# Largest Rectangular Area in a Histogram using Stack

Find the largest rectangular area possible in a given histogram where the largest rectangle can be made of a number of contiguous bars whose heights are given in an array. For simplicity, assume that all bars have the same width and the width is 1 unit.

Example:

Input: histogram = {6, 2, 5, 4, 5, 1, 6}

Output: 12

Input: histogram = {3, 5, 1, 7, 5, 9}
Output: 15

To solve the problem follow the below idea:

For every bar ‘x’, we calculate the area with ‘x’ as the smallest bar in the rectangle. If we calculate the such area for every bar ‘x’ and find the maximum of all areas, our task is done.

How to calculate the area with ‘x’ as the smallest bar?

We need to know the index of the first smaller (smaller than ‘x’) bar on the left of ‘x’ and the index of the first smaller bar on the right of ‘x’. Let us call these indexes as ‘left index’ and ‘right index’ respectively. We traverse all bars from left to right and maintain a stack of bars. Every bar is pushed to stack once. A bar is popped from the stack when a bar of smaller height is seen. When a bar is popped, we calculate the area with the popped bar as the smallest bar.

How do we get the left and right indexes of the popped bar?

The current index tells us the right index and the index of the previous item in the stack is the left index

Follow the given steps to solve the problem:

1. Create an empty stack.
2. Start from the first bar, and do the following for every bar hist[i] where ‘i‘ varies from 0 to n-1
1. If the stack is empty or hist[i] is higher than the bar at top of the stack, then push ‘i‘ to stack.
2. If this bar is smaller than the top of the stack, then keep removing the top of the stack while the top of the stack is greater.
3. Let the removed bar be hist[tp]. Calculate the area of the rectangle with hist[tp] as the smallest bar.
4. 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 the stack and do step (2.2 and 2.3) for every removed bar

Below is the implementation of the above approach.

## C++

 `// C++ program to find maximum rectangular area in ` `// linear time ` `#include ` `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 code ` `int` `main() ` `{ ` `    ``int` `hist[] = { 6, 2, 5, 4, 5, 1, 6 }; ` `    ``int` `n = ``sizeof``(hist) / ``sizeof``(hist[0]); ` ` `  `    ``// Function call ` `    ``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 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 code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int` `hist[] = { ``6``, ``2``, ``5``, ``4``, ``5``, ``1``, ``6` `}; ` ` `  `        ``// Function call ` `        ``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 calculates 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 ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``hist ``=` `[``6``, ``2``, ``5``, ``4``, ``5``, ``1``, ``6``] ` ` `  `    ``# Function call ` `    ``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 }; ` ` `  `        ``// function call ` `        ``Console.WriteLine(``"Maximum area is "` `                          ``+ getMaxArea(hist, hist.Length)); ` `    ``} ` `} ` ` `  `// This code is contributed by Shrikant13`

## Javascript

 ``

Output

`Maximum area is 12`

Time Complexity: O(N), Since every bar is pushed and popped only once
Auxiliary Space: O(N)

## Largest Rectangular Area in a Histogram by finding the next and the previous smaller element:

To solve the problem follow the below idea:

Find the previous and the next smaller element for every element of the histogram, as this would help to calculate the length of the subarray in which this current element is the minimum element. So we can create a rectangle of size (current element * length of the subarray) using this element. Take the maximum of all such rectangles.

Follow the given steps to solve the problem:

• First, we will take two arrays left_smaller[] and right_smaller[] and initialize them with -1 and n respectively
• For every element, we will store the index of the previous smaller and next smaller element in left_smaller[] and right_smaller[] arrays respectively
• Now for every element, we will calculate the 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]
• We can find the maximum of all the areas calculated in step 3 to get the desired maximum area

Below is the implementation of the above approach.

## C++

 `// C++ code for the above approach ` ` `  `#include ` `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 popped 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; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `hist[] = { 6, 2, 5, 4, 5, 1, 6 }; ` `    ``int` `n = ``sizeof``(hist) / ``sizeof``(hist[0]); ` ` `  `    ``// Function call ` `    ``cout << ``"maxArea = "` `<< getMaxArea(hist, n) << endl; ` `    ``return` `0; ` `} ` ` `  `// This code is Contributed by Arunit Kumar.`

## Java

 `// Java code for the above approach ` ` `  `import` `java.io.*; ` `import` `java.lang.*; ` `import` `java.util.*; ` ` `  `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 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 popped 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; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int` `hist[] = { ``6``, ``2``, ``5``, ``4``, ``5``, ``1``, ``6` `}; ` ` `  `        ``// Function call ` `        ``System.out.println(``"Maximum area is "` `                           ``+ getMaxArea(hist, hist.length)); ` `    ``} ` `} ` `// This code is Contributed by Arunit Kumar.`

## Python3

 `# Python3 code for the above approach ` ` `  ` `  `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 ` ` `  ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``hist ``=` `[``6``, ``2``, ``5``, ``4``, ``5``, ``1``, ``6``] ` ` `  `    ``# Function call ` `    ``print``(``"maxArea = "``, getMaxArea(hist)) ` ` `  `# This code is contributed by Arunit Kumar `

## C#

 `// C# code for the above approach ` ` `  `using` `System; ` `using` `System.Collections.Generic; ` `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<``int``> s = ``new` `Stack<``int``>(); ` ` `  `        ``// 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` `j = 0; j < n; j++) { ` `            ``left_smaller[j] = -1; ` `            ``right_smaller[j] = n; ` `        ``} ` ` `  `        ``int` `i = 0; ` `        ``while` `(i < n) { ` `            ``while` `(s.Count != 0 && 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 popped 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; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main(String[] args) ` `    ``{ ` `        ``int``[] hist = { 6, 2, 5, 4, 5, 1, 6 }; ` ` `  `        ``// Function call ` `        ``Console.WriteLine(``"Maximum area is "` `                          ``+ getMaxArea(hist, hist.Length)); ` `    ``} ` `} ` ` `  `// This code is contributed by Rajput-Ji`

## Javascript

 ``

Output

`maxArea = 12`

Time Complexity: O(N)
Auxiliary Space: O(N)

Related Articles: Divide and Conquer based O(N log N) solution

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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