package com.fishercoder.solutions;

import java.util.Stack;

/**
 * 84. Largest Rectangle in Histogram
 *
 * 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.

 Above is a histogram where width of each bar is 1, given height = [2,1,5,6,2,3].


 The largest rectangle is shown in the shaded area, which has area = 10 unit.

 For example,
 Given heights = [2,1,5,6,2,3],
 return 10.
 */
public class _84 {

  public static class Solution1 {

    /**
     * credit: https://leetcode.com/articles/largest-rectangle-histogram/#approach-5-using-stack-accepted
     * and https://discuss.leetcode.com/topic/7599/o-n-stack-based-java-solution
     */
    public int largestRectangleArea(int[] heights) {
      int len = heights.length;
      Stack<Integer> s = new Stack<>();
      int maxArea = 0;
      for (int i = 0; i <= len; i++) {
        int h = (i == len ? 0 : heights[i]);
        if (s.isEmpty() || h >= heights[s.peek()]) {
          s.push(i);
        } else {
          int tp = s.pop();
          maxArea = Math.max(maxArea, heights[tp] * (s.isEmpty() ? i : i - 1 - s.peek()));
          i--;
        }
      }
      return maxArea;
    }
  }

}

package com.fishercoder.solutions;

import java.util.Stack;

/**
 * 84. Largest Rectangle in Histogram
 *
 * 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.

 Above is a histogram where width of each bar is 1, given height = [2,1,5,6,2,3].
 The largest rectangle is shown in the shaded area, which has area = 10 unit.

 For example,
 Given heights = [2,1,5,6,2,3],
 return 10.
 */
public class _84 {

  public static class Solution1 {

    /**
     * credit: https://leetcode.com/articles/largest-rectangle-histogram/#approach-5-using-stack-accepted
     * and https://discuss.leetcode.com/topic/7599/o-n-stack-based-java-solution
     */
    public int largestRectangleArea(int[] heights) {
      int len = heights.length;
      Stack<Integer> s = new Stack<>();
      int maxArea = 0;
      for (int i = 0; i <= len; i++) {
        int h = (i == len ? 0 : heights[i]);
        if (s.isEmpty() || h >= heights[s.peek()]) {
          s.push(i);
        } else {
          int tp = s.pop();
          maxArea = Math.max(maxArea, heights[tp] * (s.isEmpty() ? i : i - 1 - s.peek()));
          i--;
        }
      }
      return maxArea;
    }
  }

}

refactor 84