Largest Rectangular Area in a Histogram
Last Updated :
18 Feb, 2025
Given a histogram represented by an array arr[], where each element of the array denotes the height of the bars in the histogram. All bars have the same width of 1 unit.
Task is to 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.
Example:
Input: arr[] = [60, 20, 50, 40, 10, 50, 60]
Output: 100
Explanation: We get the maximum are by picking bars highlighted above in green (50, and 60). The area is computed (smallest height) * (no. of the picked bars) = 50 * 2 = 100.
Input: arr[] = [3, 5, 1, 7, 5, 9]
Output: 15
Explanation: We get the maximum are by picking bars 7, 5 and 9. The area is computed (smallest height) * (no. of the picked bars) = 5 * 3 = 15.
[Naive Approach] By Finding Max Area of Rectangles all Heights - O(n^2) Time and O(1) Space
The idea is to consider each bar as the minimum height and find the maximum area. We traverse toward left of it and add its height until we see a smaller element. We do the same thing for right side of it.
So the area with current bar as minimum is going to be height of current bar multiplied by total width traversed on both left and right including the current bar. The area is the bar’s height multiplied by the total traversed width. Finally, we return the maximum of all such areas.
C++
// C++ program to find the largest rectangular area possible
// in a given histogram
#include <bits/stdc++.h>
using namespace std;
// Function to calculate the maximum rectangular
// area in the Histogram
int getMaxArea(vector<int> &arr){
int res = 0, n = arr.size();
// Consider every bar one by one
for(int i = 0; i < n; i++){
int curr = arr[i];
// Traverse left while we have a greater height bar
for(int j = i-1; j>=0 && arr[j] >= arr[i]; j--)
curr += arr[i];
// Traverse right while we have a greater height bar
for(int j = i+1; j<n && arr[j] >= arr[i]; j++)
curr += arr[i];
res = max(res, curr);
}
return res;
}
int main() {
vector<int> arr = {60, 20, 50, 40, 10, 50, 60};
cout << getMaxArea(arr);
return 0;
}
C
// Cprogram to find the largest rectangular area possible
// in a given histogram
#include <stdio.h>
// Function to calculate the maximum rectangular
// area in the Histogram
int getMaxArea(int arr[], int n) {
int res = 0;
// Consider every bar one by one
for (int i = 0; i < n; i++) {
int curr = arr[i];
// Traverse left while we have a greater height bar
for (int j = i - 1; j >= 0 && arr[j] >= arr[i]; j--) {
curr += arr[i];
}
// Traverse right while we have a greater height bar
for (int j = i + 1; j < n && arr[j] >= arr[i]; j++) {
curr += arr[i];
}
if (curr > res) {
res = curr;
}
}
return res;
}
int main() {
int arr[] = {60, 20, 50, 40, 10, 50, 60};
int n = sizeof(arr) / sizeof(arr[0]);
printf("%d\n", getMaxArea(arr, n));
return 0;
}
Java
// Java program to find the largest rectangular area possible
// in a given histogram
import java.util.*;
class GfG {
// Function to calculate the maximum rectangular
// area in the Histogram
static int getMaxArea(int[] arr) {
int res = 0, n = arr.length;
for (int i = 0; i < n; i++) {
int curr = arr[i];
// Traverse left while we have a greater height bar
for (int j = i - 1; j >= 0 && arr[j] >= arr[i]; j--)
curr += arr[i];
// Traverse right while we have a greater height bar
for (int j = i + 1; j < n && arr[j] >= arr[i]; j++)
curr += arr[i];
res = Math.max(res, curr);
}
return res;
}
public static void main(String[] args) {
int[] arr = {60, 20, 50, 40, 10, 50, 60};
System.out.println(getMaxArea(arr));
}
}
Python
# Python program to find the largest rectangular area possible
# in a given histogram
# Function to calculate the maximum rectangular
# area in the Histogram
def getMaxArea(arr):
res = 0
n = len(arr)
for i in range(n):
curr = arr[i]
# Traverse left while we have a greater height bar
j = i - 1
while j >= 0 and arr[j] >= arr[i]:
curr += arr[i]
j -= 1
# Traverse right while we have a greater height bar
j = i + 1
while j < n and arr[j] >= arr[i]:
curr += arr[i]
j += 1
res = max(res, curr)
return res
if __name__ == "__main__":
arr = [60, 20, 50, 40, 10, 50, 60]
print(getMaxArea(arr))
C#
// C# program to find the largest rectangular area possible
// in a given histogram
using System;
class GfG {
// Function to calculate the maximum rectangular
