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; } |

Excellent !

maxAreaInHist fails for H = {1,2,1,3,3,2,1}. After the while loop, if stack is not empty, you should still calculate the area for remaining elements. Your code outputs 6, whereas the actual max area is 7

[n+1] is genius

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

I have a more elegant implement about it.

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

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. ๐

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?