LeetCode 782: Transform to Chessboard Solution

Master LeetCode problem 782 (Transform to Chessboard), a hard 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.

782. Transform to Chessboard

Hard

Array Math Bit Manipulation Matrix

Problem Explanation

Explanation:

To transform the board into a chessboard, we need to determine two key things:

If the given board can be transformed into a valid chessboard.
Calculate the minimum number of swaps required to transform it into a chessboard.

Algorithm:

Check if the given board can be transformed into a valid chessboard:
- For a board to be a valid chessboard, it must have exactly 2 different types of rows and exactly 2 different types of columns.
- The number of rows and columns with the same patterns should be either n/2 or (n+1)/2.
- If these conditions are not met, the board cannot be transformed into a valid chessboard.
Calculate the minimum number of swaps required:
- To calculate the minimum number of swaps, we need to determine the number of swaps required for rows and columns separately.
- For each row and column, we calculate the number of swaps needed to convert it into the correct pattern.
- The minimum swaps for rows and columns will give us the overall minimum swaps needed.

Time Complexity: O(n^2) Space Complexity: O(n)

Solution Code

class Solution {
    public int movesToChessboard(int[][] board) {
        int n = board.length;
        int rowSum = 0, colSum = 0, rowSwap = 0, colSwap = 0;
        
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if ((board[0][0] ^ board[i][0] ^ board[0][j] ^ board[i][j]) == 1)
                    return -1;
            }
        }
        
        for (int i = 0; i < n; i++) {
            rowSum += board[0][i];
            colSum += board[i][0];
            if (board[i][0] == i % 2) rowSwap++;
            if (board[0][i] == i % 2) colSwap++;
        }
        
        if (rowSum != n / 2 && rowSum != (n + 1) / 2) return -1;
        if (colSum != n / 2 && colSum != (n + 1) / 2) return -1;
        if (n % 2 == 1) {
            if (rowSwap % 2 == 1) rowSwap = n - rowSwap;
            if (colSwap % 2 == 1) colSwap = n - colSwap;
        } else {
            rowSwap = Math.min(rowSwap, n - rowSwap);
            colSwap = Math.min(colSwap, n - colSwap);
        }
        
        return (rowSwap + colSwap) / 2;
    }
}

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