Competitive_Programming/LeetCode/1583.Count_Unhappy_Friends.cpp at master · Pankajcoder1/Competitive_Programming · GitHub

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/*
    written by Pankaj Kumar.
    country:-INDIA
    Institute: National Institute of Technology, Uttarakhand
*/
typedef long long ll ;
typedef unsigned long long ull;
typedef vector<int> vl;
typedef vector<vector<int>> vvl;
#define speed cin.tie(0);cout.tie(0);ios_base::sync_with_stdio(0);
/*  Abbrevations  */
#define ff first
#define ss second
#define mp make_pair
#define pb push_back
// loops
#define forin(arr,n) for(ll i=0;i<n;i++) cin>>arr[i];
// Some print
#define no cout<<"NO"<<endl;
#define yes cout<<"YES"<<endl;
// sort
#define all(V) (V).begin(),(V).end()
#define srt(V) sort(all(V))
#define srtGreat(V) sort(all(V),greater<ll>())
// some extra
#define printv(v) for(ll i=0;i<ll(v.size());i++){cout<<v[i]<<" ";} cout<<endl;
#define precision(x) cout<<fixed<<setprecision(x);
#define sz(V) ll(V.size())


/* ascii value
A=65,Z=90,a=97,z=122
*/

/* Some syntax
//Syntax to create a min heap for priority queue
//priority_queue <int, vector<int>, greater<int>>pq;
*/
/*  --------------------MAIN PROGRAM----------------------------*/
// to run ctrl+b
const ll INF=1e18;
const ll mod1=1e9+7;
const ll mod2=998244353;
// Techniques :
// divide into cases, brute force, pattern finding
// sort, greedy, binary search, two pointer
// transform into graph

// Experience :
// Cp is nothing but only observation and mathematics.


//Add main code here


class Solution {
public:
    int unhappyFriends(int n, vector<vector<int>>& preferences, vector<vector<int>>& pairs) {
        vector<vector<int>> rank(n,vector<int>(n,0));
        unordered_set<int> us;
        for(int i=0;i<n;i++){
            for(int j=0;j<n-1;j++){
                rank[i][preferences[i][j]]=n-1-j;
            }
        }
        for(int i=1;i<n/2;i++){
            for(int j=0;j<i;j++){
                int x=pairs[i][0],y=pairs[i][1];
                int u=pairs[j][0], v = pairs[j][1];
                if(rank[x][u]>rank[x][y] && rank[u][x]>rank[u][v]){
                    us.insert(x);
                    us.insert(u);
                }
                if(rank[y][v]>rank[y][x] && rank[v][y]>rank[v][u]){
                    us.insert(y),us.insert(v);
                }
                if(rank[y][u]>rank[y][x] && rank[u][y]>rank[u][v]){
                    us.insert(u),us.insert(y);
                }
                if(rank[x][v]>rank[x][y] && rank[v][x]>rank[v][u]){
                    us.insert(v),us.insert(x);
                }
            }
        }

        return sz(us);
    }
};

/* -----------------END OF PROGRAM --------------------*/
/*
* stuff you should look before submission
* constraint and time limit
* int overflow
* special test case (n=0||n=1||n=2)
* don't get stuck on one approach if you get wrong answer
*/