Відео

Why we multiplying k time every element like See suppose array contail [a,b,c,d] and K=2 But in solution we are multiply each element 2^k time and checking which or is maximum meand 4 to each element and checking But there can be possibility like first we multiply 2 to b then in k=2 2 to c means final array will be a, 2b,2c,d

awesome explanation

If string is 00001 Then alice can't put 'or' at last as it's Bob's chance

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Great Solution dude

Useless logic

Just small update in code in both while loops just swap two lines use below while(mp[ans%x]>0) { mp[ans%x]--; ans++; } instead of this while(mp[ans%x]>0) { ans++; mp[ans%x]--; }

Keep it up 🎉🎉🎉

Wah Shekhar bhai wah 🎉

can u please share the project

Time complexity is o(m*n) and Extra Space complexity is o(m*n) of recursion stack space and the equation is fg=fg && g1[i][j] not fg=1 && g1[i][j] in further video.

where are you from ?

In code use (1LL<<j) in place of 1<<j as it will give run time error because int can't handle j after 32

#include <bits/stdc++.h> using namespace std; void solve() { int n; cin>>n; vector<int> a(n); for(int i=0;i<n;++i) { cin>>a[i]; } long long int sum=accumulate(a.begin(),a.end(),0); int element=sum/n; long long int extra=0; for(int i=0;i<n;++i) { if(a[i]>=element) { extra+=(a[i]-element); } else { long long int need=element-a[i]; if(extra>=need) { extra-=(need); } else { cout<<"NO"<<endl; return; } } } cout<<"YES"<<endl; } int main() { int t; cin>>t; while(t--) { solve(); } return 0; } Updated code as there were some changes in code shown in video

Thank you sir for uploading the B question. It really helps us 🙏🙏

There is an update in code part use for(int i=1;i<=sqrt(x);++i) instead of for(int i=0;i<=sqrt(x);++i) as it will lead to division by zero.

Sir please upload Div2 question B