Largest Rectangular Area in a Histogram

Largest Rectangular Area in a Histogram | Set 2

Difficulty Level : Hard
Last Updated : 04 Jun, 2021

Find the largest rectangular area possible in a given histogram where the largest rectangle can be made of a number of contiguous bars. For simplicity, assume that all bars have same width and the width is 1 unit.
For example, consider the following histogram with 7 bars of heights {6, 2, 5, 4, 5, 1, 6}. The largest possible rectangle possible is 12 (see the below figure, the max area rectangle is highlighted in red)

We have discussed a Divide and Conquer based O(nLogn) solution for this problem. In this post, O(n) time solution is discussed. Like the previous post, width of all bars is assumed to be 1 for simplicity. For every bar ‘x’, we calculate the area with ‘x’ as the smallest bar in the rectangle. If we calculate such area for every bar ‘x’ and find the maximum of all areas, our task is done. How to calculate area with ‘x’ as smallest bar? We need to know index of the first smaller (smaller than ‘x’) bar on left of ‘x’ and index of first smaller bar on right of ‘x’. Let us call these indexes as ‘left index’ and ‘right index’ respectively.
We traverse all bars from left to right, maintain a stack of bars. Every bar is pushed to stack once. A bar is popped from stack when a bar of smaller height is seen. When a bar is popped, we calculate the area with the popped bar as smallest bar. How do we get left and right indexes of the popped bar – the current index tells us the ‘right index’ and index of previous item in stack is the ‘left index’. Following is the complete algorithm.
1) Create an empty stack.
2) Start from first bar, and do following for every bar ‘hist[i]’ where ‘i’ varies from 0 to n-1.
……a) If stack is empty or hist[i] is higher than the bar at top of stack, then push ‘i’ to stack.
……b) If this bar is smaller than the top of stack, then keep removing the top of stack while top of the stack is greater. Let the removed bar be hist[tp]. Calculate area of rectangle with hist[tp] as smallest bar. For hist[tp], the ‘left index’ is previous (previous to tp) item in stack and ‘right index’ is ‘i’ (current index).
3) If the stack is not empty, then one by one remove all bars from stack and do step 2.b for every removed bar.
Following is implementation of the above algorithm.

C++

// C++ program to find maximum rectangular area in
// linear time
#include<bits/stdc++.h>
using namespace std;
 
// The main function to find the maximum rectangular
// area under given histogram with n bars
int getMaxArea(int hist[], int n)
{
    // Create an empty stack. The stack holds indexes
    // of hist[] array. The bars stored in stack are
    // always in increasing order of their heights.
    stack<int> s;
 
    int max_area = 0; // Initialize max area
    int tp;  // To store top of stack
    int area_with_top; // To store area with top bar
                       // as the smallest bar
 
    // Run through all bars of given histogram
    int i = 0;
    while (i < n)
    {
        // If this bar is higher than the bar on top
        // stack, push it to stack
        if (s.empty() || hist[s.top()] <= hist[i])
            s.push(i++);
 
        // If this bar is lower than top of stack,
        // then calculate area of rectangle with stack
        // top as the smallest (or minimum height) bar.
        // 'i' is 'right index' for the top and element
        // before top in stack is 'left index'
        else
        {
            tp = s.top();  // store the top index
            s.pop();  // pop the top
 
            // Calculate the area with hist[tp] stack
            // as smallest bar
            area_with_top = hist[tp] * (s.empty() ? i :
                                   i - s.top() - 1);
 
            // update max area, if needed
            if (max_area < area_with_top)
                max_area = area_with_top;
        }
    }
 
    // Now pop the remaining bars from stack and calculate
    // area with every popped bar as the smallest bar
    while (s.empty() == false)
    {
        tp = s.top();
        s.pop();
        area_with_top = hist[tp] * (s.empty() ? i :
                                i - s.top() - 1);
 
        if (max_area < area_with_top)
            max_area = area_with_top;
    }
 
    return max_area;
}
 
// Driver program to test above function
int main()
{
    int hist[] = {6, 2, 5, 4, 5, 1, 6};
    int n = sizeof(hist)/sizeof(hist[0]);
    cout << "Maximum area is " << getMaxArea(hist, n);
    return 0;
}

