\n \n # Select only the \'tweet_id\' column from the invalid tweets DataFrame\n result_df = invalid_tweets_df[[\'tweet_id\']]\n \n return result_df\n```\n# Mysql code\n```\nSELECT tweet_id\nFROM Tweets\nWHERE CHAR_LENGTH(content) > 15;\n```\n# Explanation\n- We use the SELECT statement to retrieve the tweet_id column from the Tweets table.\n\n- The WHERE clause filters the rows where the length of the content column is strictly greater than 15. This ensures that we get the IDs of the invalid tweets.\n\nThe SQL query will return the tweet_id of the invalid tweets where the number of characters in the content of the tweet is strictly greater than 15.\n\n
| 121 | 0 |
['MySQL', 'Pandas']
| 7 |
invalid-tweets
|
Simple & Easy Approach 🔥II length() ✅ || sql
|
simple-easy-approach-ii-length-sql-by-ka-rgn6
|
Approach\n Describe your approach to solving the problem. \nSimplest approach to extract the valid tweets by comparing the length of content grater than 15.i.e.
|
kartikeymish
|
NORMAL
|
2023-05-11T09:12:14.448801+00:00
|
2023-05-11T09:12:14.448841+00:00
| 10,245 | false |
# Approach\n<!-- Describe your approach to solving the problem. -->\nSimplest approach to extract the valid tweets by comparing the length of content grater than 15.i.e. `length(content)>15` \n\n\n# Code\n```\nselect tweet_id from Tweets where length(content)>15\n```\n\n
| 72 | 0 |
['MySQL']
| 4 |
invalid-tweets
|
One line solution - LENGTH() vs CHAR_LENGTH() || SQL Gotcha
|
one-line-solution-length-vs-char_length-59d78
|
Using LENGTH() instead of CHAR_LENGTH() is incorrect when trying to get the number of characters in the string\n\n# Definitions: \n\n- LENGTH(): returns the len
|
Pacman45
|
NORMAL
|
2023-12-20T16:20:39.139829+00:00
|
2023-12-20T16:20:39.139862+00:00
| 4,384 | false |
Using ``LENGTH()`` instead of ```CHAR_LENGTH()``` is incorrect when trying to get the ```number of characters in the string```\n\n# Definitions: \n\n- LENGTH(): returns the length of a string _(in bytes)_.\n- CHAR_LENGTH(): return the length of a string _(in characters)_.\n\n# Code\n```\n# Write your MySQL query statement below\nSELECT tweet_id FROM Tweets \nWHERE CHAR_LENGTH(content) > 15;\n```\n# Important Note: \n\nUsing Length() indeed passes all the test cases in this example. But it fails in case of UTF8 where different characters take up different number of bytes. \n\nFor example: Each Chinese character is represented by 3 bytes.\n\n```\n-- Creating a table with a string containing a multibyte character\nCREATE TABLE Example (\n id INT,\n content VARCHAR(50)\n);\n\nINSERT INTO Example VALUES\n(1, \'Hello, \u4F60\u597D\'); --> (\'Hello\' -> 3 + \', \' -> 4 + \'\u4F60\u597D\' -> 6 = 13)\n\n\nSELECT LENGTH(content) AS Char_Length FROM Example;\n```\n\nOutput: \n\n```\n+------------+\n| Char_Length |\n+------------+\n| 13 |\n+------------+\n```\n\n**Correct** length of the string is **9** but using Length() here gives out **13** as the answer.
| 30 | 0 |
['MySQL']
| 2 |
invalid-tweets
|
Easy and Simple Solution | 100% Faster Approach
|
easy-and-simple-solution-100-faster-appr-q9oz
|
CodeIf you found this solution helpful, please give it an upvote! 🌟✨ It really motivates me to share more solutions like this. 😊👍
Happy coding! 💻🐱🏍
🔥 Approach
|
Yadav_Akash_
|
NORMAL
|
2025-01-29T18:42:05.591163+00:00
|
2025-01-29T18:42:05.591163+00:00
| 6,727 | false |
# Code
```mysql []
# Write your MySQL query statement below
SELECT tweet_id
FROM Tweets
WHERE CHAR_LENGTH(content)>15;
```
**If you found this solution helpful, please give it an upvote! 🌟✨ It really motivates me to share more solutions like this. 😊👍
Happy coding! 💻🐱🏍**
<img src="https://assets.leetcode.com/users/images/6e8c623a-1111-4d21-9a5b-54e00e7feabb_1738093236.165277.jpeg" width=300px>
# 🔥 Approach & Intuition
### Filtering Based on Content Length:
- We check if the `content` has more than 15 characters using `CHAR_LENGTH(content)`.
- `CHAR_LENGTH()` is preferred over `LENGTH()` because it counts characters, whereas `LENGTH()` counts bytes (important in multi-byte encodings).
### Query Execution:
- The database scans the `Tweets` table, applies the condition, and returns the `tweet_id` of invalid tweets.
- Since `tweet_id` is the **primary key**, retrieval is optimized.
---
## ⏳ Time & Space Complexity Analysis
### Time Complexity: **O(N)**
- The query applies `CHAR_LENGTH(content) > 15` as a filter.
- Since there is no index on the `content` column, the database performs a full table scan.
- For `N` rows in the `Tweets` table:
- **Best Case (O(1))**: If the table is empty.
- **Worst Case (O(N))**: When all rows must be scanned and checked.
- **Average Case (O(N))**: The query runs in **linear time**, scanning each row once.
### Space Complexity: **O(1)**
- The query does not use any extra space apart from the result storage.
- The database engine processes data **in-place** without requiring additional memory.
| 28 | 1 |
['MySQL']
| 2 |
invalid-tweets
|
Pandas vs SQL | Elegant & Short | All 30 Days of Pandas solutions ✅
|
pandas-vs-sql-elegant-short-all-30-days-n30vx
|
Complexity\n- Time complexity: O(n)\n- Space complexity: O(n)\n\n# Code\nPython []\ndef invalid_tweets(tweets: pd.DataFrame) -> pd.DataFrame:\n return tweets
|
Kyrylo-Ktl
|
NORMAL
|
2023-08-01T16:53:11.019208+00:00
|
2023-08-06T16:43:34.273770+00:00
| 3,984 | false |
# Complexity\n- Time complexity: $$O(n)$$\n- Space complexity: $$O(n)$$\n\n# Code\n```Python []\ndef invalid_tweets(tweets: pd.DataFrame) -> pd.DataFrame:\n return tweets[\n tweets[\'content\'].str.len() > 15\n ][[\'tweet_id\']]\n```\n```SQL []\nSELECT tweet_id\n FROM Tweets\n WHERE length(content) >= 16;\n```\n\n# Important!\n###### If you like the solution or find it useful, feel free to upvote for it, it will support me in creating high quality solutions)\n\n# 30 Days of Pandas solutions\n\n### Data Filtering \u2705\n- [Big Countries](https://leetcode.com/problems/big-countries/solutions/3848474/pandas-elegant-short-1-line/)\n- [Recyclable and Low Fat Products](https://leetcode.com/problems/recyclable-and-low-fat-products/solutions/3848500/pandas-elegant-short-1-line/)\n- [Customers Who Never Order](https://leetcode.com/problems/customers-who-never-order/solutions/3848527/pandas-elegant-short-1-line/)\n- [Article Views I](https://leetcode.com/problems/article-views-i/solutions/3867192/pandas-elegant-short-1-line/)\n\n\n### String Methods \u2705\n- [Invalid Tweets](https://leetcode.com/problems/invalid-tweets/solutions/3849121/pandas-elegant-short-1-line/)\n- [Calculate Special Bonus](https://leetcode.com/problems/calculate-special-bonus/solutions/3867209/pandas-elegant-short-1-line/)\n- [Fix Names in a Table](https://leetcode.com/problems/fix-names-in-a-table/solutions/3849167/pandas-elegant-short-1-line/)\n- [Find Users With Valid E-Mails](https://leetcode.com/problems/find-users-with-valid-e-mails/solutions/3849177/pandas-elegant-short-1-line/)\n- [Patients With a Condition](https://leetcode.com/problems/patients-with-a-condition/solutions/3849196/pandas-elegant-short-1-line-regex/)\n\n\n### Data Manipulation \u2705\n- [Nth Highest Salary](https://leetcode.com/problems/nth-highest-salary/solutions/3867257/pandas-elegant-short-1-line/)\n- [Second Highest Salary](https://leetcode.com/problems/second-highest-salary/solutions/3867278/pandas-elegant-short/)\n- [Department Highest Salary](https://leetcode.com/problems/department-highest-salary/solutions/3867312/pandas-elegant-short-1-line/)\n- [Rank Scores](https://leetcode.com/problems/rank-scores/solutions/3872817/pandas-elegant-short-1-line-all-30-days-of-pandas-solutions/)\n- [Delete Duplicate Emails](https://leetcode.com/problems/delete-duplicate-emails/solutions/3849211/pandas-elegant-short/)\n- [Rearrange Products Table](https://leetcode.com/problems/rearrange-products-table/solutions/3849226/pandas-elegant-short-1-line/)\n\n\n### Statistics \u2705\n- [The Number of Rich Customers](https://leetcode.com/problems/the-number-of-rich-customers/solutions/3849251/pandas-elegant-short-1-line/)\n- [Immediate Food Delivery I](https://leetcode.com/problems/immediate-food-delivery-i/solutions/3872719/pandas-elegant-short-1-line-all-30-days-of-pandas-solutions/)\n- [Count Salary Categories](https://leetcode.com/problems/count-salary-categories/solutions/3872801/pandas-elegant-short-1-line-all-30-days-of-pandas-solutions/)\n\n\n### Data Aggregation \u2705\n- [Find Total Time Spent by Each Employee](https://leetcode.com/problems/find-total-time-spent-by-each-employee/solutions/3872715/pandas-elegant-short-1-line-all-30-days-of-pandas-solutions/)\n- [Game Play Analysis I](https://leetcode.com/problems/game-play-analysis-i/solutions/3863223/pandas-elegant-short-1-line/)\n- [Number of Unique Subjects Taught by Each Teacher](https://leetcode.com/problems/number-of-unique-subjects-taught-by-each-teacher/solutions/3863239/pandas-elegant-short-1-line/)\n- [Classes More Than 5 Students](https://leetcode.com/problems/classes-more-than-5-students/solutions/3863249/pandas-elegant-short/)\n- [Customer Placing the Largest Number of Orders](https://leetcode.com/problems/customer-placing-the-largest-number-of-orders/solutions/3863257/pandas-elegant-short-1-line/)\n- [Group Sold Products By The Date](https://leetcode.com/problems/group-sold-products-by-the-date/solutions/3863267/pandas-elegant-short-1-line/)\n- [Daily Leads and Partners](https://leetcode.com/problems/daily-leads-and-partners/solutions/3863279/pandas-elegant-short-1-line/)\n\n\n### Data Aggregation \u2705\n- [Actors and Directors Who Cooperated At Least Three Times](https://leetcode.com/problems/actors-and-directors-who-cooperated-at-least-three-times/solutions/3863309/pandas-elegant-short/)\n- [Replace Employee ID With The Unique Identifier](https://leetcode.com/problems/replace-employee-id-with-the-unique-identifier/solutions/3872822/pandas-elegant-short-1-line-all-30-days-of-pandas-solutions/)\n- [Students and Examinations](https://leetcode.com/problems/students-and-examinations/solutions/3872699/pandas-elegant-short-1-line-all-30-days-of-pandas-solutions/)\n- [Managers with at Least 5 Direct Reports](https://leetcode.com/problems/managers-with-at-least-5-direct-reports/solutions/3872861/pandas-elegant-short/)\n- [Sales Person](https://leetcode.com/problems/sales-person/solutions/3872712/pandas-elegant-short-1-line-all-30-days-of-pandas-solutions/)\n
| 23 | 0 |
