LeetCode #3127 — EASY

Make a Square with the Same Color

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

You are given a 2D matrix grid of size 3 x 3 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 x 2 square where all cells are of the same color.

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

Example 1:

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

Output: true

Explanation:

It can be done by changing the color of the grid[0][2].

Example 2:

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

Output: false

Explanation:

It cannot be done by changing at most one cell.

Example 3:

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

Output: true

Explanation:

The grid already contains a 2 x 2 square of the same color.

Constraints:

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

Step 01

Brute Force Baseline

Problem summary: You are given a 2D matrix grid of size 3 x 3 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 x 2 square where all cells are of the same color. Return true if it is possible to create a 2 x 2 square of the same color, otherwise, return false. .grid-container { display: grid; grid-template-columns: 30px 30px 30px; padding: 10px; } .grid-item { background-color: black; border: 1px solid gray; height: 30px; font-size: 30px; text-align: center; } .grid-item-white { background-color: white; }

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

[["B","W","B"],["B","W","W"],["B","W","B"]]

Example 2

[["B","W","B"],["W","B","W"],["B","W","B"]]

Example 3

[["B","W","B"],["B","W","W"],["B","W","W"]]

Step 02

Core Insight

What unlocks the optimal approach

It is impossible to create <code>2 x 2</code> square with the same color by changing the color of at most one cell when the number of <code>‘W'</code> or <code>'B’</code> in all squares is 2.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3127: Make a Square with the Same Color
class Solution {
    public boolean canMakeSquare(char[][] grid) {
        final int[] dirs = {0, 0, 1, 1, 0};
        for (int i = 0; i < 2; ++i) {
            for (int j = 0; j < 2; ++j) {
                int cnt1 = 0, cnt2 = 0;
                for (int k = 0; k < 4; ++k) {
                    int x = i + dirs[k], y = j + dirs[k + 1];
                    cnt1 += grid[x][y] == 'W' ? 1 : 0;
                    cnt2 += grid[x][y] == 'B' ? 1 : 0;
                }
                if (cnt1 != cnt2) {
                    return true;
                }
            }
        }
        return false;
    }
}

// Accepted solution for LeetCode #3127: Make a Square with the Same Color
func canMakeSquare(grid [][]byte) bool {
	dirs := [5]int{0, 0, 1, 1, 0}
	for i := 0; i < 2; i++ {
		for j := 0; j < 2; j++ {
			cnt1, cnt2 := 0, 0
			for k := 0; k < 4; k++ {
				x, y := i+dirs[k], j+dirs[k+1]
				if grid[x][y] == 'W' {
					cnt1++
				} else {
					cnt2++
				}
			}
			if cnt1 != cnt2 {
				return true
			}
		}
	}
	return false
}

# Accepted solution for LeetCode #3127: Make a Square with the Same Color
class Solution:
    def canMakeSquare(self, grid: List[List[str]]) -> bool:
        for i in range(0, 2):
            for j in range(0, 2):
                cnt1 = cnt2 = 0
                for a, b in pairwise((0, 0, 1, 1, 0)):
                    x, y = i + a, j + b
                    cnt1 += grid[x][y] == "W"
                    cnt2 += grid[x][y] == "B"
                if cnt1 != cnt2:
                    return True
        return False

// Accepted solution for LeetCode #3127: Make a Square with the Same Color
/**
 * [3127] Make a Square with the Same Color
 */
pub struct Solution {}

// submission codes start here

const DELTA: [(usize, usize); 4] = [(0, 0), (0, 1), (1, 0), (1, 1)];

impl Solution {
    pub fn can_make_square(grid: Vec<Vec<char>>) -> bool {
        for i in 0..2 {
            for j in 0..2 {
                let mut white: i32 = 0;
                let mut black: i32 = 0;

                for (x, y) in DELTA {
                    match grid[i + x][j + y] {
                        'B' => black += 1,
                        'W' => white += 1,
                        _ => {}
                    }
                }

                if white.abs_diff(black) == 2 || white.abs_diff(black) == 4 {
                    return true;
                }
            }
        }

        false
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_3127() {}
}

// Accepted solution for LeetCode #3127: Make a Square with the Same Color
function canMakeSquare(grid: string[][]): boolean {
    const dirs: number[] = [0, 0, 1, 1, 0];
    for (let i = 0; i < 2; ++i) {
        for (let j = 0; j < 2; ++j) {
            let [cnt1, cnt2] = [0, 0];
            for (let k = 0; k < 4; ++k) {
                const [x, y] = [i + dirs[k], j + dirs[k + 1]];
                if (grid[x][y] === 'W') {
                    ++cnt1;
                } else {
                    ++cnt2;
                }
            }
            if (cnt1 !== cnt2) {
                return true;
            }
        }
    }
    return false;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(1)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.