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Given n non-negative integers representing the histogram's bar height where the width of each bar is 1, find the area of largest rectangle in the histogram.

For example,

Given heights = [2,1,5,6,2,3],return 10.

Solution

We can coclude that, for any bar i, the maximum rectangle is of width r - l - 1 where r is the last coordinate of the bar to the right with height h[r] >= h[i] and l is the last coordinate of the bar to the left which height h[l] >= h[i].

Therefore, for any rectangle i, if we can locate its last coordinate to the right and to the left, then we can easily calculate the target area.

So the main problem is how to calculate these two boundary, and the trick is to efficiently iterate the arrays to find the target.

When we start from the utermost left and rigght side, we can actaully reuse the value from the last iteration and thus achieve O(n) complexity.

The code is shown as below.
Java

public class Solution {
    public int largestRectangleArea(int[] heights) {
        
    if (heights == null || heights.length == 0) {
        return 0;
    }
        
    int[] left = new int[heights.length]; 
    int[] right = new int[heights.length]; 
    
    left[0] = -1;
    right[heights.length - 1] = heights.length;

    //left side
    for (int i = 1; i < heights.length; i++) {
        int p = i - 1;
        while (p >= 0 && heights[p] >= heights[i]) {
            p = left[p];
        }
        left[i] = p;
    }
    //right side
    for (int i = heights.length - 2; i >= 0; i--) {
        int p = i + 1;
        while (p < heights.length && heights[p] >= heights[i]) {
            p = right[p];
        }
        right[i] = p;
    }

    int maxArea = 0;
    for (int i = 0; i < heights.length; i++) {
        maxArea = Math.max(maxArea, heights[i] * (right[i] - left[i] - 1));
    }
    return maxArea;
    }
}