['Python', 'Python3', 'Pandas']
| 2 |
invalid-tweets
|
✅💯 Only 2 line | Easiest MySql Code || 🔥✔️✅💯
|
only-2-line-easiest-mysql-code-by-patilp-ovyi
|
we have to print those tweet ids ,having tweet > 15so, we just select those content having length is greater than 15**Code
|
patilprajakta
|
NORMAL
|
2025-01-13T16:35:42.872193+00:00
|
2025-01-13T16:35:42.872193+00:00
| 2,552 | false |
## we have to print those tweet ids ,having tweet > 15
so, we just select those content having length is greater than 15**
# Code
```mysql []
# Write your MySQL query statement below
select tweet_id from Tweets
where length(content)>15;
```
| 12 | 0 |
['MySQL']
| 0 |
invalid-tweets
|
✅💯🔥Simple My SQL Code || 🔥✔️Easy to understand🎯 || 🎓🧠Logical🔥|| Beginner friendly💀💯
|
simple-my-sql-code-easy-to-understand-lo-4s67
|
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time
|
atishayj4in
|
NORMAL
|
2024-07-22T19:35:41.528188+00:00
|
2024-07-22T19:35:41.528221+00:00
| 2,031 | false |
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\n# Write your MySQL query statement below\nselect tweet_id from tweets where length(content)>15;\n```\n\n
| 12 | 0 |
['MySQL']
| 1 |
invalid-tweets
|
Easy and Simple Solution
|
easy-and-simple-solution-by-mayankluthya-it79
|
Please Like ❤️IntuitionTo identify tweets with content longer than 15 characters, the Tweets table's content column must be evaluated for its length.Approach
Fi
|
mayankluthyagi
|
NORMAL
|
2025-01-28T01:24:13.002490+00:00
|
2025-01-28T01:24:13.002490+00:00
| 1,964 | false |
```mysql
SELECT tweet_id
FROM Tweets
WHERE LENGTH(content) > 15;
```
# Please Like ❤️
# Intuition
To identify tweets with content longer than 15 characters, the `Tweets` table's `content` column must be evaluated for its length.
# Approach
1. **Filter Rows by Content Length**: Use the `LENGTH` function to calculate the length of the `content` field.
2. **Apply Condition**: Use a `WHERE` clause to filter rows where `LENGTH(content) > 15`.
3. **Select Relevant Column**: Only retrieve the `tweet_id` of tweets meeting the condition.
# Complexity
- **Time complexity**:
$$O(n)$$, where $$n$$ is the number of rows in the `Tweets` table.
- The length of each tweet's `content` must be calculated.
- **Space complexity**:
$$O(k)$$, where $$k$$ is the number of tweets with `content` longer than 15 characters.
| 11 | 0 |
['MySQL']
| 0 |
invalid-tweets
|
✅5/50 | All SQL50 Explained
|
550-all-sql50-explained-by-piotr_maminsk-1gzp
|
Code\nmysql []\nSELECT tweet_id\nFROM Tweets\nWHERE CHAR_LENGTH(content) > 15\n;\n\nmssql [MS SQL Server]\nSELECT tweet_id\nFROM Tweets\nWHERE LEN(content) > 15
|
Piotr_Maminski
|
NORMAL
|
2024-10-04T15:52:47.349354+00:00
|
2024-10-24T19:45:18.122834+00:00
| 3,067 | false |
# Code\n```mysql []\nSELECT tweet_id\nFROM Tweets\nWHERE CHAR_LENGTH(content) > 15\n;\n```\n```mssql [MS SQL Server]\nSELECT tweet_id\nFROM Tweets\nWHERE LEN(content) > 15;\n;\n```\n```oraclesql [Oracle]\nSELECT tweet_id\nFROM Tweets\nWHERE LENGTH(content) > 15\n;\n```\n```postgresql [PostgreSQL]\nSELECT tweet_id\nFROM Tweets\nWHERE CHAR_LENGTH(content) > 15\n;\n```\n```pythondata [Pandas]\n# Time: O(n) , Space: O(n) \nimport pandas as pd\n\ndef invalid_tweets(tweets: pd.DataFrame) -> pd.DataFrame:\n return tweets[tweets[\'content\'].str.len()>15][[\'tweet_id\']]\n\n```\n### System differences\nMS SQL Server - **LEN()** is synonymous with CHAR_LENGTH() in MySQL.\nOracle - **LENGTH()** is synonymous with CHAR_LENGTH() in MySQL.\n## So why CHAR_LENGTH?\n\nLENGTH() counts in bytes. Some characters occupy more than one byte. CHAR_LENGTH() counts characters and is safer.\n\n\n\n\n\n# Order of Execution\n\n\n1. **FROM** Tweets\n2. **WHERE** The condition **CHAR_LENGTH(content)** > 15 is applied to filter rows\n3. **SELECT** tweet_id\n\n\n\n\n
| 11 | 0 |
['MySQL', 'Oracle', 'MS SQL Server', 'PostgreSQL', 'Pandas']
| 2 |
invalid-tweets
|
✅100% Beast Solution || Easy Sql Query
|
100-beast-solution-easy-sql-query-by-gou-79cv
|
Intuition\nThis SQL query retrieves tweet IDs from the Tweets table where the length of the content exceeds 15 characters.\n\n# Approach\nThe approach is to fil
|
gouravsharma2806
|
NORMAL
|
2024-02-10T06:46:42.108371+00:00
|
2024-02-10T06:46:42.108407+00:00
| 5,315 | false |
# Intuition\nThis SQL query retrieves tweet IDs from the Tweets table where the length of the content exceeds 15 characters.\n\n# Approach\nThe approach is to filter rows from the Tweets table where the length of the content, measured in characters, is greater than 15. This is achieved using the char_length() function in SQL, which returns the number of characters in a string.\n\n# Complexity\n- Time Complexity:\nThe time complexity largely depends on the indexing of the content column in the Tweets table. If properly indexed, the time complexity for filtering rows based on character length should be close to O(log n), where n is the number of rows in the Tweets table.\nThe time complexity of the char_length() function itself can be considered O(1) as it simply calculates the length of a string.\n- Space Complexity:\nThe space complexity is O(1) since the query only retrieves tweet IDs and doesn\'t depend on the size of the data being queried.\n\n# Code\n```\nselect tweet_id from Tweets where char_length(content)>15;\n\n```
| 9 | 0 |
['Database', 'MySQL']
| 1 |
invalid-tweets
|
Easy Code || MySQL
|
easy-code-mysql-by-me_avi-acwz
|
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time
|
me_avi
|
NORMAL
|
2023-10-05T18:05:48.639044+00:00
|
2023-10-05T18:05:48.639075+00:00
| 2,355 | false |
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\n# Write your MySQL query statement below\nselect tweet_id \nfrom Tweets\nwhere CHAR_LENGTH(content) > 15\n```
| 8 | 0 |
['MySQL']
| 1 |
invalid-tweets
|
Two Approches Mysql
|
two-approches-mysql-by-ganjinaveen-fs9t
|
\nselect tweet_id from Tweets where length(content)>15 \n\n\nselect tweet_id from Tweets where char_length(content)>15 \n\n# please upvote me it would encourage
|
GANJINAVEEN
|
NORMAL
|
2023-06-22T18:32:49.995637+00:00
|
2023-06-22T18:32:49.995665+00:00
| 1,731 | false |
```\nselect tweet_id from Tweets where length(content)>15 \n```\n```\nselect tweet_id from Tweets where char_length(content)>15 \n```\n# please upvote me it would encourage me alot\n
| 8 | 0 |
['MySQL']
| 1 |
invalid-tweets
|
Filter Tweets with MySQL LENGTH() Function
|
filter-tweets-with-mysql-length-function-i0vg
|
Intuition\n Describe your first thoughts on how to solve this problem. \nWe need to find the IDs of tweets with content exceeding 15 characters.\n\n# Approach\n
|
samabdullaev
|
NORMAL
|
2023-10-12T11:49:33.542577+00:00
|
2023-10-28T12:00:37.312890+00:00
| 1,288 | false |
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWe need to find the IDs of tweets with content exceeding 15 characters.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n1. Select specific columns from the table.\n\n > `SELECT` \u2192 This command retrieves data from a database\n\n2. Filter for tweets with content that is longer than 15 characters.\n\n > `WHERE` \u2192 This command filters a result set to include only records that fulfill a specified condition\n\n > `LENGTH()` \u2192 This function returns the length of a string (in bytes)\n\n > `>` \u2192 This operator compares if one value is greater than another value\n\n# Complexity\n- Time complexity: $O(n)$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n`n` is the number of rows in the table. In the worst case, it would need to examine each row to check the content length.\n\n# Code\n```\nSELECT tweet_id FROM Tweets WHERE LENGTH(content)>15\n```
| 7 | 0 |
['MySQL']
| 0 |
invalid-tweets
|
Beats 100.00%of users with Pandas
|
beats-10000of-users-with-pandas-by-suvha-u0lo
|
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time
|
suvha123
|
NORMAL
|
2023-07-31T22:57:11.635084+00:00
|
2023-07-31T22:57:11.635106+00:00
| 1,598 | false |
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nimport pandas as pd\n\ndef invalid_tweets(tweets: pd.DataFrame) -> pd.DataFrame:\n tweets = tweets[tweets[\'content\'].apply(lambda x: True if len(x)> 15 else False)]\n return tweets[[\'tweet_id\']]\n```
| 7 | 0 |
['Pandas']
| 2 |
invalid-tweets
|
📌 One line sol
|
one-line-sol-by-mayankgurjar1570-dg0y
|
\n\n# Code\n\n\nselect tweet_id from Tweets where CHAR_LENGTH(content) > 15;\n
|
mayankgurjar1570
|
NORMAL
|
2024-06-06T14:37:20.138895+00:00
|
2024-06-06T14:37:20.138930+00:00
| 2,800 | false |
\n\n# Code\n```\n\nselect tweet_id from Tweets where CHAR_LENGTH(content) > 15;\n```
| 6 | 0 |
['MySQL']
| 0 |
invalid-tweets
|
Easy solution with explanation.
