Make a Square with the Same Color

Make a Square with the Same Color - Problem

You are given a 3 × 3 matrix grid consisting only of characters 'B' and 'W'. Character 'W' represents the white color, and character 'B' represents the black color.

Your task is to change the color of at most one cell so that the matrix has a 2 × 2 square where all cells are of the same color.

Return true if it is possible to create a 2 × 2 square of the same color, otherwise return false.

Input & Output

Example 1 — Solution Exists

$ Input: grid = [["B","W","B"],["B","W","W"],["W","B","B"]]

› Output: true

💡 Note: The top-right 2×2 square is ["W","B"],["W","W"]. We can change grid[0][2] from "B" to "W" to make all cells "W".

Example 2 — Already Has Square

$ Input: grid = [["B","W","B"],["W","B","W"],["B","W","B"]]

› Output: false

💡 Note: No 2×2 square can be made uniform with at most 1 change. Each square needs 2 or more changes.

Example 3 — Already Perfect

$ Input: grid = [["B","B","W"],["B","B","W"],["W","W","W"]]

› Output: true

💡 Note: The top-left 2×2 square is already all "B": ["B","B"],["B","B"]. No changes needed.

Constraints

grid.length == 3
grid[i].length == 3
grid[i][j] is either 'W' or 'B'

Visualization

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Asked in

G Google 15 f Meta 12 a Amazon 8

The key insight is that there are only 4 possible 2×2 squares in a 3×3 grid. For each square, count how many cells need to change to make all cells the same color (try both B and W). Best approach uses early termination when a solution is found. Time: O(1), Space: O(1)

Common Approaches

✓ Two Pass

⏱️ Time: N/A Space: N/A

Brute Force Enumeration

⏱️ Time: O(1) Space: O(1)

For each of the 4 possible 2×2 squares in the 3×3 grid, count how many cells would need to be changed to make all 4 cells the same color. If any square needs 0 or 1 changes, return true.

Early Termination Check

⏱️ Time: O(1) Space: O(1)

Same approach as brute force but with early termination. As soon as we find any 2×2 square that needs at most 1 change, we immediately return true without checking remaining squares.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

Time & Space Complexity

Time Complexity

⏱️

✓ Linear Growth

Space Complexity

✓ Linear Space

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Medium Frequency

~10 min Avg. Time

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Smart Actions

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