// area in the Histogram
static int getMaxArea(int[] arr) {
int n = arr.Length;
int res = 0;
// Consider every bar one by one
for (int i = 0; i < n; i++) {
int curr = arr[i];
// Traverse left while we have a greater height bar
int j = i - 1;
while (j >= 0 && arr[j] >= arr[i]) {
curr += arr[i];
j--;
}
// Traverse right while we have a greater height bar
j = i + 1;
while (j < n && arr[j] >= arr[i]) {
curr += arr[i];
j++;
}
res = Math.Max(res, curr);
}
return res;
}
static void Main(string[] args) {
int[] arr = {60, 20, 50, 40, 10, 50, 60};
Console.WriteLine(getMaxArea(arr));
}
}
JavaScript
// JavaScript program to find the largest rectangular area possible
// in a given histogram
// Function to calculate the maximum rectangular
// area in the Histogram
function getMaxArea(arr) {
let n = arr.length;
let res = 0;
// Consider every bar one by one
for (let i = 0; i < n; i++) {
let curr = arr[i];
// Traverse left while we have a greater height bar
let j = i - 1;
while (j >= 0 && arr[j] >= arr[i]) {
curr += arr[i];
j--;
}
// Traverse right while we have a greater height bar
j = i + 1;
while (j < n && arr[j] >= arr[i]) {
curr += arr[i];
j++;
}
res = Math.max(res, curr);
}
return res;
}
// Driver code
let arr = [ 60, 20, 50, 40, 10, 50, 60 ];
console.log(getMaxArea(arr));
[Expected Approach] Precomputing (Using Two Stack) - O(n) Time and O(n) Space
The idea is based on the naive approach. Instead of linearly finding previous smaller and next smaller for every element, we find previous smaller and next smaller for the whole array in linear time.
- Build an array prevS[] in O(n) time using stack that holds index of previous smaller element for every item.
- Build another array nextS[] in O(n) time using stack that holds index of next smaller element for every item.
- Now do following for every element arr[i]. Consider arr[i] find width of the largest histogram with arr[i] being the smallest height. width = nextS[i] - prevS[i] - 1. Now find the area as arr[i] * width.
- Return the maximum of all values found in step 3.
C++
// C++ program to find the largest rectangular area possible
// in a given histogram
#include <bits/stdc++.h>
using namespace std;
// Function to find next smaller for every element
vector<int> nextSmaller(vector<int>& arr) {
int n = arr.size();
// Initialize with n for the cases when next smaller
// does not exist
vector<int> nextS(n, n);
stack<int> st;
// Traverse all array elements from left to right
for (int i = 0; i < n; ++i) {
while (!st.empty() && arr[i] < arr[st.top()]) {
// Setting the index of the next smaller element
// for the top of the stack
nextS[st.top()] = i;
st.pop();
}
st.push(i);
}
return nextS;
}
// Function to find previous smaller for every element
vector<int> prevSmaller(vector<int>& arr) {
int n = arr.size();
// Initialize with -1 for the cases when prev smaller
// does not exist
vector<int> prevS(n, -1);
stack<int> st;
// Traverse all array elements from left to right
for (int i = 0; i < n; ++i) {
while (!st.empty() && arr[i] < arr[st.top()]) {
// Setting the index of the previous smaller element
// for the top of the stack
st.pop();
}
if (!st.empty()) {
prevS[i] = st.top();
}
st.push(i);
}
return prevS;
}
// Function to calculate the maximum rectangular
// area in the Histogram
int getMaxArea(vector<int>& arr) {
vector<int> prevS = prevSmaller(arr);
vector<int> nextS = nextSmaller(arr);
int maxArea = 0;
// Calculate the area for each Histogram bar
for (int i = 0; i < arr.size(); ++i) {
int width = nextS[i] - prevS[i] - 1;
int area = arr[i] * width;
maxArea = max(maxArea, area);
}
return maxArea;
}
int main() {
vector<int> arr = {60, 20, 50, 40, 10, 50, 60};
cout << getMaxArea(arr) << endl;
return 0;
}
C
// C program to find the largest rectangular area possible
// in a given histogram
#include <stdio.h>
#include <stdlib.h>
// Stack structure
struct Stack {
int top;
int capacity;
int* items;
};
// Function to create an empty stack with dynamic memory allocation
struct Stack* createStack(int capacity) {
struct Stack* stack =
(struct Stack*)malloc(sizeof(struct Stack));
stack->capacity = capacity;
stack->top = -1;
stack->items = (int*)malloc(stack->capacity * sizeof(int));
return stack;
}
// Function to check if the stack is empty
int isEmpty(struct Stack* stack) {