|
easy-solution-with-explanation-by-sajjan-wzal
|
\n# Approach\n Describe your approach to solving the problem. \n1. SELECT tweet_id:\n\n- This part of the query selects the tweet_id column from the Tweets tabl
|
sajjan_shivani
|
NORMAL
|
2024-09-16T14:07:31.312406+00:00
|
2024-09-16T14:07:31.312437+00:00
| 3,222 | false |
\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. SELECT tweet_id:\n\n- This part of the query selects the tweet_id column from the Tweets table.\n- The tweet_id identifies each tweet in the table.\n2. FROM Tweets:\n\n- Specifies the table Tweets from which the data is being retrieved.\nWHERE LENGTH(content) > 15:\n\n- The WHERE clause filters the records, only including tweets where the length of the text in the content column is greater than 15 characters.\n- The LENGTH() function is used to calculate the number of characters in the content field.\n\n\n\n# Code\n```mysql []\n# Write your MySQL query statement below\nselect tweet_id from Tweets where LENGTH(content) > 15;\n```
| 5 | 0 |
['MySQL']
| 1 |
invalid-tweets
|
✅EASY MY SQL SOLUTION
|
easy-my-sql-solution-by-swayam28-wqtx
|
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time
|
swayam28
|
NORMAL
|
2024-08-07T11:45:48.287831+00:00
|
2024-08-07T11:45:48.287865+00:00
| 2,336 | false |
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\n# Write your MySQL query statement below\nselect tweet_id from tweets where length(content)>15\n```
| 5 | 0 |
['MySQL']
| 0 |
invalid-tweets
|
Easy SQL Query
|
easy-sql-query-by-deepakumar-developer-juhn
|
\n\n# Query\n\n# Write your MySQL query statement below\nselect tweet_id\nfrom Tweets\nwhere length(content) > 15\n
|
deepakumar-developer
|
NORMAL
|
2023-12-27T15:26:05.599235+00:00
|
2023-12-27T15:26:05.599259+00:00
| 1,445 | false |
\n\n# Query\n```\n# Write your MySQL query statement below\nselect tweet_id\nfrom Tweets\nwhere length(content) > 15\n```
| 5 | 0 |
['MySQL']
| 2 |
invalid-tweets
|
🟡Invalid Tweets | 🤯✅Beats 91.35% of users with MySQL 🤫🔥|
|
invalid-tweets-beats-9135-of-users-with-igvlw
|
Intuition\nWe need to find the IDs of tweets with content exceeding 15 characters.\n\nApproach\nSelect specific columns from the table.\n\nSELECT \u2192 This co
|
komronabdulloev
|
NORMAL
|
2023-10-12T15:32:01.470146+00:00
|
2023-10-12T15:32:01.470170+00:00
| 669 | false |
Intuition\nWe need to find the IDs of tweets with content exceeding 15 characters.\n\nApproach\nSelect specific columns from the table.\n\nSELECT \u2192 This command retrieves data from a database\n\nFilter for tweets with content that is longer than 15 characters.\n\nLENGTH() \u2192 This function returns the length of a string (in bytes)\n\nComplexity\nTime complexity: O(n)O(n)O(n)\nn is the number of rows in the table. In the worst case, it would need to examine each row to check the content length.\n\n\n\n```\nSELECT tweet_id\n FROM Tweets\n WHERE LENGTH(content) > 15;\n```
| 5 | 0 |
['MySQL']
| 2 |
invalid-tweets
|
✅ 100% EASY || FAST 🔥|| CLEAN ONE LINE SOLUTION 🌟
|
100-easy-fast-clean-one-line-solution-by-kvqj
|
IF THIS WILL BE HELPFUL TO YOU, PLEASE UPVOTE !\n\n# Code\n\n/* Write your PL/SQL query statement below */\n\nSELECT tweet_id FROM Tweets WHERE LENGTH(content)
|
kartik_ksk7
|
NORMAL
|
2023-08-03T08:57:01.990531+00:00
|
2023-08-05T05:52:17.663728+00:00
| 804 | false |
IF THIS WILL BE HELPFUL TO YOU, PLEASE UPVOTE !\n\n# Code\n```\n/* Write your PL/SQL query statement below */\n\nSELECT tweet_id FROM Tweets WHERE LENGTH(content) > 15\n```\n
| 5 | 0 |
['Database', 'Oracle']
| 1 |
invalid-tweets
|
SUPER EASY SOLUTION
|
super-easy-solution-by-himanshu__mehra-nhkk
|
Code\n\nSelect t.tweet_id\nfrom Tweets t\nwhere length(t.content) > 15; \n\nPLEASE UPVOTE TO MOTIVATE ME WRITE MORE SOLUTION\n
|
himanshu__mehra__
|
NORMAL
|
2023-06-29T19:20:26.399978+00:00
|
2023-06-29T19:20:26.400012+00:00
| 1,937 | false |
# Code\n```\nSelect t.tweet_id\nfrom Tweets t\nwhere length(t.content) > 15; \n\n```PLEASE UPVOTE TO MOTIVATE ME WRITE MORE SOLUTION\n
| 5 | 0 |
['MySQL']
| 2 |
invalid-tweets
|
MYSQL Solution ||LENGTH()
|
mysql-solution-length-by-sidsai-7lgt
|
\n# Code\n\n# Write your MySQL query statement below\nSELECT tweet_id FROM\nTweets where LENGTH(content)>15;\n
|
SidSai
|
NORMAL
|
2023-06-10T08:59:19.039893+00:00
|
2023-06-10T08:59:19.039932+00:00
| 3,395 | false |
\n# Code\n```\n# Write your MySQL query statement below\nSELECT tweet_id FROM\nTweets where LENGTH(content)>15;\n```
| 5 | 0 |
['MySQL']
| 1 |
invalid-tweets
|
✅[MySQL] Simple and Clean, beats 88%✅
|
mysql-simple-and-clean-beats-88-by-_tanm-88xh
|
Please upvote if you find this helpful. \u270C\n\n\n# Intuition\nThe problem asks us to find the IDs of invalid tweets. A tweet is considered invalid if its con
|
_Tanmay
|
NORMAL
|
2023-05-30T10:17:20.771754+00:00
|
2023-06-01T16:46:09.569589+00:00
| 678 | false |
### Please upvote if you find this helpful. \u270C\n<img src="https://assets.leetcode.com/users/images/b8e25620-d320-420a-ae09-94c7453bd033_1678818986.7001078.jpeg" alt="Cute Robot - Stable diffusion" width="200"/>\n\n# Intuition\nThe problem asks us to find the IDs of invalid tweets. A tweet is considered invalid if its content has more than 15 characters. Our first thought is to use an SQL query to filter out the tweets that have more than 15 characters in their content.\n\n# Approach\n1. Select the `tweet_id` column from the `Tweets` table.\n2. Use the `WHERE` clause to filter out rows where the length of the `content` column is strictly greater than 15.\n3. The final query should look like this:\n```SQL\nSELECT tweet_id\nFROM Tweets\nWHERE LENGTH(content) > 15;\n```\n\n# Complexity\n- Time complexity: The time complexity of this query depends on the size of the `Tweets` table and the efficiency of the database management system.\n- Space complexity: The space complexity of this query is constant as it only returns a single column with the IDs of invalid tweets.\n
| 5 | 0 |
['Database', 'MySQL']
| 0 |
invalid-tweets
|
given SQL query is to retrieve the tweet_id from the Tweets table
|
given-sql-query-is-to-retrieve-the-tweet-xadq
|
Intuition\n Describe your first thoughts on how to solve this problem. \nThe intuition behind the given SQL query is to retrieve the tweet_id from the Tweets ta
|
RinatMambetov
|
NORMAL
|
2023-04-01T16:20:35.317228+00:00
|
2023-04-01T16:20:35.317261+00:00
| 2,757 | false |
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe intuition behind the given SQL query is to retrieve the tweet_id from the Tweets table where the length of the content is greater than 15 characters. This query will help to filter out tweets with shorter content and retrieve only those tweets with longer content.\n# Approach\n<!-- Describe your approach to solving the problem. -->\nTo retrieve the tweet_id from the Tweets table where the length of the content is greater than 15 characters, we can use the SQL SELECT statement. The SELECT statement is used to retrieve data from a table or view. In this case, we need to retrieve the tweet_id column from the Tweets table. We also need to specify a condition using the WHERE clause to filter out tweets with shorter content. The length function is used to calculate the length of the content in characters.\n# Complexity\n<!-- - Time complexity: -->\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n<!-- - Space complexity: -->\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\nThe complexity of the given SQL query is relatively low as it involves only a single table and a simple WHERE condition. The complexity of the query can depend on the size of the Tweets table and the number of rows that satisfy the given condition. If the Tweets table is very large, the query might take longer to execute. However, since we are only retrieving a single column, the amount of data returned by the query would be relatively small. Therefore, the overall complexity of this query is moderate.\n# Code\n```\n# Write your MySQL query statement below\nselect tweet_id from Tweets where length(content) > 15\n```
| 5 | 0 |
['MySQL']
| 1 |
invalid-tweets
|
MYSQL : WHERE and CHAR_LENGTH() with explanation. Good Luck 👍✨
|
mysql-where-and-char_length-with-explana-y45c
|
\n# Write your MySQL query statement below\n/**\nONE TABLE : Tweets\ntweet_id is the primary key.\nTweets table contains all the tweets in a social media app.\n
|
ashlyjoseph
|
NORMAL
|
2022-07-27T00:02:30.168301+00:00
|
2022-07-27T00:02:30.168332+00:00
| 1,340 | false |
```\n# Write your MySQL query statement below\n/**\nONE TABLE : Tweets\ntweet_id is the primary key.\nTweets table contains all the tweets in a social media app.\n\nPROBLEM: find the IDs of the invalid tweets.tweet is invalid if the number of characters used in the content of the tweet is strictly greater than 15.\n\nSTEPS:\n 1. WHERE CHAR_LENGTH(content) >15\n\n\nNote: Difference blw CHAR_LENGTH() and LENGH()\n\nCHAR_LENGTH() - measured in characters of a string.\nLENGH() - measured in bytes of a string.\n*/\n\nSELECT \n tweet_id\nFROM Tweets\nWHERE CHAR_LENGTH(content) > 15 \n```\n
| 5 | 0 |
['MySQL']
| 1 |
invalid-tweets
|
WrItE tHe QUeRy On Ur OwN bY rEaDiNg ThE eXpLaNaTiOn🔥. Don't cOpY... jUsT rEaD, tHiNk & TyPe 💻😌
|
write-the-query-on-ur-own-by-reading-the-q54z
|
You just need to check the length of the tweet content.
