Given a matrix N*N, and an integer K. The matrix initially contains 0, for each i in [1,k] you are given two integers in vectors of size 2 each. These are row and columns respectively. in each time you can update that cells with that row and col to 1. And for each k you have to find the number of cells having 0; IN O (1).

Count Number of Free Cell

map<long long int,long long int>mr,mc;
      long long int cntc = 0;
      long long int cntr = 0;
      long long int i = 0;
            long long int val = (long long int)(n*n);
           
      vector<long long int>ans;
        while(k--){
            long long r = arr[i][0];
            long long c = arr[i][1];
            ++i;
            cout<<val<<endl;
            if(mr.find(r)==mr.end()){
                val -= n;
                val += cntc;
                ++cntr;
                mr[r] = 1;
            }
            if(mc.find(c)==mc.end()){
                val -= n;
                val += cntr;
                ++cntc;
                mc[c] = 1;
            }
            ans.push_back(val);
        }
        return ans;