return stack->top == -1;
}
// Function to push an element onto the stack
void push(struct Stack* stack, int value) {
if (stack->top == stack->capacity - 1) {
printf("Stack overflow\n");
return;
}
stack->items[++(stack->top)] = value;
}
// Function to pop an element from the stack
int pop(struct Stack* stack) {
if (isEmpty(stack)) {
printf("Stack underflow\n");
return -1;
}
return stack->items[(stack->top)--];
}
// Function to get the top element of the stack
int peek(struct Stack* stack) {
if (!isEmpty(stack)) {
return stack->items[stack->top];
}
return -1;
}
// Function to find the next smaller element for every element
void nextSmaller(int arr[], int n, int nextS[]) {
struct Stack* stack = createStack(n);
// Initialize with n for the cases
// when next smaller does not exist
for (int i = 0; i < n; i++) {
nextS[i] = n;
}
// Traverse all array elements from left to right
for (int i = 0; i < n; i++) {
while (!isEmpty(stack) && arr[i] < arr[peek(stack)]) {
nextS[peek(stack)] = i;
pop(stack);
}
push(stack, i);
}
}
// Function to find the previous smaller element for every element
void prevSmaller(int arr[], int n, int prevS[]) {
struct Stack* stack = createStack(n);
// Initialize with -1 for the cases when prev smaller does not exist
for (int i = 0; i < n; i++) {
prevS[i] = -1;
}
// Traverse all array elements from left to right
for (int i = 0; i < n; i++) {
while (!isEmpty(stack) && arr[i] < arr[peek(stack)]) {
pop(stack);
}
if (!isEmpty(stack)) {
prevS[i] = peek(stack);
}
push(stack, i);
}
}
// Function to calculate the maximum rectangular
// area in the Histogram
int getMaxArea(int arr[], int n) {
int* prevS = (int*)malloc(n * sizeof(int));
int* nextS = (int*)malloc(n * sizeof(int));
int maxArea = 0;
// Find previous and next smaller elements
prevSmaller(arr, n, prevS);
nextSmaller(arr, n, nextS);
// Calculate the area for each arrogram bar
for (int i = 0; i < n; i++) {
int width = nextS[i] - prevS[i] - 1;
int area = arr[i] * width;
if (area > maxArea) {
maxArea = area;
}
}
return maxArea;
}
// Driver code
int main() {
int arr[] = {60, 20, 50, 40, 10, 50, 60};
int n = sizeof(arr) / sizeof(arr[0]);
printf("%d\n", getMaxArea(arr, n));
return 0;
}
Java
// Java program to find the largest rectangular area possible
// in a given histogram
import java.util.Stack;
class GfG {
// Function to find next smaller for every element
static int[] nextSmaller(int[] arr) {
int n = arr.length;
// Initialize with n for the cases when next smaller
// does not exist
int[] nextS = new int[n];
for (int i = 0; i < n; i++) {
nextS[i] = n;
}
Stack<Integer> st = new Stack<>();
// Traverse all array elements from left to right
for (int i = 0; i < n; i++) {
while (!st.isEmpty() && arr[i] < arr[st.peek()]) {
// Setting the index of the next smaller element
// for the top of the stack
nextS[st.pop()] = i;
}
st.push(i);
}
return nextS;
}
// Function to find previous smaller for every element
static int[] prevSmaller(int[] arr) {
int n = arr.length;
// Initialize with -1 for the cases when prev smaller
// does not exist
int[] prevS = new int[n];
for (int i = 0; i < n; i++) {
prevS[i] = -1;
}
Stack<Integer> st = new Stack<>();
// Traverse all array elements from left to right
for (int i = 0; i < n; i++) {
while (!st.isEmpty() && arr[i] < arr[st.peek()]) {
st.pop();
}
if (!st.isEmpty()) {
prevS[i] = st.peek();
}
st.push(i);
}
return prevS;
}
// Function to calculate the maximum rectangular
// area in the histogram
static int getMaxArea(int[] arr) {
int[] prevS = prevSmaller(arr);
int[] nextS = nextSmaller(arr);
int maxArea = 0;
// Calculate the area for each arrogram bar
for (int i = 0; i < arr.length; i++) {
int width = nextS[i] - prevS[i] - 1;
int area = arr[i] * width;
maxArea = Math.max(maxArea, area);
}
return maxArea;
}
public static void main(String[] args) {
int[] arr = {60, 20, 50, 40, 10, 50, 60};
System.out.println(getMaxArea(arr));
}
}
Python
# Python program to find the largest rectangular area possible
# in a given histogram
# Function to find next smaller for every element
def nextSmaller(arr):
n = len(arr)
# Initialize with n for the cases when next smaller
# does not exist
nextS = [n] * n
st = []
# Traverse all array elements from left to right
for i in range(n):
while st and arr[i] < arr[st[-1]]:
