Largest Rectangle in Histogram [ LC# 84]

Given an array of integers heights representing the histogram’s bar height where the width of each bar is 1, return the area of the largest rectangle in the histogram.

Brute Force

• Enumerate every range. Compute the Area
• $T(n)=O(n^2)$; $S(n)=O(n^2)$

area[i][j] = min_heights[i:j] * ( j-i+1 )
min_heights[i:j] = min(min_heights[i-1:j-1], heights[i], heights[j])
min_heights[i:i] = heights[i]

Intuition: Monotonic Stack

• If we are at a rectangle of height $h$ and we are interested in finding the largest rectangle with minimum height $h$ in the histogram. We can extend the rectangle to both the sides till we encounter a height that is smaller.
• If we trasverse the histogram from left to right and maintain a non-decreasing monotonic stack, it will have all these states that we need.

           ┌─┐         ┌─┐    
       ┌─┐ │ │       ┌─┤ │    
   ┌─┐ │ ├─┤ │     ┌─┤ │ │    
   │ ├─┤ │ │ │   ┌─┤ │ │ │    
   │ │ │ │ │ │...│h│ │ │ ├─┐   
 ┌─┤ │ │ │ │ │   │ │ │ │ │ │   
 └─┴─┴─┴─┴─┴─┘   └─┴─┴─┴─┴─┘   
   └─────────────────────┘                
┌─┴──┐          ┌─┴─┐  ┌──┴──┐
 left            mid    right

Code

def largest_rectangle_in_historgam(heights: List[int]) -> int:
    stack = [-1] # monotonic stack : non decreasing order
    max_area = 0
    
    heights.append(0)
    for i in range(len(heights)):
        while stack and heights[stack[-1]] > heights[i]:
            mid = stack.pop()
            left = stack[-1] + 1
            right = i
            max_area = max(max_area,  heights[mid] * (right - left))
        stack.append(i)
    return max_area

Time complexity

• $T(n)=O(n)$; $S(n)=O(n)$