If it’s strictly greater than 15, it’s invalid. That’s it.
Use LENGTH() or CHAR_LENGTH() — and you’re do
|
Vaishu021103
|
NORMAL
|
2025-04-05T12:42:26.003353+00:00
|
2025-04-05T12:42:26.003353+00:00
| 243 | false |
You just need to check the length of the tweet content.
If it’s strictly greater than 15, it’s invalid. That’s it.
Use LENGTH() or CHAR_LENGTH() — and you’re done ✅
No joins, no drama — just one simple check 😎.
Try it before peeking 👀👇
Adding some lines here because I know you're gonna scroll straight to the code 😜
.
.
.
.
.
.
.
.
.
# Code
```mysql []
# Write your MySQL query statement below
select tweet_id from Tweets where length(content) > 15
```
| 4 | 0 |
['MySQL']
| 0 |
invalid-tweets
|
1683. Invalid Tweets
|
1683-invalid-tweets-by-layan_am-vsjz
|
IntuitionApproachComplexity
Time complexity:
Space complexity:
Code
|
layan_am
|
NORMAL
|
2025-01-04T21:23:11.859075+00:00
|
2025-01-04T21:23:11.859075+00:00
| 944 | false |
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```oraclesql []
/* Write your PL/SQL query statement below */
SELECT tweet_id
FROM Tweets
WHERE LENGTH(TRIM(content)) > 15;
```
| 4 | 0 |
['Database', 'Oracle']
| 1 |
invalid-tweets
|
✔✔✔easy 1 liner code - beats 95%⚡⚡⚡MS SQL Server || MySQL || Post⚡⚡⚡greSQL || Oracle
|
easy-1-liner-code-beats-95ms-sql-server-zq567
|
Code - MySQL || PostgreSQL || Oracle\n\nSELECT tweet_id FROM Tweets WHERE LENGTH(content) > 15\n\n# Code - MS SQL Server\n\nSELECT tweet_id FROM Tweets WHERE LE
|
anish_sule
|
NORMAL
|
2024-02-27T05:45:55.224397+00:00
|
2024-06-12T12:22:09.758634+00:00
| 1,517 | false |
# Code - MySQL || PostgreSQL || Oracle\n```\nSELECT tweet_id FROM Tweets WHERE LENGTH(content) > 15\n```\n# Code - MS SQL Server\n```\nSELECT tweet_id FROM Tweets WHERE LEN(content) > 15\n```\n# [for all LeetCode SQL50 solutions...](https://github.com/anish-sule/LeetCode-SQL-50)
| 4 | 0 |
['MySQL', 'Oracle', 'MS SQL Server', 'PostgreSQL']
| 0 |
invalid-tweets
|
EASY💯 QUERY 🔥|| MYSQL || LENGTH()/CHAR_LENGTH()
|
easy-query-mysql-lengthchar_length-by-ta-q32a
|
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time
|
tanish-hrk
|
NORMAL
|
2024-01-18T17:08:26.170834+00:00
|
2024-01-18T17:08:26.170866+00:00
| 6,718 | false |
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code 1\n```\n# Write your MySQL query statement below\nselect tweet_id from Tweets where length(content) > 15;\n```\n# Code 2\n```\n# Write your MySQL query statement below\nselect tweet_id from Tweets where char_length(content) > 15;\n```\n\n
| 4 | 0 |
['MySQL']
| 0 |
invalid-tweets
|
Easy SQL Query
|
easy-sql-query-by-lakshmiyadavalli-kiiq
|
\nSELECT tweet_id\nFROM Tweets\nWHERE CHAR_LENGTH(content) > 15;\n
|
lakshmiyadavalli
|
NORMAL
|
2024-01-05T05:38:02.439468+00:00
|
2024-01-05T05:38:02.439500+00:00
| 2,656 | false |
```\nSELECT tweet_id\nFROM Tweets\nWHERE CHAR_LENGTH(content) > 15;\n```
| 4 | 0 |
['MySQL']
| 0 |
invalid-tweets
|
Solution: LEN built_in function in SQL Server
|
solution-len-built_in-function-in-sql-se-ulqq
|
\n# Code\n\n/* Write your T-SQL query statement below */\n\n\nSELECT DISTINCT\n tweet_id\nFROM Tweets\nWHERE LEN(content) > 15\n
|
k_a_m_o_l
|
NORMAL
|
2023-11-14T18:16:12.891954+00:00
|
2023-11-14T18:16:12.891972+00:00
| 817 | false |
\n# Code\n```\n/* Write your T-SQL query statement below */\n\n\nSELECT DISTINCT\n tweet_id\nFROM Tweets\nWHERE LEN(content) > 15\n```
| 4 | 0 |
['MS SQL Server']
| 1 |
invalid-tweets
|
Easy SQL solution
|
easy-sql-solution-by-ankushmallick7976-p1ls
|
Intuition\n Describe your first thoughts on how to solve this problem. \nAt first convert the content from varchar to char using cast function. Inside the funct
|
ankushmallick7976
|
NORMAL
|
2023-09-05T19:39:43.101301+00:00
|
2023-09-05T19:40:49.840026+00:00
| 2,533 | false |
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nAt first convert the content from varchar to char using cast function. Inside the function "content as char" should be mentioned. After that length function is used all these things should be kept inside it.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\n# Write your MySQL query statement below\nselect tweet_id from Tweets where length(cast(content as char)) > 15;\n```
| 4 | 0 |
['Database', 'MySQL', 'Oracle', 'MS SQL Server']
| 5 |
invalid-tweets
|
1 line
|
1-line-by-kushpatel49-97fx
|
\n# Code\n\ndef invalid_tweets(tweets: pd.DataFrame) -> pd.DataFrame:\n return tweets[tweets[\'content\'].str.len() > 15][[\'tweet_id\']]\n
|
Kushpatel49
|
NORMAL
|
2023-08-02T06:35:59.466778+00:00
|
2023-08-02T06:35:59.466811+00:00
| 821 | false |
\n# Code\n```\ndef invalid_tweets(tweets: pd.DataFrame) -> pd.DataFrame:\n return tweets[tweets[\'content\'].str.len() > 15][[\'tweet_id\']]\n```
| 4 | 0 |
['Pandas']
| 0 |
invalid-tweets
|
len() Function Use in Database ... Absolute Beginner
|
len-function-use-in-database-absolute-be-nsd7
|
Code\n\nSELECT tweet_id \nFROM Tweets \nwhere len(content) > 15;\n
|
Taimoor_hussain47
|
NORMAL
|
2023-05-21T12:46:20.080019+00:00
|
2023-05-21T12:46:20.080054+00:00
| 1,605 | false |
# Code\n```\nSELECT tweet_id \nFROM Tweets \nwhere len(content) > 15;\n```
| 4 | 0 |
['Database', 'MS SQL Server']
| 1 |
invalid-tweets
|
Approach in three SQL variants | Oracle | MS SQL | MySQL
|
approach-in-three-sql-variants-oracle-ms-1kp7
|
We use the basic commands to get the required information from the table.\nThe main command in this problem is the usage of LENGTH/LEN command. Look at the comm
|
younus_baig
|
NORMAL
|
2023-05-19T02:40:25.415811+00:00
|
2023-05-19T02:40:25.415842+00:00
| 2,249 | false |
We use the basic commands to get the required information from the table.\nThe main command in this problem is the usage of LENGTH/LEN command. Look at the comments in the code to find out more.\n\nYou can read about the commands at the end for learning.\n\n# Code\n``` Oracle []\n-- Uses LENGTH()\nSELECT tweet_id FROM tweets WHERE LENGTH(content) > 15;\n```\n``` MySQL []\n-- Uses LENGTH()\nSELECT tweet_id FROM tweets WHERE LENGTH(content) > 15;\n```\n``` MS_SQL []\n-- Uses LEN()\nSELECT tweet_id FROM tweets WHERE LEN(content) > 15;\n```\n\n# Commands\n\n## SELECT\n This command is used to select data from a database. It specifies the columns that you want to retrieve from one or more tables.\n## FROM\n This command is used to specify the table or tables from which you want to retrieve data.\n## WHERE\nThis command is used to filter the data that you want to retrieve based on one or more conditions.\n## LENGTH / LEN\n This function is used to get the length of a string in Oracle and MySQL. In SQL, we use LEN()
| 4 | 0 |
['MySQL', 'Oracle', 'MS SQL Server']
| 0 |
invalid-tweets
|
Beginner friendly single line code-mysql
|
beginner-friendly-single-line-code-mysql-qab7
|
\n\n# Code\nmysql []\n# Write your MySQL query statement below\nselect tweet_id from Tweets where LENGth(content)>15\n
|
THANMAYI01
|
NORMAL
|
2024-10-06T14:55:55.254138+00:00
|
2024-10-06T14:55:55.254170+00:00
| 1,540 | false |
\n\n# Code\n```mysql []\n# Write your MySQL query statement below\nselect tweet_id from Tweets where LENGth(content)>15\n```
| 3 | 0 |
['Database', 'MySQL']
| 0 |
invalid-tweets
|
✅✅100% Beats👇 || Fully Explained ✌️||SQL
|
100-beats-fully-explained-sql-by-surbhi_-uwvd
|
Intuition\nWe want to find tweets where the length of the content is greater than 15 characters.\n\n# Approach\nUse a SELECT statement with a WHERE clause to fi
|
Surbhi_g
|
NORMAL
|
2024-03-10T05:59:29.078225+00:00
|
2024-03-10T05:59:29.078253+00:00
| 1,087 | false |
# Intuition\nWe want to find tweets where the length of the content is greater than 15 characters.\n\n# Approach\nUse a SELECT statement with a WHERE clause to filter rows where the char_length of the content column is greater than 15.\n\n# Complexity\n- Time complexity: O(n), where n is the number of rows in the Tweets table.\n\n- Space complexity: O(1), as we are only returning the tweet_id values and not storing any additional data.\n\n# Code\n```\n# Write your MySQL query statement below\nselect tweet_id from Tweets where char_length(content)>15\n```
| 3 | 0 |
