# 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.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));
}

## Input

{{1,0,1},{0,1,0},{1,0,0}}

## Output

8