I i knew striver early i would have been in job right now

notes link?

much needed session 🔥

starts at 3:10

video starts at 8:00

34:55 correct rearrangement: bx1 + (a-[a/b]b)y1 = g bx1 + ay1 - [a/b]by1 = g bx1 - [a/b]by1 + ay1 = g b(x1 - [a/b]y1) + ay1 = g by + ax = g ax + by = c

I understood all the remaining videos perfectly but faced a bit of difficulty here. Anyways, Thank you for such a detailed video, Striver 🔥🫡.

awesome

1:26:19

1:10:35

GOAT STRIVER bolthe

#striver_king of codding world....... love you

36:18 stivers words come true srinjoy joined google 4 months back

1:10:10

Very good video bhaiya. Learnt a lot of stuff from here.👍

Ok it's average,not so great

Thanks a lot man❤ i can apply this logic elsewhere! I got idea by you logic 😅

4:12 start

😂😂😂😂what the fuck shit way of teaching

24:19

Very Informative

1:56:00 this solution is getting TLE In python, ig codeforces is impossible in python

Understood

bezoor

Wtf are you teaching my friend

2024 People Like

19:00 auto it = st.begin(); std::advance(it, 2); // Advance iterator to point to the third element st.erase(st.begin(), it);

1:01:20 good one

56:30 good resume

50:00

Out put will be different

Y u does not run the program

understood 👍👍👍❤❤

One of the best segment tree video. I was able to clear infosys oa round because of this video. Thank you so much striver bhaiya ♥️

1:54:44 king fss!!!

#include<iostream> #include<bits/stdc++.h> using namespace std; class SGTree{ vector<int> open; vector<int> close; vector<int> complete; public: SGTree(int n){ open.resize(4*n+1); close.resize(4*n+1); complete.resize(4*n+1); } void build(int ind, int low, int high, string& s){ if(low==high){ if(s[low]=='('){ open[ind]=1; close[ind]=0; } else{ open[ind]=0; close[ind]=1; } complete[ind]=0; return; } int mid=low+((high-low)>>1); build(2*ind+1, low,mid, s); build(2*ind+2, mid+1, high, s); open[ind]=open[2*ind+1]+open[2*ind+2]-min(open[2*ind+1], close[2*ind+2]); close[ind]=close[2*ind+1]+close[2*ind+2]-min(open[2*ind+1], close[2*ind+2]); complete[ind]=complete[2*ind+1]+complete[2*ind+2]+min(open[2*ind+1], close[2*ind+2]); } vector<int> query(int ind, int low, int high, int l, int r, string& s){ if(l> high || r< low) return vector<int>{0,0,0}; if(high<=r && low>=l) return vector<int>{open[ind], close[ind], complete[ind]}; int mid=low+((high-low)>>1); vector<int> left=query(2*ind+1, low, mid, l, r, s); vector<int> right=query(2*ind+2, mid+1, high, l, r, s); return vector<int>{left[0]+right[0]-min(left[0], right[1]), left[1]+right[1]-min(left[0], right[1]), left[2]+right[2]+min(left[0], right[1])}; } void print(){ for(int i=0;i<open.size();i++){ cout<<"| "<<open[i]<<" "<<close[i]<<" "<<complete[i]<<"| "; } } }; int main(){ string s; cin>>s; int n; cin>>n; SGTree sg1(s.size()); sg1.build(0, 0, s.size()-1, s); //sg1.print(); while(n--){ int l, r; cin>>l>>r; vector<int> ans=sg1.query(0, 0, s.size()-1, l-1, r-1, s); cout<<ans[2]*2<<endl; } }can someone tell me where can i optimize, because i am getting tle in test case 13

GOD!

46:00:00

#include <bits/stdc++.h> using namespace std; #define nl ' ' int main() { int t; cin >> t; while (t--) { int n; cin >> n; int a = 1, b = 1, c = 1; for (int i = 2; i*i <= n; i++) { if (n % i == 0) { a = i; break; } } n = n/a; for (int i = a+1; i*i <= n; i++) { if (n % i == 0) { b = i; break; } } c = n/b; if (a != 1 && b != 1 && c != 1 && a!=b && b!=c && c!=a) { cout << "YES " << a << " " << b << " " << c << nl; } else { cout << "NO "; } } return 0; } my code

I love you sir. I am gay. will you marry me ?

I Did not understand the multiple possible answers wala part

nice, never seen such awesome teaching in paid course of fraz

yes, not ly fraz course paid covered

36:13

Great work strive Raj bro gained more knowledge 😊 Thank you bro🎉

23:00 ,

Learned a lot , finally completed. 12th July 2024.

39:20 Great lecture 9/7/2024

these are all not working priority_queue<pair<int,int>> pal; pal.push({3,4}); pal.push({3,3}); pal.push({7,10}); list and deque ------ cout<<mal.at(mal.begin());