HackerEarth Unsafe elements in a matrix problem solution

In this HackerEarth Unsafe elements in a matrix problem solution, You have a matrix S consisting of N rows and M columns. Let u be the maximum element of the matrix and v be the smallest element of the matrix. If any element whose value is equal to u or v is called unsafe elements and they disfigure the complete row and column of the matrix. More formally, if any element is equal to u or v and contains cell number (x, y), that is, S[x][y] = u or S[x][y] = v are unsafe so that they also disfigure all the cells that have row x or y column and also are unsafe.

Your task is to find the number of safe elements.

HackerEarth Unsafe elements in a matrix problem solution.

#include<bits/stdc++.h>
using namespace std;
#define FIO ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define mod 1000000007
#define test ll t; cin>>t; while(t--)
typedef long long int ll;
int main() {
    FIO;
    test{
        ll n,m,u=LONG_MAX,v=LONG_MIN,i,j;
        cin>>n>>m;
        vector<vector<ll>>inp(n,vector<ll>(m));
        for( i=0;i<n;i++){
            for( j=0;j<m;j++){
                cin>>inp[i][j];
                u=min(u,inp[i][j]);
                v=max(v,inp[i][j]);
            }
        }
        ll r=0,c=0;
        for(i=0;i<n;i++){
            if(*max_element(inp[i].begin(),inp[i].end())!=v && *min_element(inp[i].begin(),inp[i].end())!=u){
                r++;
            }
        }
        for(i=0;i<m;i++){
            ll x=LONG_MAX,y=LONG_MIN;
            for(j=0;j<n;j++){
                x=min(x,inp[j][i]);
                y=max(y,inp[j][i]);
            }
            if(x!=u && y!=v){
                c++;
            }
        }
        cout<<r*c<<endl;
    }
    return 0;
}

Second solution

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

int main(){
    ios::sync_with_stdio(0), cin.tie(0);
    int t;
    cin >> t;
    while(t--){
        int n, m;
        cin >> n >> m;
        vector<int> rmx(n, INT_MIN), cmx(m, INT_MIN), rmn(n, INT_MAX), cmn(m, INT_MAX);
        for(int i = 0; i < n; i++)
            for(int j = 0; j < m; j++){
                int x;
                cin >> x;
                rmx[i] = max(rmx[i], x);
                cmx[j] = max(cmx[j], x);
                rmn[i] = min(rmn[i], x);
                cmn[j] = min(cmn[j], x);
            }
        int mx = *max_element(rmx.begin(), rmx.end()), mn = *min_element(rmn.begin(), rmn.end());
        int r = 0, c = 0;
        for(int i = 0; i < n; i++)
            r += rmn[i] != mn && rmx[i] != mx;
        for(int i = 0; i < m; i++)
            c += cmn[i] != mn && cmx[i] != mx;
        cout << c * r << 'n';
    }
}

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