# Setting the index of the next smaller element
# for the top of the stack
nextS[st.pop()] = i
st.append(i)
return nextS
# Function to find previous smaller for every element
def prevSmaller(arr):
n = len(arr)
# Initialize with -1 for the cases when prev smaller
# does not exist
prevS = [-1] * n
st = []
# Traverse all array elements from left to right
for i in range(n):
while st and arr[i] < arr[st[-1]]:
st.pop()
if st:
prevS[i] = st[-1]
st.append(i)
return prevS
# Function to calculate the maximum rectangular
# area in the Histogram
def getMaxArea(arr):
prevS = prevSmaller(arr)
nextS = nextSmaller(arr)
maxArea = 0
# Calculate the area for each arrogram bar
for i in range(len(arr)):
width = nextS[i] - prevS[i] - 1
area = arr[i] * width
maxArea = max(maxArea, area)
return maxArea
if __name__ == "__main__":
arr = [60, 20, 50, 40, 10, 50, 60]
print(getMaxArea(arr))
C#
// C# program to find the largest rectangular area possible
// in a given histogram
using System;
using System.Collections.Generic;
class GfG {
// Function to find next smaller for every element
static int[] nextSmaller(int[] arr) {
int n = arr.Length;
// Initialize with n for the cases when next smaller
// does not exist
int[] nextS = new int[n];
for (int i = 0; i < n; i++) {
nextS[i] = n;
}
Stack<int> st = new Stack<int>();
// Traverse all array elements from left to right
for (int i = 0; i < n; i++) {
while (st.Count > 0 && arr[i] < arr[st.Peek()]) {
// Setting the index of the next smaller element
// for the top of the stack
nextS[st.Pop()] = i;
}
st.Push(i);
}
return nextS;
}
// Function to find previous smaller for every element
static int[] prevSmaller(int[] arr) {
int n = arr.Length;
// Initialize with -1 for the cases when prev smaller
// does not exist
int[] prevS = new int[n];
for (int i = 0; i < n; i++) {
prevS[i] = -1;
}
Stack<int> st = new Stack<int>();
// Traverse all array elements from left to right
for (int i = 0; i < n; i++) {
while (st.Count > 0 && arr[i] < arr[st.Peek()]) {
st.Pop();
}
if (st.Count > 0) {
prevS[i] = st.Peek();
}
st.Push(i);
}
return prevS;
}
// Function to calculate the maximum rectangular
// area in the Histogram
static int getMaxArea(int[] arr) {
int[] prevS = prevSmaller(arr);
int[] nextS = nextSmaller(arr);
int maxArea = 0;
// Calculate the area for each arrogram bar
for (int i = 0; i < arr.Length; i++) {
int width = nextS[i] - prevS[i] - 1;
int area = arr[i] * width;
maxArea = Math.Max(maxArea, area);
}
return maxArea;
}
static void Main() {
int[] arr = {60, 20, 50, 40, 10, 50, 60};
Console.WriteLine(getMaxArea(arr));
}
}
JavaScript
// JavaScript program to find the largest rectangular area possible
// in a given histogram
// Function to find next smaller for every element
function nextSmaller(arr){
const n = arr.length;
// Initialize with n for the cases when next smaller
// does not exist
const nextS = new Array(n).fill(n);
const stack = [];
// Traverse all array elements from left to right
for (let i = 0; i < n; i++) {
while (stack.length
&& arr[i] < arr[stack[stack.length - 1]]) {
// Setting the index of the next smaller element
// for the top of the stack
nextS[stack.pop()] = i;
}
stack.push(i);
}
return nextS;
}
// Function to find previous smaller for every element
function prevSmaller(arr) {
const n = arr.length;
// Initialize with -1 for the cases when prev smaller
// does not exist
const prevS = new Array(n).fill(-1);
const stack = [];
// Traverse all array elements from left to right
for (let i = 0; i < n; i++) {
while (stack.length
&& arr[i] < arr[stack[stack.length - 1]]) {
stack.pop();
}
if (stack.length) {
prevS[i] = stack[stack.length - 1];
}
stack.push(i);
}
return prevS;
}
// Function to calculate the maximum rectangular
// area in the Histogram
function getMaxArea(arr) {
const prevS = prevSmaller(arr);
const nextS = nextSmaller(arr);
let maxArea = 0;
// Calculate the area for each arrogram bar
for (let i = 0; i < arr.length; i++) {
const width = nextS[i] - prevS[i] - 1;
const area = arr[i] * width;
maxArea = Math.max(maxArea, area);
}
return maxArea;
}
// Driver code
const arr = [60, 20, 50, 40, 10, 50, 60];
console.log(getMaxArea(arr));
[Further Optimized] Using Single Stack - O(n) Time and O(n) Space
This approach is mainly an optimization over the previous approach.