['Database', 'MySQL']
| 0 |
invalid-tweets
|
Beginner-friendly solution using MySQL || 1 line approach
|
beginner-friendly-solution-using-mysql-1-b2e6
|
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time
|
truongtamthanh2004
|
NORMAL
|
2024-03-07T16:50:22.326363+00:00
|
2024-03-07T16:50:22.326384+00:00
| 381 | false |
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\n# Write your MySQL query statement below\nselect tweet_id from Tweets where (LENGTH(content) > 15)\n```
| 3 | 0 |
['MySQL']
| 0 |
invalid-tweets
|
Very simple MySQL solution (Beats 85.01% of users)
|
very-simple-mysql-solution-beats-8501-of-icx5
|
Intuition\n Describe your first thoughts on how to solve this problem. \nThe goal is to identify tweets with content lengths exceeding 15 characters, marking th
|
mastan_sayyad
|
NORMAL
|
2024-03-06T16:27:07.393226+00:00
|
2024-03-06T16:34:49.344053+00:00
| 1,539 | false |
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe goal is to identify tweets with content lengths exceeding 15 characters, marking them as invalid. \n# Approach\n<!-- Describe your approach to solving the problem. -->\nIt **selects** the **tweet_id** from the **"Tweets"** table where the length of the content is greater than 15.\n\n- **LENGTH()** or **length()** is a character function used to find the length of the data.\n- We can also use **char_length()**\n\n# Code\n```\n# Write your MySQL query statement below\nselect tweet_id from Tweets where length(content)>15;\n```
| 3 | 0 |
['MySQL']
| 0 |
invalid-tweets
|
100 % FAST && EASY SQL Query || Well - Explained || Beats 99.9 %
|
100-fast-easy-sql-query-well-explained-b-2m1v
|
Intuition\n Describe your first thoughts on how to solve this problem. \nSELECT --> for show the table...\nDISTINCT tweet_id --> consider tweet ids...\nFROM -->
|
ganpatinath07
|
NORMAL
|
2024-01-27T15:23:43.070233+00:00
|
2024-01-27T15:23:43.070255+00:00
| 1,465 | false |
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nSELECT --> for show the table...\nDISTINCT tweet_id --> consider tweet ids...\nFROM --> for association...\nTweets --> Table name...\nWHERE --> Clause...\nLENGTH(content) > 15 --> Condition given in problem...\nLENGTH is a function to find length...\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nUsing Fundamentals of SQL...\n\n# Complexity\n- Time complexity: O(n)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(1)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nSELECT DISTINCT tweet_id FROM Tweets\nWHERE LENGTH(content) > 15;\n```
| 3 | 0 |
['Database', 'MySQL']
| 2 |
invalid-tweets
|
Pandas | SQL | Easy | Invalid Tweets
|
pandas-sql-easy-invalid-tweets-by-khosiy-ii3t
|
see the Successfully Accepted Submission\n\nimport pandas as pd\n\ndef invalid_tweets(tweets: pd.DataFrame) -> pd.DataFrame:\n \n # First, we retrieve the
|
Khosiyat
|
NORMAL
|
2023-09-21T13:28:16.997265+00:00
|
2023-10-01T17:42:28.718703+00:00
| 368 | false |
[see the Successfully Accepted Submission](https://leetcode.com/submissions/detail/988713025/)\n```\nimport pandas as pd\n\ndef invalid_tweets(tweets: pd.DataFrame) -> pd.DataFrame:\n \n # First, we retrieve the column `content`.\n content=tweets[\'content\']\n # In the next step, we pply string methods to the elements (strings) in the `content` Series.\n tweet_string=content.str\n # Then, we calculate the length (number of characters) of each element (string) in the `content` Series.\n tweet_length=tweet_string.len()\n # After that, we filter the tweet which is greater than 15 characters\n longer_tweet=tweet_length > 15\n # Then, we access the "tweet_id" column of the tweets DataFrame \n invalid_tweet_id = tweets[longer_tweet][\'tweet_id\']\n # Finally, we create pandas DataFrame\n invalid_tweet=pd.DataFrame(invalid_tweet_id)\n \n return invalid_tweet\n```\n\n**SQL**\n[see the Successfully Accepted Submission](https://leetcode.com/submissions/detail/1061726158/)\n\n```\nSELECT tweet_id \nFROM Tweets \nWHERE char_length(content)>15;\n```\n\n\n```\n-- Select the \'tweet_id\' column\nSELECT tweet_id\n\n-- Retrieve data from the \'Tweets\' table\nFROM Tweets\n\n-- Apply a filter to select rows where the length of the \'content\' column is greater than 15 characters\n-- This filter selects tweets with content longer than 15 characters\nWHERE char_length(content) > 15;\n```\n\n\n\n\n\n# Pandas & SQL | SOLVED & EXPLAINED LIST\n\n**EASY**\n\n- [Combine Two Tables](https://leetcode.com/problems/combine-two-tables/solutions/4051076/pandas-sql-easy-combine-two-tables/)\n\n- [Employees Earning More Than Their Managers](https://leetcode.com/problems/employees-earning-more-than-their-managers/solutions/4051991/pandas-sql-easy-employees-earning-more-than-their-managers/)\n\n- [Duplicate Emails](https://leetcode.com/problems/duplicate-emails/solutions/4055225/pandas-easy-duplicate-emails/)\n\n- [Customers Who Never Order](https://leetcode.com/problems/customers-who-never-order/solutions/4055429/pandas-sql-easy-customers-who-never-order-easy-explained/)\n\n- [Delete Duplicate Emails](https://leetcode.com/problems/delete-duplicate-emails/solutions/4055572/pandas-sql-easy-delete-duplicate-emails-easy/)\n- [Rising Temperature](https://leetcode.com/problems/rising-temperature/solutions/4056328/pandas-sql-easy-rising-temperature-easy/)\n\n- [ Game Play Analysis I](https://leetcode.com/problems/game-play-analysis-i/solutions/4056422/pandas-sql-easy-game-play-analysis-i-easy/)\n\n- [Find Customer Referee](https://leetcode.com/problems/find-customer-referee/solutions/4056516/pandas-sql-easy-find-customer-referee-easy/)\n\n- [Classes More Than 5 Students](https://leetcode.com/problems/classes-more-than-5-students/solutions/4058381/pandas-sql-easy-classes-more-than-5-students-easy/)\n- [Employee Bonus](https://leetcode.com/problems/employee-bonus/solutions/4058430/pandas-sql-easy-employee-bonus/)\n\n- [Sales Person](https://leetcode.com/problems/sales-person/solutions/4058824/pandas-sql-easy-sales-person/)\n\n- [Biggest Single Number](https://leetcode.com/problems/biggest-single-number/solutions/4063950/pandas-sql-easy-biggest-single-number/)\n\n- [Not Boring Movies](https://leetcode.com/problems/not-boring-movies/solutions/4065350/pandas-sql-easy-not-boring-movies/)\n- [Swap Salary](https://leetcode.com/problems/swap-salary/solutions/4065423/pandas-sql-easy-swap-salary/)\n\n- [Actors & Directors Cooperated min Three Times](https://leetcode.com/problems/actors-and-directors-who-cooperated-at-least-three-times/solutions/4065511/pandas-sql-easy-explained-step-by-step-actors-directors-cooperated-min-three-times/)\n\n- [Product Sales Analysis I](https://leetcode.com/problems/product-sales-analysis-i/solutions/4065545/pandas-sql-easy-product-sales-analysis-i/)\n\n- [Project Employees I](https://leetcode.com/problems/project-employees-i/solutions/4065635/pandas-sql-easy-project-employees-i/)\n- [Sales Analysis III](https://leetcode.com/problems/sales-analysis-iii/solutions/4065755/sales-analysis-iii-pandas-easy/)\n\n- [Reformat Department Table](https://leetcode.com/problems/reformat-department-table/solutions/4066153/pandas-sql-easy-explained-step-by-step-reformat-department-table/)\n\n- [Top Travellers](https://leetcode.com/problems/top-travellers/solutions/4066252/top-travellers-pandas-easy-eaxplained-step-by-step/)\n\n- [Replace Employee ID With The Unique Identifier](https://leetcode.com/problems/replace-employee-id-with-the-unique-identifier/solutions/4066321/pandas-sql-easy-replace-employee-id-with-the-unique-identifier/)\n- [Group Sold Products By The Date](https://leetcode.com/problems/group-sold-products-by-the-date/solutions/4066344/pandas-sql-easy-explained-step-by-step-group-sold-products-by-the-date/)\n\n- [Customer Placing the Largest Number of Orders](https://leetcode.com/problems/customer-placing-the-largest-number-of-orders/solutions/4068822/pandas-sql-easy-explained-step-by-step-customer-placing-the-largest-number-of-orders/)\n\n- [Article Views I](https://leetcode.com/problems/article-views-i/solutions/4069777/pandas-sql-easy-article-views-i/)\n\n- [User