HackerEarth Handshakes in a party problem solution

In this HackerEarth Handshakes in a party problem solution, You are organizing a new year’s party and you have invited a lot of guests. Now, you want each of the guests to handshake with every other guest to make the party interactive. Your task is to know what will be the minimum time by which every guest meets others. One person can handshake with only one other person at once following the handshake should be of 3 seconds sharp.

You have prepared some data for each guest which is the order in which the Guest(i) will meet others. If the data is inconsistent, then print -1. Otherwise, print the minimum time in seconds.

HackerEarth Handshakes in a party problem solution.

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
const ll mod = 1e9+7;
const int maxn = 1010;

int n;
int id[maxn][maxn];
int k = 1;

vector<int> g[maxn*maxn];
int indeg[maxn*maxn];

int tot;

int main() {
    ios_base::sync_with_stdio(false); cin.tie(0);  cout.tie(0);
    cin>>n;
    tot=n*(n-1)/2;
    
    for (int i=0; i<n; i++) {
        int prev = 0;
        for (int j=1; j<n; j++) {
            int x; 
            cin>>x;
            --x;

            // assign a unique id to the G(i)-G(x) handshake
            if (!id[i][x]) {
                id[i][x]=id[x][i] = k;
                k++;
            }

            // create handshake sequence graph 
            if (prev) {
                g[prev].push_back(id[i][x]);
                indeg[id[i][x]]++;
            }
            prev = id[i][x];
        }
    }


    // add all handshakes which doesn't need prior handshake as source nodes in queue.
    queue<int> Q;
    int count = 0;
    for (int i=1; i<k; i++) {
        if (indeg[i]==0) {
            Q.push(i);
            count++;
        }
    }

    ll seconds=0;
    // multi-point BFS to traverse and visit every node in directed graph
    while(!Q.empty()) {
        int len = Q.size();
        queue<int> nQ;
        while(!Q.empty()) {
            int front_node = Q.front();
            Q.pop();
            for (int x: g[front_node]) {
                if (--indeg[x] == 0) {
                    nQ.push(x);
                    count++;
                }
            }
        }
        Q = nQ;
        seconds++;
    }

    // count denotes the number of visited nodes in the graph
    // if the visited nodes are equal to total handshakes in the party, then the data is valid
    if (count == tot) 
    cout << seconds*3 << endl;
    else
    cout << -1 << endl;
    return 0;
}

Second solution

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;

const int N = 1e3 + 14;
vector<int> g[N * N];
int id[N][N], in[N * N], d[N * N];

int main() {
    ios::sync_with_stdio(0), cin.tie(0);
    int n;
    cin >> n;
    int cnt = 0;
    for (int i = 0; i < n; ++i) {
        int p = -1;
        for (int j = 0; j < n - 1; ++j) {
            int o;
            cin >> o;
            --o;
            if (!id[i][o])
                id[i][o] = id[o][i] = cnt++;
            if (p != -1) {
                g[id[i][o]].push_back(p);
                ++in[p];
            }
            p = id[i][o];
        }
    }
    memset(d, 63, sizeof d);
    queue<int> q;
    for (int i = 0; i < cnt; ++i)
        if (!in[i]) {
            d[i] = 0;
            q.push(i);
        }
    while (!q.empty()) {
        int v = q.front();
        q.pop();
        for (auto u : g[v])
            if (!--in[u]) {
                d[u] = d[v] + 1;
                q.push(u);
            }
    }
    int ans = *max_element(d, d + cnt);
    cout << (ans > 1e9 ? -1 : ans * 3 + 3);
}

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