Flipped Matrix Prequel in C++

Suppose we have one binary matrix. We have to find the maximum number of 1s we can get if we flip a row and then flip a column.

So, if the input is like

1	0	1
0	1	0
1	0	0

then the output will be 8

To solve this, we will follow these steps −

n := size of rows in matrix
m := size of columns in matrix
ret := 0
Define an array row of size n
Define an array col of size n
total := 0
for initialize i := 0, when i < n, update (increase i by 1), do −
- for initialize j := 0, when j < m, update (increase j by 1), do −
  - row[i] := row[i] + matrix[i, j]
  - col[j] := col[j] + matrix[i, j]
  - total := total + matrix[i, j]
for initialize i := 0, when i < n, update (increase i by 1), do −
- for initialize j := 0, when j < m, update (increase j by 1), do −
  - cand := total - row[i] - col[j] + ((m - row[i]) + (n - col[j]))
  - if matrix[i, j] is non-zero, then −
    - cand := cand + 2
  - Otherwise
    - cand := cand - 2
  - ret := maximum of ret and cand
return ret

Let us see the following implementation to get better understanding −

Example

Live Demo

#include <bits/stdc++.h>
using namespace std;
class Solution {
   public:
   int solve(vector<vector<int>> &matrix) {
      int n = matrix.size();
      int m = matrix[0].size();
      int ret = 0;
      vector<int> row(n);
      vector<int> col(m);
      int total = 0;
      for (int i = 0; i < n; i++) {
         for (int j = 0; j < m; j++) {
            row[i] += matrix[i][j];
            col[j] += matrix[i][j];
            total += matrix[i][j];
         }
      }
      for (int i = 0; i < n; i++) {
         for (int j = 0; j < m; j++) {
            int cand = total - row[i] - col[j] + (m - row[i]) + (n -
            col[j]);
            if (matrix[i][j]) {
               cand += 2;
            }else {
               cand -= 2;
            }
            ret = max(ret, cand);
         }
      }
      return ret;
   }
};
main() {
   Solution ob;
   vector<vector<int>> v = {{1,0,1},{0,1,0},{1,0,0}};
   cout << (ob.solve(v));
}