Activity for the Past 30 Days I](https://leetcode.com/problems/user-activity-for-the-past-30-days-i/solutions/4069797/pandas-sql-easy-user-activity-for-the-past-30-days-i/)\n\n- [Find Users With Valid E-Mails](https://leetcode.com/problems/find-users-with-valid-e-mails/solutions/4069810/pandas-sql-easy-find-users-with-valid-e-mails/)\n\n- [Patients With a Condition](https://leetcode.com/problems/patients-with-a-condition/solutions/4069817/pandas-sql-easy-patients-with-a-condition/)\n\n- [Customer Who Visited but Did Not Make Any Transactions](https://leetcode.com/problems/customer-who-visited-but-did-not-make-any-transactions/solutions/4072542/pandas-sql-easy-customer-who-visited-but-did-not-make-any-transactions/)\n\n\n- [Bank Account Summary II](https://leetcode.com/problems/bank-account-summary-ii/solutions/4072569/pandas-sql-easy-bank-account-summary-ii/)\n\n- [Invalid Tweets](https://leetcode.com/problems/invalid-tweets/solutions/4072599/pandas-sql-easy-invalid-tweets/)\n\n- [Daily Leads and Partners](https://leetcode.com/problems/daily-leads-and-partners/solutions/4072981/pandas-sql-easy-daily-leads-and-partners/)\n\n- [Recyclable and Low Fat Products](https://leetcode.com/problems/recyclable-and-low-fat-products/solutions/4073003/pandas-sql-easy-recyclable-and-low-fat-products/)\n\n- [Rearrange Products Table](https://leetcode.com/problems/rearrange-products-table/solutions/4073028/pandas-sql-easy-rearrange-products-table/)\n- [Calculate Special Bonus](https://leetcode.com/problems/calculate-special-bonus/solutions/4073595/pandas-sql-easy-calculate-special-bonus/)\n\n- [Count Unique Subjects](https://leetcode.com/problems/number-of-unique-subjects-taught-by-each-teacher/solutions/4073666/pandas-sql-easy-count-unique-subjects/)\n\n- [Count Unique Subjects](https://leetcode.com/problems/number-of-unique-subjects-taught-by-each-teacher/solutions/4073666/pandas-sql-easy-count-unique-subjects/)\n\n- [Fix Names in a Table](https://leetcode.com/problems/fix-names-in-a-table/solutions/4073695/pandas-sql-easy-fix-names-in-a-table/)\n- [Primary Department for Each Employee](https://leetcode.com/problems/primary-department-for-each-employee/solutions/4076183/pandas-sql-easy-primary-department-for-each-employee/)\n\n- [The Latest Login in 2020](https://leetcode.com/problems/the-latest-login-in-2020/solutions/4076240/pandas-sql-easy-the-latest-login-in-2020/)\n\n- [Find Total Time Spent by Each Employee](https://leetcode.com/problems/find-total-time-spent-by-each-employee/solutions/4076313/pandas-sql-easy-find-total-time-spent-by-each-employee/)\n\n- [Find Followers Count](https://leetcode.com/problems/find-followers-count/solutions/4076342/pandas-sql-easy-find-followers-count/)\n- [Percentage of Users Attended a Contest](https://leetcode.com/problems/percentage-of-users-attended-a-contest/solutions/4077301/pandas-sql-easy-percentage-of-users-attended-a-contest/)\n\n- [Employees With Missing Information](https://leetcode.com/problems/employees-with-missing-information/solutions/4077308/pandas-sql-easy-employees-with-missing-information/)\n\n- [Average Time of Process per Machine](https://leetcode.com/problems/average-time-of-process-per-machine/solutions/4077402/pandas-sql-easy-average-time-of-process-per-machine/)\n
| 3 | 0 |
['MySQL']
| 0 |
invalid-tweets
|
Solution using LENGTH
|
solution-using-length-by-exhe-kjv4
|
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time
|
exhe
|
NORMAL
|
2023-09-07T12:48:24.442738+00:00
|
2023-09-07T12:48:24.442761+00:00
| 866 | false |
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\n# Write your MySQL query statement below\nSELECT tweet_id FROM Tweets WHERE LENGTH(content) > 15;\n```
| 3 | 0 |
['MySQL']
| 0 |
invalid-tweets
|
Very easy approach, single line statement with easy approach
|
very-easy-approach-single-line-statement-9by6
|
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \nStep 1 : Select \'conte
|
inform2verma
|
NORMAL
|
2023-08-13T16:13:57.132940+00:00
|
2023-08-13T16:13:57.132967+00:00
| 1,463 | false |
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nStep 1 : Select \'content\' column from tweets dataframe, for this use:\ntweets[\'content\']\n\nStep 2: Find length of content which is greater than 15,for this use: tweets[\'content\'].apply(len)>15\nThe apply() function is particularly useful for performing element-wise operations on Series or DataFrame columns, especially when you want to transform or manipulate data using a custom function. It\'s often used to avoid explicit loops, which can be slower and less concise.\n\nStep 3: Filter data and find only \'tweet_id\' column. But the thing you remember here is that if you will return this like :\n\ntweets[tweets[\'content\'].apply(len)>15][\'tweet_id\'],\nthen the above command will return series, like \n<class \'pandas.core.series.Series\'> \nnot dataframe, but in this question we need to find dataframe,so convert the result into dataframe using :\npd.DataFrame(tweets[tweets[\'content\'].apply(len)>15][\'tweet_id\'])\n\n# Complexity\n- Time complexity: O(n+k)\nThe total time complexity of the code is O(n + k), where n is the total number of rows in the DataFrame and k is the number of rows that satisfy the condition.\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(k)\nThe new DataFrame df is created to store the selected \'tweet_id\' values, which takes O(k) space, where k is the number of rows that satisfy the condition.\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nimport pandas as pd\n\ndef invalid_tweets(tweets: pd.DataFrame) -> pd.DataFrame:\n \n df=pd.DataFrame(tweets[tweets[\'content\'].apply(len)>15][\'tweet_id\'])\n\n #result=df[[\'tweet_id\']]\n\n return df\n```
| 3 | 0 |
['Pandas']
| 0 |
invalid-tweets
|
Python easy solutions
|
python-easy-solutions-by-purvi85-yl0r
|
Code\n\nimport pandas as pd\n\ndef invalid_tweets(tweets: pd.DataFrame) -> pd.DataFrame:\n return tweets[tweets[\'content\'].str.len() > 15][[\'tweet_id\']]\
|
Purvi85
|
NORMAL
|
2023-08-02T06:36:10.486595+00:00
|
2023-08-02T06:36:10.486619+00:00
| 701 | false |
# Code\n```\nimport pandas as pd\n\ndef invalid_tweets(tweets: pd.DataFrame) -> pd.DataFrame:\n return tweets[tweets[\'content\'].str.len() > 15][[\'tweet_id\']]\n```
| 3 | 0 |
['Pandas']
| 0 |
invalid-tweets
|
Beats 100.00% Apply Mask
|
beats-10000-apply-mask-by-jmp23-of72
|
Approach\n Describe your approach to solving the problem. \nCreate a mask where the length of content field is greater than 15. \n\n# Code\n\nimport pandas as p
|
jmp23
|
NORMAL
|
2023-07-31T20:39:11.981879+00:00
|
2023-07-31T20:39:11.981914+00:00
| 1,417 | false |
# Approach\n<!-- Describe your approach to solving the problem. -->\nCreate a mask where the length of content field is greater than 15. \n\n# Code\n```\nimport pandas as pd\n\ndef invalid_tweets(tweets: pd.DataFrame) -> pd.DataFrame:\n\n mask = tweets[\'content\'].str.len() > 15\n\n return tweets.loc[mask][[\'tweet_id\']]\n```
| 3 | 0 |
['Pandas']
| 0 |
invalid-tweets
|
Easy C++ Solution ✅ ✅ || 🔥🔥🔥
|
easy-c-solution-by-sh_77074-tkxi
|
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time
|
sh_77074
|
NORMAL
|
2023-07-20T08:22:30.820300+00:00
|
2023-07-20T08:22:30.820326+00:00
| 573 | false |
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\n# Write your MySQL query statement below\nSelect tweet_id from Tweets \nWhere length(Tweets.content) > 15; \n```\n\n
| 3 | 0 |
['MySQL']
| 0 |
invalid-tweets
|
Easy MYSQL Solution
|
easy-mysql-solution-by-2005115-ma3p
|
\n\n# Code\n\n# Write your MySQL query statement below\nselect tweet_id from Tweets where LENGTH(content)>15;\n
|
2005115
|
NORMAL
|
2023-07-17T03:00:26.489834+00:00
|
2023-07-17T03:00:57.944780+00:00
| 2,334 | false |
\n\n# Code\n```\n# Write your MySQL query statement below\nselect tweet_id from Tweets where LENGTH(content)>15;\n```
| 3 | 0 |
['MySQL']
| 0 |
invalid-tweets
|
Simply MySQL solution with CHAR_LENGTH method
|
simply-mysql-solution-with-char_length-m-q35f
|
