# LeetCode – Maximal Rectangle (Java)

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

Analysis

This problem can be converted to the "Largest Rectangle in Histogram" problem.

Java 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];   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; }```
Category >> Algorithms >> Interview
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• Yinglei Zhang

This is pretty awesome. Finding connections between different problems is better than solving the problem itself.

• Simon

I have a more elegant implement about it.

• Kyle

It’s very cool that you converted a problem to another. Brilliant!!

• Kyle

It will not. Because height[] has one additional element at the end which is 0 ( int[][] height = new int[m][n + 1];) . So even if the elements are in increasing order, zero will be at the end, so non increasing. 🙂

• Guangyu Wang

Hi, I have a question about the maxAreaInHist() method: if the heights are in ascending order, the max will remain zero all the time. This is a bug right?