Minimum Number of Flips to Make Binary Grid Palindromic II Python/Java solution

Most implementations iterate over the top-left quadrant of the grid and process symmetric cell groups. Python, Java, C++, and JavaScript solutions all follow the same logic: compute mirrored indices, count ones in the group, and add the minimum flips needed to make them equal.

What is the best approach for Minimum Number of Flips to Make Binary Grid Palindromic II?

The mirror-based symmetric grouping approach is the most efficient. Each cell belongs to a group of up to four symmetric positions formed by horizontal and vertical reflection. By counting how many ones exist in each group and flipping the minority, you minimize operations. The algorithm runs in O(m*n) time with O(1) extra space.

Is Minimum Number of Flips to Make Binary Grid Palindromic II asked at Google/Amazon/Meta?

Matrix symmetry and minimum flip problems frequently appear in interviews at companies like Google, Amazon, and Meta. Variations involving palindromic grids, mirrored pairs, or minimum operations on matrices are common because they test indexing logic and algorithmic reasoning.

What data structure is used in Minimum Number of Flips to Make Binary Grid Palindromic II?

The problem primarily uses a 2D array (matrix). The algorithm relies on index calculations to find mirrored positions and counts values inside small symmetric groups. No additional data structures beyond simple counters are required.

What is the time complexity of Minimum Number of Flips to Make Binary Grid Palindromic II?

The optimal solution runs in O(m*n) time because every grid cell is processed once as part of a symmetric group. Space complexity remains O(1) since the algorithm only keeps counters while scanning the matrix.

Description Hints 2 Solution Video 19:47

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

Q: Is Minimum Number of Flips to Make Binary Grid Palindromic II easy or hard?

The problem is rated Medium on LeetCode with an acceptance rate around 25%. The challenge comes from correctly identifying all symmetric groups and handling edge cases such as the middle row, middle column, and center cell when dimensions are odd.

Q: How to solve Minimum Number of Flips to Make Binary Grid Palindromic II in O(n)?

Treat symmetric cells as groups defined by reflections across the grid center. For each group of four cells, count the number of ones and flip the minority so all values match. Handle middle row and column pairs separately when dimensions are odd. Since every cell is visited once, the complexity becomes O(m*n).

Medium25.5% acceptance1 company

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.

Example 1:

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

Output: 3

Explanation:

Example 2:

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

Output: 2

Explanation:

Example 3:

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

Output: 2

Explanation:

Constraints:

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

Asked at 1 company

Google

Topics

Hints open one at a time

Hint 1

For each (x, y), find (m - 1 - x, y), (m - 1 - x, n - 1 - y), and (x, n - 1 - y); they should be the same.

Hint 2

Note that we need to specially handle the middle row (column) if the number of rows (columns) is odd.

19:47

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1class Solution:
2    def minFlips(self, grid: List[List[int]]) -> int:
3        

Starter codePython 3Spaces: 4

Examples

Case 1Case 2Case 3

Input

[[1,0,0],[0,1,0],[0,0,1]]

Input

[[0,1],[0,1],[0,0]]

Input

[[1],[1]]

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