# 85 Maximal Rectangle

### Problem

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

For example, given the following matrix:

``````1 0 1 0 0
1 0 1 1 1
1 1 1 1 1
1 0 0 1 0
``````

Return 6.

### Solutions

``````class Solution {
public int maximalRectangle(char[][] matrix) {
int m = matrix.length;
int n = m == 0 ? 0 : matrix[0].length;
int[][] height = new int[m][n + 1];//last element needs to be 0

int maxArea = 0;
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (matrix[i][j] == '0') {
height[i][j] = 0;
} else {
height[i][j] = i == 0 ? 1 : height[i - 1][j] + 1;
}
}
}

for (int i = 0; i < m; i++) {
int area = maxAreaInHist(height[i]);
if (area > maxArea) {
maxArea = area;
}
}

return maxArea;
}
private int maxAreaInHist(int[] height) {
Stack<Integer> stack = new Stack<Integer>();

int i = 0;
int max = 0;

while (i < height.length) {
if (stack.isEmpty() || height[stack.peek()] <= height[i]) {
stack.push(i++);
} else {
int t = stack.pop();
max = Math.max(max, height[t]
* (stack.isEmpty() ? i : i - stack.peek() - 1));
}
}

return max;
}
}
``````