LeetCode 3239: Minimum Number of Flips to Make Binary Grid Palindromic I Solution

Master LeetCode problem 3239 (Minimum Number of Flips to Make Binary Grid Palindromic I), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

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

Medium

Array Two Pointers Matrix

Problem Explanation

Explanation

To solve this problem, we can iterate through all possible rows and columns to determine the minimum number of flips needed to make each row or column palindromic. We can create helper functions to check if a row or column is palindromic and another function to calculate the number of flips needed for a given row or column. By iterating through all rows and columns and keeping track of the minimum flips needed, we can find the overall minimum number of flips required.

We can check if a row or column is palindromic by comparing the values from the start and end simultaneously until they meet at the middle.
To calculate the number of flips needed for a row or column, we can count the number of differences between the current value and the corresponding value from the other end.
By iterating through all rows and columns, we can find the minimum number of flips needed.

Time Complexity: The time complexity of this approach is O(m * n) where m is the number of rows and n is the number of columns in the grid.

Space Complexity: The space complexity is O(1) since we are using a constant amount of extra space.

Solution Code

class Solution {
    public int minFlips(int[][] grid) {
        int m = grid.length;
        int n = grid[0].length;
        
        int minFlips = Integer.MAX_VALUE;
        
        for (int i = 0; i < m; i++) {
            minFlips = Math.min(minFlips, getRowFlips(grid, i));
        }
        
        for (int j = 0; j < n; j++) {
            minFlips = Math.min(minFlips, getColFlips(grid, j));
        }
        
        return minFlips == Integer.MAX_VALUE ? -1 : minFlips;
    }
    
    private int getRowFlips(int[][] grid, int row) {
        int flips = 0;
        int left = 0, right = grid[row].length - 1;
        
        while (left < right) {
            if (grid[row][left] != grid[row][right]) {
                flips++;
            }
            left++;
            right--;
        }
        
        return flips;
    }
    
    private int getColFlips(int[][] grid, int col) {
        int flips = 0;
        int top = 0, bottom = grid.length - 1;
        
        while (top < bottom) {
            if (grid[top][col] != grid[bottom][col]) {
                flips++;
            }
            top++;
            bottom--;
        }
        
        return flips;
    }
}

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