# LC 3127. Make a Square with the Same Color

### [Problem link](https://leetcode.com/problems/make-a-square-with-the-same-color/)

###### tags: `leedcode` `easy` `c++`

You are given a 2D matrix <code>grid</code> of size <code>3 x 3</code> consisting only of characters <code>'B'</code> and <code>'W'</code>. Character <code>'W'</code> represents the white color<!-- notionvc: 06a49cc0-a296-4bd2-9bfe-c8818edeb53a -->, and character <code>'B'</code> represents the black color<!-- notionvc: 06a49cc0-a296-4bd2-9bfe-c8818edeb53a -->.

Your task is to change the color of **at most one** cell<!-- notionvc: c04cb478-8dd5-49b1-80bb-727c6b1e0232 --> so that the matrix has a <code>2 x 2</code> square where all cells are of the same color.<!-- notionvc: adf957e1-fa0f-40e5-9a2e-933b95e276a7 -->

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

**Example 1:**
Input: <span class="example-io">grid = [["B","W","B"],["B","W","W"],["B","W","B"]]

Output: <span class="example-io">true

Explanation:

It can be done by changing the color of the <code>grid[0][2]</code>.

**Example 2:**
Input: <span class="example-io">grid = [["B","W","B"],["W","B","W"],["B","W","B"]]

Output: <span class="example-io">false

Explanation:

It cannot be done by changing at most one cell.

**Example 3:**
Input: <span class="example-io">grid = [["B","W","B"],["B","W","W"],["B","W","W"]]

Output: <span class="example-io">true

Explanation:

The <code>grid</code> already contains a <code>2 x 2</code> square of the same color.<!-- notionvc: 9a8b2d3d-1e73-457a-abe0-c16af51ad5c2 -->

**Constraints:**

- <code>grid.length == 3</code>
- <code>grid[i].length == 3</code>
- <code>grid[i][j]</code> is either <code>'W'</code> or <code>'B'</code>.

## Solution 1
#### C++

class Solution {
public:
    bool canMakeSquare(vector<vector<char>>& grid) {
        for (int i = 0; i < 2; i++) {
            for (int j = 0; j < 2; j++) {
                unordered_map<char, int> umap;
                umap[grid[i][j]]++;
                umap[grid[i][j + 1]]++;
                umap[grid[i + 1][j]]++;
                umap[grid[i + 1][j + 1]]++;
                
                if (umap['B'] != 2 && umap['W'] != 2) {
                    return true;
                }
            }
        }
        return false;
    }
};

>### Complexity
>n = grid.length
>| | Time Complexity | Space Complexity |
>| ----------- | --------------- | ---------------- |
>| Solution 1 | O(n^2) | O(1) |

## Note
x
[](https://)