วีดีโอ

very nice explanation thanks :)

thank you ! keep up your good work sir !

you are the best problem solver in youtube who can explain , i hope u keep going !

very clear and intutive. Thanks man❤

thx , u made it simple

Welcome Back Brother. Just love your explanations

not explained well

Thanks man .. I finally got it accepted .. thanks alot

the problem was so easy, i thought of using the inclusion - exclusion technique,couldn't do it.

your explanation is amazing!! LUB

Bhaiya why have you now stopped uploading videos? Are you on a break ? Please reply or give update regarding it.

Vai your are great.I love your videos.Take love from Bangladesh.Keep up the good work

In case, people don't get the flag approach, this maybe the better code 👇👇👇👇 int n; cin>>n; string s; cin>>s; int cnt=1; int ans=1; loop(n-1){ if(s[i]==s[i+1]) { cnt++; ans=max(ans,cnt); } else cnt=1; } cout<<ans+1<<nl;

no one explained this is question in such a detail fabulous

thank you very much for this amazing explanation !! i really appreciate it

good explanation

#include <bits/stdc++.h> using namespace std; int main() { ios::sync_with_stdio(false); cin.tie(0), cout.tie(0); int T = 1; cin >> T; while (T--) { int n; cin >> n; vector<int> a(n); for (int i = 0; i < n; i++) cin >> a[i]; long long s = accumulate(a.begin(), a.end(), 0ll), cur = 0; long long ans = 1; for (int i = 0; i < n - 1; i++) { cur += a[i], s -= a[i]; ans = max(ans, __gcd(s, cur)); } cout << ans << " "; } }

great explanation

thanks man this is the best solution for this problem

oh my god .. how you have come up with this solution ..please tell

void levelSabkeNiklenge(){ int n, k; cin >> n >> k; vector<int> a(n), b(n); for(int i=0; i<n; i++) cin >> a[i]; for(int i=0; i<n; i++) cin >> b[i]; int ans = 0, maxx = b[0], finalAns = 0; for(int i=0; i<min(k,n); i++){ int rem = k-i-1; maxx = max(maxx, b[i]); ans += a[i] + (rem * maxx); finalAns = max(finalAns, ans); ans -= (rem * maxx); } cout << finalAns << endl; }

Please bring more videos

Thanks a lot! Finally understand the solution ;)

gave a sub

DAMN DAMN DAMN very well done

mindblowing, thanks dude

BEAUTIFUL detailed explanation.Loved it. Thank you so much bro.❤

thanks

thanks buddy

it is very good and simple approach, thanks sir

awesome explanation thankyou so much

Bro if the rem =2 and the all values comes out to be 1(all odd numbers in arr) So if we insert 1 in set and after that we make the rem = 4 and we get the all values to be 0 and we insert 0 in set or (may be that it is guaranteed that we will get 1 only ) I don't know the bracket part is it im predicting right or if we get 0 So the set size becomes 2 right and we'll output rem = 4 are ans which is wrong .Im pretty much confused here what if I would have make the set empty after every time rm*2 . Is it ok or m i wrong would u please clear my doubt

nice explanation

Thanks for clear explanation its simple but i dont know why i dont get it before

why can't we use this, sort array than try for all prefix element and same number of suffix element sum and check sum=max(sum, abs(pre[i]-pre[n-i-1])) please some one explain this...

great explaination

good explanation.Thanks❤

Thank you! why in the code the while loop condition is l<r-1, and not l<=r? and also why mid = l+(r-l)/2? and why l = mid, r = mid, not l = mid+1, and r = mid - 1? because I'm confused

Very Clear Explanation🖤

You explain great bro 😍

Epic

very well explained.

You explain really well. Thanks a lot.

god job👏

understood better than the editorial one

why LLONG_MAX/2 and NOT LLONG_MAX I am getting worong answer for LLONG_MAX

Thank you, Bhaiya!

bhai kuch smjh nhi aya

WoW! What a explanation.