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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'`

.

```
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[0])
dp = [[0]*(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
```

```
class Solution {
public:
int maximalSquare(vector<vector<char>>& matrix) {
if (matrix.empty()) {
return 0;
}
int m = matrix.size(), n = matrix[0].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;
}
};
```

```
public int maximalSquare(char[][] a) {
if(a.length == 0) return 0;
int m = a.length, n = a[0].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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