Make a Square with the Same Color - Practice Coding Problems

Make a Square with the Same Color - Problem

You're given a 3×3 grid filled with black ('B') and white ('W') cells, like a mini chessboard pattern. Your challenge is to determine if you can create a 2×2 square where all four cells have the same color by changing at most one cell.

Think of it as a puzzle game where you need to find if there's already a monochrome 2×2 square, or if you can create one with just a single color flip. There are exactly four possible 2×2 squares in a 3×3 grid:

Top-left: positions (0,0), (0,1), (1,0), (1,1)
Top-right: positions (0,1), (0,2), (1,1), (1,2)
Bottom-left: positions (1,0), (1,1), (2,0), (2,1)
Bottom-right: positions (1,1), (1,2), (2,1), (2,2)

Return true if it's possible to achieve this goal, false otherwise.

Input & Output

example_1.py — Basic Example

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

› Output: true

💡 Note: We can change the cell at (0,1) from 'W' to 'B' to create a 2×2 black square in the top-left corner: positions (0,0), (0,1), (1,0), (1,1) would all be 'B'.

example_2.py — Already Valid

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

› Output: true

💡 Note: The top-right 2×2 square at positions (0,1), (0,2), (1,1), (1,2) contains ['W','B','W','W']. We can change the 'B' at (0,2) to 'W' to make all four cells white with just 1 change.

example_3.py — Impossible Case

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

› Output: false

💡 Note: This is a perfect checkerboard pattern. Every 2×2 square contains exactly 2 black and 2 white cells, so each would need 2 changes to become monochrome. Since we can change at most 1 cell, it's impossible.

Visualization

Tap to expand

Understanding the Visualization

Identify Candidates

There are exactly 4 possible 2×2 squares in a 3×3 grid

Count Colors

For each 2×2 area, count black vs white tiles

Calculate Changes

Changes needed = min(blacks, whites) to make it monochrome

Find Solution

If any area needs ≤1 change, we have a solution!

Key Takeaway

🎯 Key Insight: In a 3×3 grid, systematically check all 4 possible 2×2 squares. If any square has 3 or 4 cells of the same color, you can make it monochrome with ≤1 change!

Time & Space Complexity

Time Complexity

⏱️

O(1)

We check exactly 4 squares with 4 cells each = 16 operations, which is constant

✓ Linear Growth

Space Complexity

O(1)

Only using a few variables to count colors, no additional data structures needed

✓ Linear Space

Constraints

The grid is always exactly 3 × 3
Each cell contains only 'B' (black) or 'W' (white)
You can change at most 1 cell

Asked in

G Google 15 a Amazon 12 f Meta 8 ⊞ Microsoft 6

This problem is optimally solved by enumerating all four possible 2×2 squares in the 3×3 grid. For each square, count the black and white cells, then calculate the minimum changes needed as min(black_count, white_count). If any square needs ≤1 change, return true. The time complexity is O(1) since we always check exactly 16 cells total.

Common Approaches

Approach	Time	Space	Notes
✓ Enumerate All 2×2 Squares (Optimal)	O(1)	O(1)	Check all four possible 2×2 squares and see if any can be made monochrome with ≤1 change

Enumerate All 2×2 Squares (Optimal) — Algorithm Steps

Define the 4 possible 2×2 square positions in the 3×3 grid
For each 2×2 square, count black and white cells
Calculate changes needed: min(black_count, white_count)
If any square needs ≤1 change, return true
If all squares need >1 change, return false

Visualization

Tap to expand

Step-by-Step Walkthrough

Square 1: Top-Left

Check positions (0,0), (0,1), (1,0), (1,1)

Square 2: Top-Right

Check positions (0,1), (0,2), (1,1), (1,2)

Square 3: Bottom-Left

Check positions (1,0), (1,1), (2,0), (2,1)

Square 4: Bottom-Right

Check positions (1,1), (1,2), (2,1), (2,2)

Code -

solution.c — C

#include <stdio.h>
#include <stdbool.h>

bool canMakeSquare(char grid[3][3]) {
    // All possible 2x2 square starting positions
    int squares[4][2] = {{0,0}, {0,1}, {1,0}, {1,1}};
    
    for (int s = 0; s < 4; s++) {
        int startRow = squares[s][0];
        int startCol = squares[s][1];
        int blackCount = 0;
        int whiteCount = 0;
        
        // Count colors in this 2x2 square
        for (int i = 0; i < 2; i++) {
            for (int j = 0; j < 2; j++) {
                if (grid[startRow + i][startCol + j] == 'B') {
                    blackCount++;
                } else {
                    whiteCount++;
                }
            }
        }
        
        // Changes needed to make it all one color
        int changesNeeded = (blackCount < whiteCount) ? blackCount : whiteCount;
        
        // If we can make it monochrome with ≤1 change
        if (changesNeeded <= 1) {
            return true;
        }
    }
    
    return false;
}

int main() {
    char grid[3][3];
    for (int i = 0; i < 3; i++) {
        for (int j = 0; j < 3; j++) {
            scanf(" %c", &grid[i][j]);
        }
    }
    printf("%s\n", canMakeSquare(grid) ? "true" : "false");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(1)

We check exactly 4 squares with 4 cells each = 16 operations, which is constant

✓ Linear Growth

Space Complexity

O(1)

Only using a few variables to count colors, no additional data structures needed

✓ Linear Space

Constraints

The grid is always exactly 3 × 3
Each cell contains only 'B' (black) or 'W' (white)
You can change at most 1 cell

28.5K Views

Medium Frequency

~12 min Avg. Time

890 Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Input & Output

Visualization

Time & Space Complexity

Related Problems

Constraints

Common Approaches

Enumerate All 2×2 Squares (Optimal) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Constraints

Select Compiler