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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);
}

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