This explanation is way better than any other video!

Bhai chess khela karo 3:50 par knight usi row me kyu nhi aa skti 💀💀💀💀

8:15 teri maaki .... chup hoja sale 😂😂😂

you just yapped ... yet i understood the approach Thanks

i am here only to see the implementation of yes cases

Thanks for your easy-to-understand explanation!

AMAZING PROBLEM WITH AMAZING SOLUTION

nunnu touching solution

in first example why can z[i] = 11 , it can contain all of the letters starting from index 0 = 'a'.

isse acha parha ta hi nhi bhai bakwas

I am getting RE for solution with O(n*x) space and O(n*x) time. But it is working on gfg. I have tried with the tabulation as well #include <bits/stdc++.h> using namespace std; #define int long long #define ll long long int mod = 1e9+7; vector<vector<int>>dp; int rec(int n, int sum , int arr[]){ if(sum == 0){ return 0; } if(n==0){ return 1e8; } if(dp[n][sum] != -1){ return dp[n][sum]; } if(arr[n-1] <= sum){ return dp[n][sum] = min(1+rec(n,sum-arr[n-1],arr) , rec(n-1,sum,arr)); }else{ return dp[n][sum] = rec(n-1,sum,arr); } } int32_t main() { ios::sync_with_stdio(false); cin.tie(nullptr); int n;cin >> n; int k;cin >> k; dp.resize(n+1,vector<int>(k+1,-1)); cout << rec(n,k,arr) << endl; return 0; }

#include <iostream> #include <unordered_map> using namespace std; unordered_map<long long, int> collatz_cache; int collatz_length(long long n) { if (n == 1) return 1; if (collatz_cache.find(n) != collatz_cache.end()) return collatz_cache[n]; if (n % 2 == 0) { collatz_cache[n] = 1 + collatz_length(n / 2); } else { collatz_cache[n] = 1 + collatz_length(3 * n + 1); } return collatz_cache[n]; } void weirdAlgo(long long n) { while (n != 1) { cout << n << " "; if (collatz_cache.find(n) != collatz_cache.end()) { n = collatz_length(n); } else if (n % 2 == 0) { n = n / 2; } else { n = 3 * n + 1; } } cout << "1"; return; } int main() { long long n; cin >> n; weirdAlgo(n); return 0; } the code shown in the video is showing tle. here is the optimized solution of the given problem

Please enable Arabic translation please

this can be achieved only with lower_bound or upper_bound also??

Thanks a lot sir❤

Helpful <3

yeh kesa intuition hua ,,, adjacent sides share hone ki baat kha gyi ...

Nice solution for those who understand some concepts and are not blindly attempting this code just to make a contest submission

Kya overlap kr rha hai kis-se?

bakwas

very well explained , thanks !

thanks.Very good explanation.

The explanation was amazing

i think the test cases are updated now and this is not working for the test case n=10, k=2, [5, 3, 5, 6, 9, 3, 7, 6, 6, 1] Its giving output as [2, 2, 1, 3, 0, -2, -5, -6, -11] but the correct output is [2, 2, 1, 3, 6, 4, 1, 0, 5]. Can you please tell how to solve this.

worst method i have an easy approach for this

thank you

#include<bits/stdc++.h> using namespace std; #define inf LLONG_MAX #define pb push_back #define rep(i,a,b) for(int i = a; i < b; i++) #define all(x) x.begin(),x.end() #define in(a) for(int i = 0; i<a.size(); i++) cin>>a[i]; #define out(a) for(int i = 0; i<a.size(); i++) cout<<a[i]<<" "; typedef vector<int> vi; #define sqrt(x) sqrtl(x) #define ret(a) cout<<a<<" "; return #define ppb pop_back #define mp make_pair #define set_bits(x) __builtin_popcountll(x) #define sz(x) ((int)(x).size()) #define next " " void printbinary(int n){ if(n > 0) { printbinary(n / 2);} else {return;} cout<<(n & 1);return;} const int N = 1e5 + 10; const int M = 1e9 + 7; vector<int> primes(N); vector<int> factorial(N); vector<int> inversefactorial(N); void fastio(){ ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0); } int32_t main(){ fastio(); int t; cin>>t; while(t--){ int n; cin>>n; map<int,int> m; vector<int> v(n); for(int i = 0 ; i < n ; ++i){ cin>>v[i]; m[v[i]] = i + 1; } int count = 0; for(int i = 1; i <= 2*n ; i++){ for(int j = i + 1 ; j <= 2*n ; j++){ if((i*j) > (2*n)){ break; } int product = i * j; auto index1 = m.find(i); auto index2 = m.find(j); if((index1 == m.end()) or (index2 == m.end())){ continue; } if((*index1).second + (*index2).second == product){ count++; } } } cout<<count<<endl; } } Exact same code but giving tle can you tell me why?????

very nice explanation and very reusable logic

abbay bhai line kiu ni bani

You need to work on your explanation. Wasted my time.

TLE

bkl delete kidher hai

Very nice explanation sir, Thank you!

launch and texts file does what exactly?

what happen if we take example : A-> abcde, B-> bcdef and n= 5, k = 5 then we can make A->B but for this logic we can't ???

bro jab dp use krni thi to trie ka kya kam? nodes to banaye nhi

can we do th thing like a condition by traversing a simple djkt and push the node with cost and also with cost/2 then which ever gives the minimum take it

thanks

Thanks for this awesome and concise explanation...

Very nice explanation sir, Thank you!

Very nice explanation sir, Thank you!

sir can you share link of code

kzbin.info/www/bejne/lXWrc6OQnsp2rbc

Bhai, sahi se samajh mein nahi aaya pahle to, but idea lag gaya, aur cout n times karke complexity high ho jayegi isliye use string mein append karke ek saath print karo, printing one time is more feasible

#include <iostream> #include <vector> using namespace std; typedef long long ll; void dfs(ll node, vector<ll> &vis, vector<vector<ll>> &adj, vector<ll> &ans) { vis[node] = 1; ans.push_back(node); for (auto it : adj[node]) { if (!vis[it]) { dfs(it, vis, adj, ans); } } } void solve() { ll n; cin >> n; vector<ll> v(n); for (auto &it : v) { cin >> it; } if(n==1 and v[0]==1){ cout<<2<<" "<<1<<endl; return; } vector<vector<ll>> adj(n + 2); for (int i = 0; i < n + 2; i++) { adj[i].clear(); } vector<ll> vis(n + 2, 0); for (int i = 1; i < n; i++) { adj[i].push_back(i + 1); } for (int i = 1; i < n; i++) { if (v[i] == 0) { adj[i + 1].push_back(n + 1); } else { adj[n + 1].push_back(i + 1); } } vector<ll> ans; dfs(1, vis, adj, ans); for(int i=1;i<=n+1;i++){ if(vis[i]==0){ cout<<-1<<endl; return; } } for (auto it : ans) { cout << it << " "; } cout << endl; } int main() { ll test; cin >> test; while (test--) { solve(); } return 0; } hwats wrong in my logic....?

chutiya explaination

bro kya fook ke aaya he

worst explanation possible.

Informative video

one way can be to formulate it as perfix sum problem sum(l,r) = psum(r) - psum(l-1) now it is good subarray if psum(r) - psum(l-1) = r - l + 1 psum(r) - r = psum(l-1) - (l-1) which gives indeed the prefix sums with same values