Find largest rectangle containing only zeros in an N×N binary matrix

Question

Given an NxN binary matrix (containing only 0\'s or 1\'s), how can we go about finding largest rectangle containing all 0\'s?

Example:

      I
    0          


        

Answer 1

Solution with space complexity O(columns) [Can be modified to O(rows) also] and time complexity O(rows*columns)

public int maximalRectangle(char[][] matrix) {
    int m = matrix.length;
    if (m == 0)
        return 0;
    int n = matrix[0].length;
    int maxArea = 0;
    int[] aux = new int[n];
    for (int i = 0; i < n; i++) {
        aux[i] = 0;
    }
    for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++) {
            aux[j] = matrix[i][j] - '0' + aux[j];
            maxArea = Math.max(maxArea, maxAreaHist(aux));
        }
    }
    return maxArea;
}

public int maxAreaHist(int[] heights) {
    int n = heights.length;
    Stack stack = new Stack();
    stack.push(0);
    int maxRect = heights[0];
    int top = 0;
    int leftSideArea = 0;
    int rightSideArea = heights[0];
    for (int i = 1; i < n; i++) {
        if (stack.isEmpty() || heights[i] >= heights[stack.peek()]) {
            stack.push(i);
        } else {
            while (!stack.isEmpty() && heights[stack.peek()] > heights[i]) {
                top = stack.pop();
                rightSideArea = heights[top] * (i - top);
                leftSideArea = 0;
                if (!stack.isEmpty()) {
                    leftSideArea = heights[top] * (top - stack.peek() - 1);
                } else {
                    leftSideArea = heights[top] * top;
                }
                maxRect = Math.max(maxRect, leftSideArea + rightSideArea);
            }
            stack.push(i);
        }
    }
    while (!stack.isEmpty()) {
        top = stack.pop();
        rightSideArea = heights[top] * (n - top);
        leftSideArea = 0;
        if (!stack.isEmpty()) {
            leftSideArea = heights[top] * (top - stack.peek() - 1);
        } else {
            leftSideArea = heights[top] * top;
        }
        maxRect = Math.max(maxRect, leftSideArea + rightSideArea);
    }
    return maxRect;
}

But I get Time Limite exceeded excpetion when I try this on LeetCode. Is there any less complex solution?