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Leetcode 1812. Determine Color of a Chessboard Square

Problem Statement

You are given a coordinate coordinates in the form of a string that represents the position of a square on a standard 8x8 chessboard. The string will consist of two characters:

• The first character will be a lowercase English letter representing the column (a through h).
• The second character will be a digit representing the row (1 through 8).

You need to return a boolean value indicating whether the square is white or not. A white square on a standard chessboard alternates with black squares: starting from the bottom-left corner (a1), which is a black square.

Clarifying Questions

1. Q: Can input coordinates be invalid, or should we assume it is always valid as per the problem constraints?
   • A: You can assume the input coordinates will always be valid as per the problem constraints.
2. Q: Should the result be returned as true for white and false for black?
   • A: Yes, true should be returned for a white square and false for a black square.

Strategy

The chessboard’s alternating color pattern can be visualized with this insight:

• The column represented by letters a to h can be converted to a numerical index starting from 1 to 8.
• Combine the numeric values of the row and column indices, which represents the sum of the row and column indices.
• If the sum is even, the square is black; if it’s odd, the square is white.

Here’s how we can implement this strategy:

1. Convert the column letter to its corresponding numerical index.
2. Convert the row character to an integer.
3. Determine the color based on the parity (odd/even) of the sum of the indices.

Code

public class Solution {
    public boolean squareIsWhite(String coordinates) {
        // Convert the column character to a number (a -> 1, b -> 2, ..., h -> 8)
        int column = coordinates.charAt(0) - 'a' + 1;
        // Convert the row character to a number (1 -> 1, 2 -> 2, ..., 8 -> 8)
        int row = coordinates.charAt(1) - '0';

        // Return true if the sum of column + row is odd (white), otherwise false (black)
        return (column + row) % 2 != 0;
    }
}

Time Complexity

The solution has a constant time complexity since the operations performed (character conversion, addition, and modulo) all take constant time:

• Time Complexity: O(1)
• Space Complexity: O(1)

This solution efficiently determines the color of the chessboard square with minimal computational resources.

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