Largest Rectangular Area in a Histogram | Set 2

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, 2, 6}. The largest possible rectangle possible is 12 (see the below figure, the max area rectangle is highlighted in red)


We strongly recommend that you click here and practice it, before moving on to the solution.

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 C++ implementation of the above algorithm.

// C++ program to find maximum rectangular area in linear time
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; // Initalize 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[] <= hist[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'
            tp =;  // 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 - - 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 =;
        area_with_top = hist[tp] * (s.empty() ? i : i - - 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;


Maximum area is 12

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


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.

Company Wise Coding Practice    Topic Wise Coding Practice

Writing code in comment? Please use, generate link and share the link here.

  • Prasenjit Mondal

    //helper function to calculate maximum rectangle in the histogram
    //o(n) solution
    void findMaxRectangleHist(ll *arr, ll size){
    stack sk;
    ll index = 0;
    ll max_area = LLONG_MIN;

    //last index is used to store the last height index in the stack which is greater than the left adjacent height in the array and
    //greater than height at current index
    ll last_index = 0;

    //push start index
    //get height at index by arr[index]

    while(index arr[]){
    }else if(!sk.empty() && arr[index] < arr[]){
    //pop out previous index as long as height is greater than the height at current index and stack is not empty
    while(!sk.empty() && arr[index] max_area){
    max_area = new_area;

    //store last index
    //it is used to replace the heigth at this position in the array
    last_index =;

    //pop indices

    //replace the height in the array at this position
    //since max rectangle is posible by constructing rectangle starting from this index to current index
    arr[last_index] = arr[index];


    //if stack is not empty pop previous index and update max rectangle area
    ll new_area = (index – * arr[];

    if(new_area > max_area){
    max_area = new_area;


    cout << max_area << endl;

  • SKS

    Please check when all values are equal. I think u need some correction..

  • fajsf

    I also think there is a problem with this linear algorithm,
    if you have 3,2,3, the largest possible is 6(2+2+2), but according to the algorithm we will only get 4.

    Am I misunderstand it?

    • dragonmigo

      According to the algorithm, you will get 6 as follows:

      1) 3

      push onto stack

      2) 2

      22, so push 3 onto stack.

      Now stack looks like 2 3

      Now, pop 3 off stack, area = 3 * (3 – 1 – 1) = 3

      pop 2 off stack, area = 2 * (3) = 6

      (notice s.empty() ? i : i – – 1)

    • Puneet Jaiswal

      no this algo gives 6 for input 3,2,3.

  • muhammad

    This is O(n) * (max height).

    So the solution goes like this. You first find the max rect
    area for all height 1 rectangles, then height 2 rectangles or so on..
    subtract array heights by 1 at each step, and keep track of max area at every

    Int maxArea=0;

    Int curArea=0;

    Int h=1;

    Bool nextavail=true;

    while(nextavail )


    Int curlen =0;

    Int maxlen= 0;

    //find max length from all possible rectangles.

    For( int i=0;I < Array.len; i++)


    if( Array[i]==


    Maxlen= maxof (curLen ,maxlen);



    Else curLen ++;



    //get max


    maxArea= maxof
    (curArea, maxArea);

    //subtract 1 from


    For( int i=0;I





    Return maxArea;

  • Aalien

    Will this algorithm work for sorted array? Example: {1,2,3,4,5,6}

  • innosam

    {6, 2, 5, 4, 5, 2, 6}.
    Isn’t the largest rectangle possible here is 14(2×7).

    • Sammy Mohamed

      I thought of almost exactly the same counter example. When we pop off the 6 in the beginning and push 2 back on, we aren’t keeping track of how far to the left of 2 the maximum rectangle containing it can be.

  • darkpassenger

    @geeksforgeeks its not working for the i/p like hist[] = {6, 2};
    its printing 6 but the correct o/p is 4. plz check it ….

    • kartik

      6 is the correct answer. The maximum area is formed by a single bar.

  • gladiator

    shouldn’t there be while loop in the else condition becoz we have to pop the item until stack is empty || hist[top]<=hist[i]???

    • Linfuzki

      it does have a loop for the “else” condition by not increasing i;

    According to this page there is a recursive version too – would be nice to have that too

  • Gupt

    Didn’t get the approach, please explain with example.

    /* Paste your code here (You may delete these lines if not writing code) */
  • _naive_

    Sir, Explanation is not proper. Please re explain. Thank you.

    • GeeksforGeeks

      @_naive_: We have updated the explanation for better understanding.

      • brijesh

        can this code work if we have a histogram array with zero entry?????thanks …