Java

//Java program to find maximum rectangular area in linear time
 
import java.util.Stack;
 
public class RectArea
{
    // The main function to find the maximum rectangular area under given
    // histogram with n bars
    static int getMaxArea(int hist[], int n)
    {
        // Create an empty stack. The stack holds indexes of hist[] array
        // The bars stored in stack are always in increasing order of their
        // heights.
        Stack<Integer> s = new Stack<>();
         
        int max_area = 0; // Initialize max area
        int tp;  // To store top of stack
        int area_with_top; // To store area with top bar as the smallest bar
      
        // Run through all bars of given histogram
        int i = 0;
        while (i < n)
        {
            // If this bar is higher than the bar on top stack, push it to stack
            if (s.empty() || hist[s.peek()] <= hist[i])
                s.push(i++);
      
            // If this bar is lower than top of stack, then calculate area of rectangle
            // with stack top as the smallest (or minimum height) bar. 'i' is
            // 'right index' for the top and element before top in stack is 'left index'
            else
            {
                tp = s.peek();  // store the top index
                s.pop();  // pop the top
      
                // Calculate the area with hist[tp] stack as smallest bar
                area_with_top = hist[tp] * (s.empty() ? i : i - s.peek() - 1);
      
                // update max area, if needed
                if (max_area < area_with_top)
                    max_area = area_with_top;
            }
        }
      
        // Now pop the remaining bars from stack and calculate area with every
        // popped bar as the smallest bar
        while (s.empty() == false)
        {
            tp = s.peek();
            s.pop();
            area_with_top = hist[tp] * (s.empty() ? i : i - s.peek() - 1);
      
            if (max_area < area_with_top)
                max_area = area_with_top;
        }
      
        return max_area;
 
    }
     
    // Driver program to test above function
    public static void main(String[] args)
    {
        int hist[] = { 6, 2, 5, 4, 5, 1, 6 };
        System.out.println("Maximum area is " + getMaxArea(hist, hist.length));
    }
}
//This code is Contributed by Sumit Ghosh

Python3

# Python3 program to find maximum
# rectangular area in linear time
 
def max_area_histogram(histogram):
     
    # This function calulates maximum
    # rectangular area under given
    # histogram with n bars
 
    # Create an empty stack. The stack
    # holds indexes of histogram[] list.
    # The bars stored in the stack are
    # always in increasing order of
    # their heights.
    stack = list()
 
    max_area = 0 # Initialize max area
 
    # Run through all bars of
    # given histogram
    index = 0
    while index < len(histogram):
         
        # If this bar is higher
        # than the bar on top
        # stack, push it to stack
 
        if (not stack) or (histogram[stack[-1]] <= histogram[index]):
            stack.append(index)
            index += 1
 
        # If this bar is lower than top of stack,
        # then calculate area of rectangle with
        # stack top as the smallest (or minimum
        # height) bar.'i' is 'right index' for
        # the top and element before top in stack
        # is 'left index'
        else:
            # pop the top
            top_of_stack = stack.pop()
 
            # Calculate the area with
            # histogram[top_of_stack] stack
            # as smallest bar
            area = (histogram[top_of_stack] *
                   ((index - stack[-1] - 1)
                   if stack else index))
 
            # update max area, if needed
            max_area = max(max_area, area)
 
    # Now pop the remaining bars from
    # stack and calculate area with
    # every popped bar as the smallest bar
    while stack:
         
        # pop the top
        top_of_stack = stack.pop()
 
        # Calculate the area with
        # histogram[top_of_stack]
        # stack as smallest bar
        area = (histogram[top_of_stack] *
              ((index - stack[-1] - 1)
                if stack else index))
 
        # update max area, if needed
        max_area = max(max_area, area)
 
    # Return maximum area under
    # the given histogram
    return max_area
 
# Driver Code
hist = [6, 2, 5, 4, 5, 1, 6]
print("Maximum area is",
       max_area_histogram(hist))
 
# This code is contributed
# by Jinay Shah

C#

// C# program to find maximum
// rectangular area in linear time
using System;
using System.Collections.Generic;
 
class GFG
{
// The main function to find the
// maximum rectangular area under
// given histogram with n bars
public static int getMaxArea(int[] hist,
                             int n)
{
    // Create an empty stack. The stack
    // holds indexes of hist[] array
    // The bars stored in stack are always
    // in increasing order of their heights.
    Stack<int> s = new Stack<int>();
 
    int max_area = 0; // Initialize max area
    int tp; // To store top of stack
    int area_with_top; // To store area with top
                       // bar as the smallest bar
 
