Відео

Thompson Dorothy Harris Edward Wilson Susan

anyone please share 3 or 4

Remaining 2 questions plzz

long long cntGood(string s) { // code here long long a = 0; int z = 0; for (int i = 0; i < s.size(); i++) { int v = s[i] - '0'; if (v % 2 != 0) { a += i + 1; a -= z; } if (v == 0) z++; } return a; }

long long minCost(int n, vector<int> &a, vector<int> &b) { // code here map<int, int> count; vector<int> swap_a, swap_b; for (int i = 0; i < n; i++) { count[a[i]]++; count[b[i]]--; } for (auto it: count) { if (it.second % 2 != 0) return -1; } for (int i=0; i < n; i++) { if (count[a[i]] > 0) { swap_a.push_back(a[i]); count[a[i]] -= 2; } if (count[b[i]] < 0) { swap_b.push_back(b[i]); count[b[i]] += 2; } } sort(swap_a.begin(), swap_a.end()); sort(swap_b.rbegin(), swap_b.rend()); long long cost = 0; int m = swap_a.size(); for (int i = 0; i < m; i++) { cost += min(swap_a[i], swap_b[i]); } return cost; }

Give the code in comment

other??

Bhai 2nd bhejo

something is wrong, its not working can anyone help

Bhai python solutions kha milenge ? Anyone knows?

Bhai python solutions kha milenge ? Anyone knows?

Bhai kisi ko pata h python solutions kha milenge ???

is there python solution for this?

3rd question solution please

answers in comment bro plz its not visible

4th answer is also same as 3rd answer

Bhai 2nd ya 3rd ya 4th ka answer bhi daal do

code bhej do comment pe

Bhai 3rd ka bhej

share the code in c++

Key Idea of the Problem: - You are given two paths. One leads to eternal happiness, and the other leads to certain doom. - Out of ( N + M ) people, ( N ) people are guaranteed to always tell the truth, and the other ( M ) people may or may not tell the truth (i.e., they can lie). - You need to ensure you can find the correct path by asking ( X ) people (and you don't know in advance who are the truth-tellers or liars). Approach: 1. If ( N <= M ): - If the number of truth-tellers ( N ) is less than or equal to the number of people who may lie ( M ), it's impossible to guarantee the correct path. - This is because the liars may outnumber or equal the truth-tellers, and no matter who you ask, you can't reliably distinguish the truth from the lies. In this case, the output should be `-1`. 2. If ( N > M ): - If the number of truth-tellers ( N ) is greater than the number of potential liars ( M ), it is possible to determine the correct path by asking enough people. - To guarantee finding the correct path, you need to ask enough people such that even if the maximum number of liars ( M ) give false answers, the majority of responses will still indicate the correct path. - The minimum number of people ( X ) you need to ask is ( 2M + 1 ). This is because even if all ( M ) liars lie and disagree with each other, there will still be a majority who are truth-tellers giving consistent answers.

code

code

Can you please explain the logic sir ?

Thanks ❤❤❤ bhaiya

//author: cp_god //date: 14th august 2024 #include<bits/stdc++.h> #define int long long using namespace std; int32_t main(){ int t; cin>>t; while(t--){ int n,m; cin>>n>>m; if(m>=n) cout<<-1<<endl; else{ cout<<m*2+1<<endl; } } } truth tellers code

This showing runtime

Bro please give the source code

Sourcecode please

Send code

4th wala bej do

Send code

bhai improve kaise kare ek bhi question nahi hua mera

can you make a video explaining the solution? ngl this was a pretty goog dp solution

Good, but I am not able to solve any of these

How the other country guysbare doing da they just finished everything in just 25 minutes

3rd Any Idea Bruh?

the background music is irritating

aree bahi eska solution dedo

Bro please send code in cmt because my code is working correct but showing time limit exceed please share ur code if u can in cmt section

#include <bits/stdc++.h> using namespace std; int main() { int foo; cin >> foo; vector<int> bar; for (int _ = 0; _ < foo; ++_) { int baz; cin >> baz; vector<int> qux(baz); for (int i = 0; i < baz; ++i) { cin >> qux[i]; } vector<int> quux(baz, 0); vector<int> corge(baz, 0); quux[0] = qux[0]; for (int grault = 1; grault < baz; ++grault) { quux[grault] = quux[grault - 1] + qux[grault]; } corge[baz - 1] = qux[baz - 1]; for (int grault = baz - 2; grault >= 0; --grault) { corge[grault] = corge[grault + 1] + qux[grault]; } int waldo = 0; for (int fred = 0; fred < baz; ++fred) { if (qux[fred] == 0) { if (corge[fred] == quux[fred]) { waldo += 2; } else if (abs(corge[fred] - quux[fred]) == 1) { waldo += 1; } } } bar.push_back(waldo); } for (const int& plugh : bar) { cout << plugh << endl; } return 0; }

Code bhej de bhai

bro code ki link send karo

Code in comment please

bhai code in comment

❤❤❤

bouncing ball

Bro once explain afterwards please

bro send ans