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# Maximal Square LeetCode Solution

## Problem – Maximal Square LeetCode Solution

Given an `m x n` binary `matrix` filled with `0`‘s and `1`‘s, find the largest square containing only `1`‘s and return its area.

Example 1: ``````Input: matrix = [["1","0","1","0","0"],["1","0","1","1","1"],["1","1","1","1","1"],["1","0","0","1","0"]]
Output: 4
``````

Example 2: ``````Input: matrix = [["0","1"],["1","0"]]
Output: 1
``````

Example 3:

```Input: matrix = [["0"]]
Output: 0
```

Constraints:

• `m == matrix.length`
• `n == matrix[i].length`
• `1 <= m, n <= 300`
• `matrix[i][j]` is `'0'` or `'1'`.

## Maximal Square LeetCode Solution in Python

``````class Solution:
def maximalSquare(self, matrix: List[List[str]]) -> int:
if matrix is None or len(matrix) < 1:
return 0

rows = len(matrix)
cols = len(matrix)

dp = [*(cols+1) for _ in range(rows+1)]
max_side = 0

for r in range(rows):
for c in range(cols):
if matrix[r][c] == '1':
dp[r+1][c+1] = min(dp[r][c], dp[r+1][c], dp[r][c+1]) + 1 # Be careful of the indexing since dp grid has additional row and column
max_side = max(max_side, dp[r+1][c+1])

return max_side * max_side
``````

## Maximal Square LeetCode Solution in C++

``````class Solution {
public:
int maximalSquare(vector<vector<char>>& matrix) {
if (matrix.empty()) {
return 0;
}
int m = matrix.size(), n = matrix.size(), sz = 0, pre;
vector<int> cur(n, 0);
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
int temp = cur[j];
if (!i || !j || matrix[i][j] == '0') {
cur[j] = matrix[i][j] - '0';
} else {
cur[j] = min(pre, min(cur[j], cur[j - 1])) + 1;
}
sz = max(cur[j], sz);
pre = temp;
}
}
return sz * sz;
}
};
``````

## Maximal Square LeetCode Solution in Java

``````public int maximalSquare(char[][] a) {
if(a.length == 0) return 0;
int m = a.length, n = a.length, result = 0;
int[][] b = new int[m+1][n+1];
for (int i = 1 ; i <= m; i++) {
for (int j = 1; j <= n; j++) {
if(a[i-1][j-1] == '1') {
b[i][j] = Math.min(Math.min(b[i][j-1] , b[i-1][j-1]), b[i-1][j]) + 1;
result = Math.max(b[i][j], result); // update result
}
}
}
return result*result;
}
``````
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