85. 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.

Analysis

link:

https://leetcode.com/problems/maximal-rectangle/discuss/29055/My-java-solution-based-on-Maximum-Rectangle-in-Histogram-with-explanation

https://piazza.com/class/j0eqhhdregb3i?cid=2138

Solution from Forum

public int maximalRectangle(char[][] matrix) {
    if(matrix == null || matrix.length == 0 || matrix[0].length == 0) return 0;

    int[] height = new int[matrix[0].length];
    for(int i = 0; i < matrix[0].length; i ++){
        if(matrix[0][i] == '1') height[i] = 1;
    }
    int result = largestInLine(height);
    for(int i = 1; i < matrix.length; i ++){
        resetHeight(matrix, height, i);
        result = Math.max(result, largestInLine(height));
    }

    return result;
}

private void resetHeight(char[][] matrix, int[] height, int idx){
    for(int i = 0; i < matrix[0].length; i ++){
        if(matrix[idx][i] == '1') height[i] += 1;
        else height[i] = 0;
    }
}    

public int largestInLine(int[] height) {
    if(height == null || height.length == 0) return 0;
    int len = height.length;
    Stack<Integer> s = new Stack<Integer>();
    int maxArea = 0;
    for(int i = 0; i <= len; i++){
        int h = (i == len ? 0 : height[i]);
        if(s.isEmpty() || h >= height[s.peek()]){
            s.push(i);
        }else{
            int tp = s.pop();
            maxArea = Math.max(maxArea, height[tp] * (s.isEmpty() ? i : i - 1 - s.peek()));
            i--;
        }
    }
    return maxArea;
}