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  1. Prepare
  2. Algorithms
  3. Graph Theory
  4. Jogging Cats
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Jogging Cats

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18 Discussions

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  • yashparihar729
     + 0 comments

    Here is my solution in java, javascript, python C, C++, Csharp HackerRank Jogging Cats Problem Solution

  • mineman1012221
     + 0 comments

    Here is the solution of Jogging Cats Click Here

  • thecodingsoluti2
     + 0 comments

    Here is problem solution - https://programs.programmingoneonone.com/2021/07/hackerrank-jogging-cats-problem-solution.html

  • eiduladh_a_wish1
     + 1 comment

    The scottish fold munchkin is a pretty healthy breed, but there are some hereditary health problems in the breed: Osteochondrodysplasia, a developmental abnormality that affects cartilage and bone development. Polycystic kidney disease (PKD) Cardiomyopathy, a form of heart disease.

  • konetikirankumar
     + 1 comment

    c++

    #include <iostream>
#include <cmath>
#include <cstring>
#include <queue>
#include <vector>
#include <cstdio>
#include <map>
#include <stack>
#include <set>
#include <algorithm>
#define ll long long
using namespace std;
const int Maxn = 100010 , Maxm = 11, Mo = 1e9 + 7;
const ll oo = 1ll << 60;
#define PB push_back

int T, cs = 1;
int n , m , k;
vector<int> e[Maxn];
int cnt[Maxn];
int main(){
    cin >> n >> m;
    for (int i = 1, u, v; i <= m; i++){
        cin >> u >> v;
        e[u].PB(v);
        e[v].PB(u);
    }
    ll ans = 0;
    for (int u = 1; u <= n; u++){
        vector<int> all;
        for (int i = 0; i < e[u].size(); i++){
            int v = e[u][i];
            for (int k =0; k < e[v].size(); k++){
                int t = e[v][k];
                if (t == u) continue;
                if (cnt[t] == 0) all.PB(t);
                ans += cnt[t];
                cnt[t] ++;                
            }
        }
        for (int i = 0; i < all.size(); i++) cnt[all[i]] = 0;
    }
    cout << ans / 4 << endl;
}