Transform To Chessboard - Leetcode Solution

The Transform To Chessboard problem on LeetCode asks us to find the minimum number of moves required to transform a given n x n binary matrix (containing 0's and 1's) into a chessboard pattern, where each row and column has an equal number of 0's and 1's.

To solve this problem, we can follow the below steps:

Step 1: Count the number of 1's and 0's in the matrix.

Before transforming the matrix, we need to check whether it is possible to convert it into a chessboard pattern or not. To do this, we need to count the number of 1's and 0's in the matrix. If the difference between the count of 1's and 0's is more than 1 (i.e., the difference is greater than 1 if n is even, or greater than 2 if n is odd), then it is not possible to form a chessboard pattern.

Step 2: Check row and column constraints.

In a chessboard pattern, each row and column should have an equal number of 0's and 1's. Thus, we can check the row and column constraints to see if the matrix can be transformed into a chessboard pattern. To do this, we can count the number of contiguous blocks of 0's and 1's in each row and column. If there are more than two blocks of 0's or 1's, then the matrix cannot be transformed into a chessboard pattern.

Step 3: Find the minimum number of moves to transform the matrix.

If the matrix satisfies the above two conditions, it is possible to transform it into a chessboard pattern by swapping the rows and columns. We can find the minimum number of moves required to transform the matrix by checking the two possible chessboard patterns that can be formed.

In one pattern, the first element of the first row and the first element of the first column can be either 0 or 1. In the other pattern, these elements should be the opposite of the first pattern. We can calculate the number of moves required to transform the matrix into each pattern, and return the minimum number of moves.

The time complexity of this solution is O(n^2), as we need to traverse the entire matrix.

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