# 85. Maximal Rectangle (Hard)

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

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
```

## Solutions

``````public class Solution {

// This problem is pretty complicated and similar with 84. Largest Rectangle in Histogram. But
// the former is more complicated. You need to calculate the histogram bar by yourself and
// as for problem 84, this information is given by problem.

public int maximalRectangle(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());

int area = height * width;

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

stack.push(j);
}
}

return max;
}
}
``````