# 221. Maximal Square (Medium)

https://leetcode.com/problems/maximal-square/

Given a 2D binary matrix filled with 0's and 1's, find the largest square containing only 1's and return its area.

Example:

```Input:

1 0 1 0 0
1 0 1 1 1
1 1 1 1 1
1 0 0 1 0

Output: 4
```

## Solutions

``````class Solution {
public int maximalSquare(char[][] matrix) {
if (matrix == null || matrix.length == 0 || matrix.length == 0) {
return 0;
}

int cLen = matrix.length;    // column length
int rLen = matrix.length;       // row length

// height array
int[] heights = new int[cLen + 1];
Arrays.fill(heights, 0);

int max = 0;

for (int i = 0; i < rLen; i++) {

// used to keep increasing heights
Stack<Integer> stack = new Stack<>();

for (int j = 0; j < cLen + 1; j++) {
// The quantity of consecutive 1s in same column counting from current row to above
if (j < cLen && matrix[i][j] == '1') {
heights[j] += 1;
}

// 0 cut off the continuity of 1s in same column
if (j < cLen && matrix[i][j] == '0') {
heights[j] = 0;
}

// do not push in less taller height
if (stack.isEmpty() || heights[stack.peek()] <= heights[j]) {
stack.push(j);

continue;
}

while (!stack.isEmpty() && heights[stack.peek()] > heights[j]) {
int height = heights[stack.pop()];
int width = stack.isEmpty() ? j : (j - 1 - stack.peek());

// Square requires same height and width, so here we choose the smallest one as the edge.
int min = Math.min(height, width);

int area = min * min;

max = Math.max(max, area);
}

stack.push(j);
}
}

return max;
}
}
``````