MySQL code:\n\n# Write your MySQL query statement below\nselect tweet_id from Tweets where CHAR_LENGTH(content) > 15;\n
|
sopheary
|
NORMAL
|
2023-07-12T17:17:02.482247+00:00
|
2023-07-12T17:17:02.482270+00:00
| 2,142 | false |
MySQL code:\n```\n# Write your MySQL query statement below\nselect tweet_id from Tweets where CHAR_LENGTH(content) > 15;\n```
| 3 | 0 |
['MySQL']
| 1 |
invalid-tweets
|
sql || simple Straight forward approach
|
sql-simple-straight-forward-approach-by-78msz
|
\n# Write your MySQL query statement below\n\nselect tweet_id from Tweets\nwhere CHAR_LENGTH(content) > 15;\n\n\n\n\nplease upvote!! if you like.\ncomment belo
|
haneelkumar
|
NORMAL
|
2023-07-01T13:16:53.329716+00:00
|
2023-07-01T13:16:53.329744+00:00
| 509 | false |
```\n# Write your MySQL query statement below\n\nselect tweet_id from Tweets\nwhere CHAR_LENGTH(content) > 15;\n```\n\n\n\n**please upvote!! if you like.**\ncomment below\uD83D\uDC47
| 3 | 0 |
['MySQL']
| 1 |
invalid-tweets
|
MySQL Solution for Invalid Tweets Problem
|
mysql-solution-for-invalid-tweets-proble-bvy7
|
Intuition\n Describe your first thoughts on how to solve this problem. \nThe query aims to identify invalid tweets by checking if the length of the content exce
|
Aman_Raj_Sinha
|
NORMAL
|
2023-06-01T03:42:08.157428+00:00
|
2023-06-01T03:42:08.157464+00:00
| 3,010 | false |
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe query aims to identify invalid tweets by checking if the length of the content exceeds 15 characters.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. The SELECT clause specifies the tweet_id column to be retrieved.\n1. The FROM clause specifies the table name "Tweets" from which the data will be selected.\n1. The WHERE clause applies the condition by checking if the length of the content column is greater than 15 using the LENGTH() function.\n1. The result of the query will be the tweet_id values for the invalid tweets that satisfy the condition.\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nThe time complexity of this query depends on the size of the Tweets table and the available indexes. With proper indexing, the query can perform the filtering efficiently in O(log n) or O(n) time, where n is the number of rows in the table.\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\nThe space complexity of this query is O(1) as it does not require any additional space proportional to the input size. The result will be returned as a result set without consuming significant additional memory.\n\n# Code\n```\n# Write your MySQL query statement below\nSELECT tweet_id\nFROM Tweets\nWHERE LENGTH(content) > 15;\n```
| 3 | 0 |
['MySQL']
| 0 |
invalid-tweets
|
SQL✅ MySQL✅|| Easiest
|
sql-mysql-easiest-by-jasurbekaktamov081-j39l
|
Intuition\n# Please UPVOTE \uD83D\uDC4D\n\n\n\n# Code\n\nselect\ntweet_id\nfrom Tweets\nwhere length(content) > 15\n
|
jasurbekaktamov081
|
NORMAL
|
2023-05-19T10:37:44.541750+00:00
|
2023-05-19T10:37:44.541797+00:00
| 33 | false |
# Intuition\n# ***Please UPVOTE \uD83D\uDC4D***\n\n\n\n# Code\n```\nselect\ntweet_id\nfrom Tweets\nwhere length(content) > 15\n```
| 3 | 0 |
['MySQL']
| 0 |
invalid-tweets
|
[MySQL] Simple approach beat 100%
|
mysql-simple-approach-beat-100-by-dolong-14go
|
Intuition\n Describe your first thoughts on how to solve this problem. \nThere is a function in MySQL called length to get the length of string\n\n# Approach\n
|
dolong2110
|
NORMAL
|
2022-12-10T10:24:41.687232+00:00
|
2022-12-10T10:24:41.687257+00:00
| 855 | false |
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThere is a function in MySQL called length to get the length of string\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nO(N) ~ O(N * L) - of course we have to iterate to every row in table. In each row we have to check the length of the tweets\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\nO(1)\n\n# Code\n```\n# Write your MySQL query statement below\nselect tweet_id \nfrom Tweets \nwhere length(content) > 15;\n```\n
| 3 | 0 |
['MySQL']
| 1 |
invalid-tweets
|
MySQL simple solution
|
mysql-simple-solution-by-autofrankie-ifv3
|
SELECT t.tweet_id from Tweets t\nWHERE LENGTH(t.content) > 15;
|
autofrankie
|
NORMAL
|
2021-01-05T02:00:06.105366+00:00
|
2021-01-05T02:00:06.105410+00:00
| 1,381 | false |
SELECT t.tweet_id from Tweets t\nWHERE LENGTH(t.content) > 15;
| 3 | 0 |
[]
| 1 |
invalid-tweets
|
Simple Query ✅ | With Proper Explanation 🤝
|
simple-query-with-proper-explanation-by-54nyy
|
IntuitionThe problem requires us to find tweets that have more than 15 characters in the content column. In SQL, we can determine the length of a string using t
|
NadeemMohammed
|
NORMAL
|
2025-03-30T05:37:28.466218+00:00
|
2025-03-30T05:37:28.466218+00:00
| 501 | false |
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
The problem requires us to find tweets that have more than 15 characters in the content column. In SQL, we can determine the length of a string using the LENGTH() function. Our goal is to filter out tweets where the character count exceeds 15 and return their tweet_id.
# Approach
<!-- Describe your approach to solving the problem. -->
1. Use the LENGTH() function:
* The `LENGTH(content)` function calculates the number of characters in the content column of the Tweets table.
2. Apply the WHERE condition:
* We filter rows where the length of content is greater than 15 using `WHERE LENGTH(content) > 15.`
3. Select only the tweet_id column:
* Since the problem requires us to return only the tweet_id, we select it explicitly.
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
If you found my solution helpful, I’d really appreciate it if you could give it an upvote! 🙌 Your support keeps me motivated to share more simple and efficient solutions. Thank you! 😊
# Code
```postgresql []
-- Write your PostgreSQL query statement below
SELECT tweet_id
FROM Tweets
WHERE LENGTH(content) > 15
```
| 2 | 0 |
['MySQL', 'PostgreSQL']
| 0 |
invalid-tweets
|
Solution that works for all DBMS (using LIKE)
|
solution-that-works-for-all-dbms-using-l-zmka
|
Explaination
Content's length is strictly greater than 15 -> must be at least 16 characters -> solution: 16 underscore wildcards with % in both ends
Code
|
DmCZxtEfdY
|
NORMAL
|
2025-03-24T15:59:52.766005+00:00
|
2025-03-24T15:59:52.766005+00:00
| 360 | false |
# Explaination
- Content's length is strictly greater than 15 -> must be at least 16 characters -> solution: 16 underscore wildcards with % in both ends
# Code
```
select tweet_id from Tweets where content like '%________________%'
```
| 2 | 0 |
['MySQL', 'Oracle', 'MS SQL Server']
| 0 |
invalid-tweets
|
Difference btw LENGTH() & CHAR_LENGTH() ?
|
difference-btw-length-char_length-by-amo-v08g
|
ApproachLENGTH(string)
Returns the number of bytes in the given string.
If the string contains multi-byte characters, the length will be greater than the number
|
AmoghaMayya
|
NORMAL
|
2025-03-02T08:55:49.619501+00:00
|
2025-03-02T08:55:49.619501+00:00
| 418 | false |
# Approach
<!-- Describe your first thoughts on how to solve this problem. -->
**LENGTH(string)**
Returns the number of bytes in the given string.
If the string contains multi-byte characters, the length will be greater than the number of characters.
**CHAR_LENGTH(string) / CHARACTER_LENGTH(string)**
Returns the number of characters in the given string.
This is useful for handling multi-byte characters correctly.
# Code
```postgresql []
select tweet_id
from tweets
where char_length(content) > 15;
```
| 2 | 0 |
['PostgreSQL']
| 0 |
invalid-tweets
|
Find Invalid Tweets by Length
|
find-invalid-tweets-by-length-by-noobml-4y4d
|
IntuitionWe need to find tweets where the length of the content is greater than 15. Since thecontentcolumn is a string, we can use string length functions to fi
|
NoobML
|
NORMAL
|
2025-02-24T10:22:17.180723+00:00
|
2025-02-24T10:22:17.180723+00:00
| 593 | false |
### **Intuition**
We need to find tweets where the length of the content is greater than 15. Since the `content` column is a string, we can use string length functions to filter out invalid tweets.
