# 3239. Minimum Number of Flips to Make Binary Grid Palindromic I

## Description

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 either all rows palindromic or all columns palindromic.

Example 1:

Input: grid = [[1,0,0],[0,0,0],[0,0,1]]

Output: 2

Explanation:

Flipping the highlighted cells makes all the rows palindromic.

Example 2:

Input: grid = [[0,1],[0,1],[0,0]]

Output: 1

Explanation:

Flipping the highlighted cell makes all the columns palindromic.

Example 3:

Input: grid = [[1],[0]]

Output: 0

Explanation:

All rows are already palindromic.

Constraints:

• m == grid.length
• n == grid[i].length
• 1 <= m * n <= 2 * 105
• 0 <= grid[i][j] <= 1

## Solutions

### Solution 1: Counting

We separately count the number of flips for rows and columns, denoted as $\textit{cnt1}$ and $\textit{cnt2}$, respectively. Finally, we take the minimum of the two.

The time complexity is $O(m \times n)$, where $m$ and $n$ are the number of rows and columns of the matrix $\textit{grid}$, respectively.

• class Solution {
public int minFlips(int[][] grid) {
int m = grid.length, n = grid[0].length;
int cnt1 = 0, cnt2 = 0;
for (var row : grid) {
for (int j = 0; j < n / 2; ++j) {
if (row[j] != row[n - j - 1]) {
++cnt1;
}
}
}
for (int j = 0; j < n; ++j) {
for (int i = 0; i < m / 2; ++i) {
if (grid[i][j] != grid[m - i - 1][j]) {
++cnt2;
}
}
}
return Math.min(cnt1, cnt2);
}
}

• class Solution {
public:
int minFlips(vector<vector<int>>& grid) {
int m = grid.size(), n = grid[0].size();
int cnt1 = 0, cnt2 = 0;
for (const auto& row : grid) {
for (int j = 0; j < n / 2; ++j) {
if (row[j] != row[n - j - 1]) {
++cnt1;
}
}
}
for (int j = 0; j < n; ++j) {
for (int i = 0; i < m / 2; ++i) {
if (grid[i][j] != grid[m - i - 1][j]) {
++cnt2;
}
}
}
return min(cnt1, cnt2);
}
};

• class Solution:
def minFlips(self, grid: List[List[int]]) -> int:
m, n = len(grid), len(grid[0])
cnt1 = cnt2 = 0
for row in grid:
for j in range(n // 2):
if row[j] != row[n - j - 1]:
cnt1 += 1
for j in range(n):
for i in range(m // 2):
if grid[i][j] != grid[m - i - 1][j]:
cnt2 += 1
return min(cnt1, cnt2)

• func minFlips(grid [][]int) int {
m, n := len(grid), len(grid[0])
cnt1, cnt2 := 0, 0
for _, row := range grid {
for j := 0; j < n/2; j++ {
if row[j] != row[n-j-1] {
cnt1++
}
}
}
for j := 0; j < n; j++ {
for i := 0; i < m/2; i++ {
if grid[i][j] != grid[m-i-1][j] {
cnt2++
}
}
}
return min(cnt1, cnt2)
}

• function minFlips(grid: number[][]): number {
const [m, n] = [grid.length, grid[0].length];
let [cnt1, cnt2] = [0, 0];
for (const row of grid) {
for (let j = 0; j < n / 2; ++j) {
if (row[j] !== row[n - 1 - j]) {
++cnt1;
}
}
}
for (let j = 0; j < n; ++j) {
for (let i = 0; i < m / 2; ++i) {
if (grid[i][j] !== grid[m - 1 - i][j]) {
++cnt2;
}
}
}
return Math.min(cnt1, cnt2);
}