    // Run through all bars of
    // given histogram
    int i = 0;
    while (i < n)
    {
        // If this bar is higher than the
        // bar on top stack, push it to stack
        if (s.Count == 0 || hist[s.Peek()] <= hist[i])
        {
            s.Push(i++);
        }
 
        // If this bar is lower than top of stack,
        // then calculate area of rectangle with
        // stack top as the smallest (or minimum 
        // height) bar. 'i' is 'right index' for
        // the top and element before top in stack
        // is 'left index'
        else
        {
            tp = s.Peek(); // store the top index
            s.Pop(); // pop the top
 
            // Calculate the area with hist[tp]
            // stack as smallest bar
            area_with_top = hist[tp] *
                           (s.Count == 0 ? i : i - s.Peek() - 1);
 
            // update max area, if needed
            if (max_area < area_with_top)
            {
                max_area = area_with_top;
            }
        }
    }
 
    // Now pop the remaining bars from
    // stack and calculate area with every
    // popped bar as the smallest bar
    while (s.Count > 0)
    {
        tp = s.Peek();
        s.Pop();
        area_with_top = hist[tp] *
                       (s.Count == 0 ? i : i - s.Peek() - 1);
 
        if (max_area < area_with_top)
        {
            max_area = area_with_top;
        }
    }
 
    return max_area;
 
}
 
// Driver Code
public static void Main(string[] args)
{
    int[] hist = new int[] {6, 2, 5, 4, 5, 1, 6};
    Console.WriteLine("Maximum area is " +
                       getMaxArea(hist, hist.Length));
}
}
 
// This code is contributed by Shrikant13

Output

Maximum area is 12

Time Complexity: Since every bar is pushed and popped only once, the time complexity of this method is O(n).

Another Efficient Approach : By finding next smaller element and previous smaller element for every element in O(n) time complexity and O(n) auxiliary space .

Step 1 : First we will take two arrays left_smaller[] and right_smaller[] and initialize it with -1 and n respectively.

Step 2 : For every element we will store the index of previous smaller and next smaller element in left_smaller[] and right_smaller[] arrays respectively.

(It will take O(n) time).

Step 3 : Now for every element we will calculate area by taking this ith element as the smallest in the range left_smaller[i] and right_smaller[i] and multiplying it with the difference of left_smaller[i] and right_smaller[i].

Step 4 : We can find the maximum of all the area calculated in step 3 to get the desired maximum area.

C++

#include <bits/stdc++.h>
using namespace std;
 
 
//Function to find largest rectangular area possible in a given histogram.
int getMaxArea(int arr[], int n)
{
    // Your code here
    //we create an empty stack here.
    stack<int> s;
    //we push -1 to the stack because for some elements there will be no previous
    //smaller element in the array and we can store -1 as the index for previous smaller.
    s.push(-1);
    int area = arr[0];
    int i = 0;
    //We declare left_smaller and right_smaller array of size n and initialize them with -1 and n as their default value.
    //left_smaller[i] will store the index of previous smaller element for ith element of the array.
    //right_smaller[i] will store the index of next smaller element for ith element of the array.
    vector<int> left_smaller(n, -1), right_smaller(n, n);
    while(i<n){
        while(!s.empty()&&s.top()!=-1&&arr[s.top()]>arr[i]){
            //if the current element is smaller than element with index stored on the
            //top of stack then, we pop the top element and store the current element index
            //as the right_smaller for the poped element.
            right_smaller[s.top()] = i;
            s.pop();
        }
        if(i>0&&arr[i]==arr[i-1]){
            //we use this condition to avoid the unnecessary loop to find the left_smaller.
            //since the previous element is same as current element, the left_smaller will always be the same for both.
            left_smaller[i] = left_smaller[i-1];
        }else{
            //Element with the index stored on the top of the stack is always smaller than the current element.
            //Therefore the left_smaller[i] will always be s.top().
            left_smaller[i] = s.top();
        } 
        s.push(i);
        i++;
    }
    for(int j = 0; j<n; j++){
        //here we find area with every element as the smallest element in their range and compare it with the previous area.
        // in this way we get our max Area form this.
        area = max(area, arr[j]*(right_smaller[j]-left_smaller[j]-1));
    }
    return area;
}
 
int main()
 {
    int hist[] = {6, 2, 5, 4, 5, 1, 6};
    int n = sizeof(hist)/sizeof(hist[0]);
      cout << "maxArea = " << getMaxArea(hist, n) << endl;
    return 0;
}
 