### **Approach**
1. Use `.str.len()` to compute the length of each tweet in the `content` column.
2. Create a boolean mask to filter tweets where the length is **greater than 15**.
3. Select only the `tweet_id` column to return the required result as a **DataFrame**.
### **Complexity**
- **Time Complexity:** $$O(n)$$
We iterate through all tweets once to check their lengths
- **Space Complexity:** $$O(1)$$
We modify the existing DataFrame without using extra space
# Code
```pythondata []
import pandas as pd
def invalid_tweets(tweets: pd.DataFrame) -> pd.DataFrame:
tweet = tweets[tweets['content'].str.len() > 15][['tweet_id']]
return tweet
```
| 2 | 0 |
['Pandas']
| 0 |
invalid-tweets
|
Easy Solution using wildcard approach
|
easy-solution-using-wildcard-approach-by-kety
|
Write your MySQL query statement belowselect tweet_id from tweets where content LIKE "________________%"
//Used 16 underscores
|
ishikag88
|
NORMAL
|
2025-01-14T08:50:20.378507+00:00
|
2025-01-14T08:50:20.378507+00:00
| 383 | false |
# Write your MySQL query statement below
select tweet_id from tweets where content LIKE "________________%"
//Used 16 underscores
| 2 | 0 |
['MySQL']
| 1 |
invalid-tweets
|
Easy solution of select query using MySQL
|
easy-solution-of-select-query-using-mysq-h6xs
|
IntuitionApproachComplexity
Time complexity:
Space complexity:
Code
|
Siddarth9911
|
NORMAL
|
2024-12-19T12:32:21.157523+00:00
|
2024-12-19T12:32:21.157523+00:00
| 705 | false |
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```mysql []
select tweet_id
from Tweets
where length(content)>15
```
| 2 | 0 |
['MySQL']
| 0 |
invalid-tweets
|
1 line solution beats 91.78%
|
1-line-solution-beats-9178-by-victorhpxa-q4f9
|
\n\n# Code\n\n# Write your MySQL query statement below\nSELECT tweet_id FROM Tweets WHERE length(content) > 15;\n
|
VictorhpXavier
|
NORMAL
|
2024-07-30T20:31:00.330569+00:00
|
2024-07-30T20:31:00.330600+00:00
| 19 | false |
\n\n# Code\n```\n# Write your MySQL query statement below\nSELECT tweet_id FROM Tweets WHERE length(content) > 15;\n```
| 2 | 0 |
['MySQL']
| 0 |
invalid-tweets
|
Easy and Simple my sql solution 💯
|
easy-and-simple-my-sql-solution-by-amitm-ntm4
|
\nselect tweet_id\nfrom tweets\nwhere length(content)>15\n
|
amitmungare27
|
NORMAL
|
2024-01-28T11:43:54.200458+00:00
|
2024-01-28T11:43:54.200491+00:00
| 1,131 | false |
```\nselect tweet_id\nfrom tweets\nwhere length(content)>15\n```
| 2 | 0 |
['Database', 'MySQL']
| 0 |
invalid-tweets
|
SQL Query
|
sql-query-by-khushbu_01-0ktj
|
```\nSELECT tweet_id\nFROM Tweets\nWHERE CHAR_LENGTH(content) > 15
|
Khushbu_01
|
NORMAL
|
2024-01-03T17:56:21.842178+00:00
|
2024-01-03T17:56:21.842213+00:00
| 1,054 | false |
```\nSELECT tweet_id\nFROM Tweets\nWHERE CHAR_LENGTH(content) > 15
| 2 | 0 |
[]
| 0 |
invalid-tweets
|
3 simple one-line solutions in Pandas
|
3-simple-one-line-solutions-in-pandas-by-bd4j
|
Code\n\nimport pandas as pd\n#find the IDs of the invalid tweets\ndef invalid_tweets(tweets: pd.DataFrame) -> pd.DataFrame:\n return tweets[tweets.content.ma
|
memoryofayllwros
|
NORMAL
|
2023-12-04T14:43:20.324692+00:00
|
2023-12-04T14:43:20.324716+00:00
| 427 | false |
# Code\n```\nimport pandas as pd\n#find the IDs of the invalid tweets\ndef invalid_tweets(tweets: pd.DataFrame) -> pd.DataFrame:\n return tweets[tweets.content.map(len) > 15][[\'tweet_id\']];\n#or \n return tweets[tweets.content.str.len()>15][[\'tweet_id\']];\n#or\n return tweets[tweets.content.apply(lambda x: len(x) > 15)][[\'tweet_id\']];\n```
| 2 | 0 |
['Pandas']
| 0 |
invalid-tweets
|
MYSQL || Beats 99% in Time && 100%in space complexity
|
mysql-beats-99-in-time-100in-space-compl-cyj1
|
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time
|
krishnawable
|
NORMAL
|
2023-10-28T09:32:24.317394+00:00
|
2023-10-28T09:32:24.317411+00:00
| 4,187 | false |
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\n# Write your MySQL query statement below\nselect tweet_id from Tweets where length(content) >15;\n```
| 2 | 0 |
['MySQL']
| 1 |
invalid-tweets
|
1683. Easy Sql Query
|
1683-easy-sql-query-by-shatrughan10-41tr
|
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time
|
Shatrughan10
|
NORMAL
|
2023-10-11T10:16:00.024777+00:00
|
2023-10-11T10:16:00.024809+00:00
| 7,664 | false |
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\n# Write your MySQL query statement below\nselect tweet_id from Tweets where length(content)>15\n```
| 2 | 0 |
['MySQL']
| 0 |
invalid-tweets
|
Simple 95% 3 line solution
|
simple-95-3-line-solution-by-31b4-jinl
|
ce complexity here, e.g. O(n) -->\n\n# Code\n\nSelect tweet_id\nfrom Tweets\nwhere LENGTH(content)>15\n\n\n
|
31b4
|
NORMAL
|
2023-10-09T08:34:07.333753+00:00
|
2023-10-09T08:34:07.333775+00:00
| 2,259 | false |
ce complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nSelect tweet_id\nfrom Tweets\nwhere LENGTH(content)>15\n```\n\n
| 2 | 0 |
['MySQL']
| 0 |
invalid-tweets
|
Simple One Line Solution || Beat 100% ||
|
simple-one-line-solution-beat-100-by-yas-hog7
|
\n# Complexity\n- Time complexity:O(n)\n Add your time complexity here, e.g. O(n) \n\n- Space complexity:O(1)\n Add your space complexity here, e.g. O(n) \n\n#
|
yashmenaria
|
NORMAL
|
2023-10-06T09:24:52.796965+00:00
|
2023-10-06T09:24:52.796985+00:00
| 1 | false |
\n# Complexity\n- Time complexity:O(n)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:O(1)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\n# Write your MySQL query statement below\nSelect tweet_id from Tweets where length(content)>15\n```
| 2 | 0 |
['MySQL']
| 0 |
invalid-tweets
|
✅Super simple solution🔥
|
super-simple-solution-by-uscutkarsh-0nrv
|
Intuition\nuse char_length() instead of length()\n\n\n# Code\n\n# Write your MySQL query statement below\nselect tweet_id\nfrom Tweets\nwhere char_length(conten
|
uscutkarsh
|
NORMAL
|
2023-10-05T05:40:18.856962+00:00
|
2023-10-05T05:40:18.856987+00:00
| 1,202 | false |
# Intuition\nuse char_length() instead of length()\n\n\n# Code\n```\n# Write your MySQL query statement below\nselect tweet_id\nfrom Tweets\nwhere char_length(content) > 15\n```
| 2 | 0 |
['MySQL']
| 0 |
invalid-tweets
|
easy mysql solution
|
easy-mysql-solution-by-joshi_y-sunx
|
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time
|
Joshi_Y
|
NORMAL
|
2023-09-16T11:40:12.666510+00:00
|
2023-09-16T11:40:12.666530+00:00
| 3 | false |
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\n# Write your MySQL query statement below\nselect tweet_id from tweets\nwhere case when length(content)>15 then tweet_id else null\nend;\n```
| 2 | 0 |
['MySQL']
| 0 |
invalid-tweets
|
Easiest SQL Solution || Beats 100.00% || MySQL, MS SQL Server, Oracle 🔥
|
easiest-sql-solution-beats-10000-mysql-m-gxv8
|
Intuition\nThe intuition behind solving this problem is straightforward. We need to identify and filter out tweets that have content with a character count grea
|
muhit-khan
|
NORMAL
|
2023-09-02T15:03:19.854444+00:00
|
2023-09-02T15:04:48.421112+00:00
| 239 | false |
# Intuition\nThe intuition behind solving this problem is straightforward. We need to identify and filter out tweets that have content with a character count greater than 15. In SQL, we can achieve this by using a simple condition in our query.\n\n\n# Approach\nOur approach to solving this problem is to write an SQL query that selects the `tweet_id` from the `Tweets` table where the length of the `content` column is greater than 15 characters. This condition ensures that we are only retrieving the IDs of tweets that meet the criteria of having content longer than 15 characters.\n\n# Complexity \n- Time complexity: The time complexity of this solution is $$O(n)$$, where n is the number of tweets in the `Tweets` table. We need to scan through each tweet to check its content length.\n- Space complexity: The space complexity is $$O(1)$$, as we are not using any additional data structures that scale with the input size. We are simply querying the database without significant memory usage.\n\n# Code\n\n#### SQL Query for Oracle\n```\nSELECT tweet_id FROM Tweets WHERE LENGTH(content)>15;\n```\n#### SQL Query for MS SQL Server\n```\nSELECT tweet_id FROM Tweets WHERE LEN(content)>15;\n```\n#### SQL Query for MySQL\n```\nSELECT tweet_id FROM Tweets WHERE CHAR_LENGTH(content)>15;\n```
| 2 | 0 |
['MySQL', 'Oracle', 'MS SQL Server']
| 0 |
invalid-tweets
|
⚡ Pandas Simple and Fast Solution 🚀
|
pandas-simple-and-fast-solution-by-pnira-xwlo
|
\uD83D\uDD3C IF YOU FIND THIS POST HELPFUL PLEASE UPVOTE \uD83D\uDC4D\n\nimport pandas as pd\n\ndef invalid_tweets(tweets: pd.DataFrame) -> pd.DataFrame:\n \
|
pniraj657
|
NORMAL
|
2023-08-12T15:25:12.506133+00:00
|
2023-08-12T15:25:12.506150+00:00
| 125 | false |
**\uD83D\uDD3C IF YOU FIND THIS POST HELPFUL PLEASE UPVOTE \uD83D\uDC4D**\n```\nimport pandas as pd\n\ndef invalid_tweets(tweets: pd.DataFrame) -> pd.DataFrame:\n \n res_df = tweets[(tweets[\'content\'].str.len() > 15)]\n \n return res_df[[\'tweet_id\']]\n```\n**Thank you for reading! \uD83D\uDE04 Comment if you have any questions or feedback.**
| 2 | 0 |
['Python']
| 0 |
invalid-tweets
|
Easy Pandas Solution
|
easy-pandas-solution-by-vinit_k_p-8n6q
|
\n\n# Code\n\nimport pandas as pd\n\ndef invalid_tweets(tweets: pd.DataFrame) -> pd.DataFrame:\n df = tweets[tweets[\'content\'].str.len() > 15]\n df = df
|
Vinit_K_P
|
NORMAL
|
2023-07-31T19:28:26.892609+00:00
|
2023-07-31T19:28:26.892629+00:00
| 713 | false |
\n\n# Code\n```\nimport pandas as pd\n\ndef invalid_tweets(tweets: pd.DataFrame) -> pd.DataFrame:\n df = tweets[tweets[\'content\'].str.len() > 15]\n df = df[[\'tweet_id\']]\n return df\n```
| 2 | 0 |
['Pandas']
| 0 |
invalid-tweets
|
Invalid Tweets 🧑💻🧑💻 || MySql solution code 💁💁...
|
invalid-tweets-mysql-solution-code-by-ja-7qpy
|
Code\n\n# Write your MySQL query statement below\nselect tweet_id from Tweets where length(Tweets.content) > 15; \n
|
Jayakumar__S
|
NORMAL
|
2023-07-16T19:11:06.721965+00:00
|
2023-07-16T19:11:06.721992+00:00
| 774 | false |
# Code\n```\n# Write your MySQL query statement below\nselect tweet_id from Tweets where length(Tweets.content) > 15; \n```
| 2 | 0 |
['MySQL']
| 0 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.