3240. Minimum Number of Flips to Make Binary Grid Palindromic II

You are given an m x n binary matrix grid.

A row or column is considered palindromic if its values read the same forward and backward.

You can flip any number of cells in grid from 0 to 1, or from 1 to 0.

Return the minimum number of cells that need to be flipped to make all rows and columns palindromic, and the total number of 1‘s in grid divisible by 4.

class Solution {
    int cost(vector<int> freq) {
        int ma = 0, buc[2]{};
        for(auto& f : freq) ma = max(ma, ++buc[f]);
        return freq.size() - ma;
    }
public:
    int minFlips(vector<vector<int>>& grid) {
        int n = grid.size(), m = grid[0].size(), res = 0;
        for(int i = 0; i < n / 2; i++) {
            for(int j = 0; j < m / 2; j++) {
                res += cost({grid[i][j], grid[i][m-j-1], grid[n-i-1][j], grid[n-i-1][m-j-1]});
            }
        }
        vector<vector<int>> cnt(2,vector<int>(2));
        if(n & 1) {
            for(int l = 0, r = m - 1; l < r; l++,r--) {
                int a = grid[n/2][l], b = grid[n/2][r];
                if(a > b) swap(a,b);
                cnt[a][b]++;
            }
        }
        if(m & 1) {
            for(int l = 0, r = n - 1; l < r; l++,r--) {
                int a = grid[l][m/2], b = grid[r][m/2];
                if(a > b) swap(a,b);
                cnt[a][b]++;
            }
        }
        if(cnt[1][1] % 2 == 0) {
            res += cnt[0][1];
        } else {
            res += cnt[0][1];
            if(cnt[0][1] == 0) res += 2;
        }
        if(n & 1 and m & 1 and grid[n/2][m/2]) res++;
        return res;
    }
};