How do you solve Transform to Chessboard on LeetCode?

Transform to Chessboard (LeetCode #782) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: A valid chessboard must have an equal number of 0's and 1's, or differ by one.

What is the brute force solution for Transform to Chessboard?

The brute force approach for Transform to Chessboard is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves generating all possible configurations of the board by swapping rows and columns, checking each configuration to see if it matches a valid chessboard pattern. This method is straightforward but inefficient due to the large number of possible configurations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Transform to Chessboard?

Transform to Chessboard uses the following concepts: Array, Math, Bit Manipulation, Matrix. The recommended patterns to study are: Array, Matrix.

What are the interview tips for Transform to Chessboard?

Always validate input constraints before diving into the logic. Explain your thought process clearly as you work through the problem. Consider edge cases, such as very small boards or boards that are already valid.

What are common mistakes when solving Transform to Chessboard?

Failing to check the initial counts of 0's and 1's before proceeding with swaps. Not considering both possible chessboard patterns when counting mismatches.

#782

Transform to Chessboard

Hard

ArrayMathBit ManipulationMatrixArrayMatrix

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal solution leverages the properties of a chessboard pattern and uses counting to determine how many swaps are necessary. By checking the counts of 0's and 1's and their positions, we can efficiently compute the minimum swaps required without generating all configurations.

⚙️

Algorithm

4 steps

1Step 1: Count the number of 1's in the first row and first column to determine the expected pattern.
2Step 2: Validate that the number of 1's and 0's are appropriate for a chessboard configuration.
3Step 3: Count the number of row and column mismatches against the expected pattern.
4Step 4: Return the minimum number of swaps needed based on the mismatches.

solution.py20 lines

1# Full working Python code
2
3def minSwapsToChessboard(board):
4    n = len(board)
5    rowCount = [0] * n
6    colCount = [0] * n
7    for i in range(n):
8        for j in range(n):
9            if board[i][j] == 1:
10                rowCount[i] += 1
11                colCount[j] += 1
12    if not (n // 2 <= rowCount[0] <= (n + 1) // 2) or not (n // 2 <= colCount[0] <= (n + 1) // 2):
13        return -1
14    rowSwaps = colSwaps = 0
15    for i in range(n):
16        if (board[i][0] == 1) != (i % 2):
17            rowSwaps += 1
18        if (board[0][i] == 1) != (i % 2):
19            colSwaps += 1
20    return (rowSwaps + colSwaps) // 2

ℹ

Complexity note: This complexity is achieved because we only need to iterate through the board a limited number of times, specifically to count and validate the rows and columns.

1A valid chessboard must have an equal number of 0's and 1's, or differ by one.
2The first row and column define the expected pattern for the rest of the board.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.