Video

Where is he saving the system-build code?

best c++ lecture ever been through .

Bhaiya, I think it would be helpful if you could pause slightly between your sentences. Sometimes, the continuous flow makes it a bit challenging to absorb everything you're saying.

We can use pair<int, int>

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great

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i think we calculated lowest prime, not highest prime

Sir just for fun there were a 2 luv bhaiyas teaching the time complexity in this video..

similar problem on Leetcode : Rotting Oranges

/bin/sh: g++-14: command not found Does anyone have any solution to this problem?

Your explanation is brilliant

Find my solution below class Solution { public: void dfs(int i, int j, int n, int m, vector<vector<int>> &vis, vector<vector<char>> &grid) { //condition checks on the entering node if(i<0 || j<0) return; if(i>=n || j>=m) return; if(grid[i][j]=='0') { vis[i][j]=1; return; } if(vis[i][j]==1) return; //operation on current node vis[i][j]=1; //apply dfs on the child nodes on all four directions ' dfs(i,j+1,n,m,vis,grid); dfs(i,j-1,n,m,vis,grid); dfs(i+1,j,n,m,vis,grid); dfs(i-1,j,n,m,vis,grid); } int numIslands(vector<vector<char>>& grid) { //cout<<'working'<<endl; int noIslands = 0; int n= grid.size(); int m= grid[0].size(); vector<vector<int>> vis(n, vector<int>(m, 0)); for(int i=0;i<n;i++){ for(int j=0;j<m;j++){ if( grid[i][j]!='0' && vis[i][j]!=1){ dfs(i,j,n,m,vis,grid); noIslands++; } } } return noIslands; } };

those who getting wrong answers try this one: #include<numeric> using namespace std; const int Mod=1e9+7; void dfs(int v,vector<int>g[],vector<bool>&vis,vector<int>&subtree_sum,vector<int>&val) { subtree_sum[v]+=val[v-1]; vis[v]=true; for(auto c:g[v]) { if(vis[c]) { continue; } dfs(c,g,vis,subtree_sum,val); subtree_sum[v]+=subtree_sum[c]; } } int Solution::deleteEdge(vector<int> &A, vector<vector<int> > &B) { int n=A.size(); vector<int>g[n+1],subtree_sum(n+1,0); vector<bool>vis(n+1,false); for(auto e:B) { g[e[0]].push_back(e[1]); g[e[1]].push_back(e[0]); } dfs(1,g,vis,subtree_sum,A); long long ans=0; for(int i=2; i<=n; i++) { int x=(subtree_sum[i]); int y=(subtree_sum[1]-x); ans=max(ans,(x*1LL*y)); } return ans%Mod; }

Code with Harry light pro max😅

2 - 3 ghante soch ne laga fir samga

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We can do this in O(n), without using stacks too. Just loop from the right and have a variable storing the max number

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thank u cp master 🪔

For the second question: 1. find the number (lim) greater than C with all bits set. 2. a) iterate through lim to 1, Let each number as A; b) in each iteration find B . wich is equal to A^C. c) find maximum product. Code for clear understanding: #include<bits/stdc++.h> using namespace std; int main(){ int c, a, b; long long ans = 1; cin >> c; int bits = 0, n; n = c; while(n > 0){ bits++; n >>= 1; } int lim = pow(2, bits) - 1; for(int i = lim; i > 0; --i){ b = c ^ i; ans = max(ans , b * 1LL * i); } cout << ans; return 0; }

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bhai thida level kam karke padhalee,samaj me aane ke liye 2 baar dekhna padta hai

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best video on recursion

have you covered sliding window in ur series?

pdf kaha milengi eski

In this playlist , any sliding window concept video ???

Sir mai c++ start kya hu , ye saficient rhega for cp?

iam just at ep 22 and i already have lost 55 braincells :"), still holding upppp

very clear and crisp explanation...♥♥

Thank you so much!😉😉

Can you told me how you create "model-bw_dru.h5" , model-bw_smn.h5 and "model-bw_tkdl.h5" models..... If any source code is available???...

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🔥🔥bhaiya. Kya mast samjhaya hai, ek no.

why you didnt used pow in gpl question what is the problem with that as I have used it and also my submission is partially correct #include<bits/stdc++.h> using namespace std; int main(){ int t; cin>>t; while(t--){ long int s; cin>>s; string p; cin>>p; long long int code=0; for(long int i= s-1;i>=0;--i){ long int n=0,o=0; o=s-i-1; n=p[i]-'0'; code += n*pow(2,o); } cout<<code<<endl; } }

Bhai kaha ho ab ? 😢

Thanks for this ❤

The selection of the similar problems to explain the difference is very well thought!

The important difference nobody tells you! 🙌🏻🙌🏻🙌🏻 very well explained

superb... explanation... thnkss..

Thanks bro, and @YouKnowWho

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#include <bits/stdc++.h> using namespace std; const int N = 1e7 + 10; int hsh[N][26]; int main() { int t; cin >> t; while (t--) { for (int i = 0; i < N; i++) { for (int j = 0; j < 26; j++) { hsh[i][j] = 0; } } int n, q; cin >> n >> q; string s; cin >> s; for (int i = 0; i <n; i++) { hsh[i+1][s[i] - 'a']++; } for (int i = 0; i < 26; i++) { for (int j = 1; j <= n; j++) { hsh[j][i] += hsh[j - 1][i]; } } while (q--) { int l, r; cin >> l >> r; int oddcnt = 0; for (int i = 0; i < 26; i++) { int charCt = hsh[r][i] - hsh[l - 1][i]; if (charCt % 2 != 0) { oddcnt++; } } if (oddcnt > 1) { cout << "NO" << endl; } else { cout << "YES" << endl; } } } }

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