//This code is Contributed by Arunit Kumar.

Java

import java.util.*;
import java.lang.*;
import java.io.*;
 
public class RectArea
{
     
    //Function to find largest rectangular area possible in a given histogram.
    public static int getMaxArea(int arr[], int n)
    {
        // your code here
        //we create an empty stack here.
        Stack<Integer> s = new Stack<>();
        //we push -1 to the stack because for some elements there will be no previous
        //smaller element in the array and we can store -1 as the index for previous smaller.
        s.push(-1);
        int max_area = arr[0];
        //We declare left_smaller and right_smaller array of size n and initialize them with -1 and n as their default value.
        //left_smaller[i] will store the index of previous smaller element for ith element of the array.
        //right_smaller[i] will store the index of next smaller element for ith element of the array.
        int left_smaller[] = new int[n];
        int right_smaller[] = new int[n];
        for (int i = 0; i < n; i++){
            left_smaller[i] = -1;
            right_smaller[i] = n;
        }
 
        int i = 0;
        while (i < n)
        {
            while(!s.empty()&&s.peek()!=-1&&arr[i]<arr[s.peek()]){
                //if the current element is smaller than element with index stored on the
                //top of stack then, we pop the top element and store the current element index
                //as the right_smaller for the poped element.
                right_smaller[s.peek()] = (int)i;
                s.pop();
            }
            if(i>0&&arr[i]==arr[(i-1)]){
                //we use this condition to avoid the unnecessary loop to find the left_smaller.
                //since the previous element is same as current element, the left_smaller will always be the same for both.
                left_smaller[i] = left_smaller[(int)(i-1)];
            }else{
                //Element with the index stored on the top of the stack is always smaller than the current element.
                //Therefore the left_smaller[i] will always be s.top().
                left_smaller[i] = s.peek();
            }
            s.push(i);
            i++;
        }
 
 
        for(i = 0; i<n; i++){
            //here we find area with every element as the smallest element in their range and compare it with the previous area.
            // in this way we get our max Area form this.
            max_area = Math.max(max_area, arr[i]*(right_smaller[i] - left_smaller[i] - 1));
        }
 
        return max_area;
    }
     
    public static void main(String[] args)
    {
        int hist[] = { 6, 2, 5, 4, 5, 1, 6 };
        System.out.println("Maximum area is " + getMaxArea(hist, hist.length));
    }
}
//This code is Contributed by Arunit Kumar.

Python3

#this is single line comment
"""
this
is
multiline
comment
"""
def getMaxArea(arr):
  s = [-1]
  n = len(arr)
  area = 0
  i = 0
  left_smaller = [-1]*n
  right_smaller = [n]*n
  while i < n:
      while s and (s[-1] != -1) and (arr[s[-1]] > arr[i]):
          right_smaller[s[-1]] = i
          s.pop()
      if((i > 0) and (arr[i] == arr[i-1])):
          left_smaller[i] = left_smaller[i-1]
      else:
          left_smaller[i] = s[-1]
      s.append(i)
      i += 1
  for j in range(0, n):
      area = max(area, arr[j]*(right_smaller[j]-left_smaller[j]-1))
  return area
 
hist = [6, 2, 5, 4, 5, 1, 6]
print("maxArea = ", getMaxArea(hist))
 
#This code is contributed by Arunit Kumar

Output

maxArea = 12

Time Complexity : O(n)

Space Complexity : O(n)

References
http://www.informatik.uni-ulm.de/acm/Locals/2003/html/histogram.html
http://www.informatik.uni-ulm.de/acm/Locals/2003/html/judge.html
Thanks to Ashish Anand for suggesting initial solution. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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