# Question

Given a 2D binary matrix filled with 0’s and 1’s, find the largest rectangle 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: 6
```

# Solution

Reuse the \(O(n)\) solution from Largest Rectangle in Histogram. Maintain an array of size of width of the array that stores the height of 1’s at each index at row `i`

. Therefore, we can run the stack solution to find the largest rectangle at row `i`

. Then for next row, if an index `j`

is ‘1’, then we increment the height of the 1s; otherwise we reset the height to 0.

The time complexity if \(O(mn)\).

```
class Solution {
public int maximalRectangle(char[][] matrix) {
if(matrix.length == 0 || matrix[0].length == 0)
return 0;
int x = matrix.length, y = matrix[0].length;
int maxArea = 0;
int[] h = new int[y];
Arrays.fill(h, 0);
Stack<Integer> stack = new Stack<>();
stack.push(-1);
for(int i = 0; i < x; i++) {
for(int j = 0; j < y; j++) {
if(matrix[i][j] == '1')
h[j]++;
else
h[j] = 0;
while(stack.peek() != -1 && h[stack.peek()] > h[j]) {
maxArea = Math.max(maxArea, h[stack.pop()] * (j - stack.peek() - 1));
}
stack.push(j);
}
while(stack.peek() != -1) {
maxArea = Math.max(maxArea, h[stack.pop()] * (y - stack.peek() - 1));
}
}
return maxArea;
}
}
```