///just get the smallest child for every node than loop over them to get the answer 
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ii pair<int,int>

#include <bits/stdc++.h>

using namespace std ;

#define mem(x,val) memset((x),(val),sizeof(x))

#define all(x) x.begin(),x.end()

I would love to see a stack based dfs solution to this problem? If anyone solved by that method please share.

Thanks a lot for the link!

Too bad the website only accepts C, C++, and PASCAL (!!??) submissions. The solution in Python is, as always, compact and elegant:

# https://szkopul.edu.pl/problemset/problem/4CirgBfxbj9tIAS2C7DWCCd7/site/?key=statement

#include <iostream>
#include <utility>
#include <queue>
#include <climits>
#include <cstring>

WhY wrong answer ? can you help me!

https://ideone.com/QYs4qD

#include<bits/stdc++.h>
 
using namespace std;
#define ll long long
#define ii pair<int,int>

Here is my python code,

import sys
sys.setrecursionlimit(10000000)
look_up = {}

I came up with a O(n) solution to the problem, can anyone help me prove my hypothesis. I have an intuition for my problem as follows:

    for (ll i = 0; i < n; ++i) {

Can anyone find issue in my code:

Hi Guys What is Problem With this?

My code is giving WA for 2 testcases ,can anyone tell me what's wrong in my code?

#include<bits/stdc++.h>

Why is this on the Binary Search section ?

Mine and most solutions I saw did not use it.

Is there a solution that uses it ?

#include <iostream>

using namespace std;

#include <algorithm>

#include <vector>

I was using an unordered hash map for storing the graph as adjacency list and I got some wrong answers. The problem was, I did not observed that in some test cases their might be some cities w...

What am i doing wrong here? Getting TLE

#include<bits/stdc++.h>
#define lli long long int
#define pb push_back

If a component contains cycle then the minimum number of separated cities after the reform of this component is zero, one otherwise. Because a component without cycle forms a tree and root of ...