When we compute next smaller element, we pop an item from the stack and mark current item as next smaller of it. One important observation here is the item below every item in stack is the previous smaller element. So we do not need to explicitly compute previous smaller.
Below are the detailed steps of implementation.
- Create an empty stack.
- Start from the first bar, and do the following for every bar arr[i] where 'i' varies from 0 to n-1
- If the stack is empty or arr[i] is higher than the bar at top of the stack, then push 'i' to stack.
- 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.
- Let the removed bar be arr[tp]. Calculate the area of the rectangle with arr[tp] as the smallest bar.
- For arr[tp], the 'left index' is previous (previous to tp) item in stack and 'right index' is 'i' (current index).
- 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
C++
// C++ program to find the largest rectangular area possible
// in a given histogram
#include <bits/stdc++.h>
using namespace std;
// Function to calculate the maximum rectangular area
int getMaxArea(vector<int>& arr) {
int n = arr.size();
stack<int> s;
int res = 0;
int tp, curr;
for (int i = 0; i < n; i++) {
while (!s.empty() && arr[s.top()] >= arr[i]) {
// The popped item is to be considered as the
// smallest element of the Histogram
tp = s.top();
s.pop();
// For the popped item previous smaller element is
// just below it in the stack (or current stack top)
// and next smaller element is i
int width = s.empty() ? i : i - s.top() - 1;
res = max(res, arr[tp] * width);
}
s.push(i);
}
// For the remaining items in the stack, next smaller does
// not exist. Previous smaller is the item just below in
// stack.
while (!s.empty()) {
tp = s.top(); s.pop();
curr = arr[tp] * (s.empty() ? n : n - s.top() - 1);
res = max(res, curr);
}
return res;
}
int main() {
vector<int> arr = {60, 20, 50, 40, 10, 50, 60};
cout << getMaxArea(arr);
return 0;
}
C
// C program to find the largest rectangular area possible
// in a given histogram
#include <stdio.h>
#include <stdlib.h>
// Stack structure
struct Stack {
int top;
int capacity;
int* array;
};
// Function to create a stack
struct Stack* createStack(int capacity) {
struct Stack* stack = (struct Stack*)
malloc(sizeof(struct Stack));
stack->capacity = capacity;
stack->top = -1;
stack->array = (int*)malloc(stack->capacity * sizeof(int));
return stack;
}
int isEmpty(struct Stack* stack) {
return stack->top == -1;
}
void push(struct Stack* stack, int item) {
stack->array[++stack->top] = item;
}
int pop(struct Stack* stack) {
return stack->array[stack->top--];
}
int peek(struct Stack* stack) {
return stack->array[stack->top];
}
// Function to calculate the maximum rectangular area
int getMaxArea(int arr[], int n) {
struct Stack* s = createStack(n);
int res = 0, tp, curr;
// Traverse all bars of the arrogram
for (int i = 0; i < n; i++) {
// Process the stack while the current element
// is smaller than the element corresponding to
// the top of the stack
while (!isEmpty(s) && arr[peek(s)] >= arr[i]) {
tp = pop(s);
// Calculate width and update result
int width = isEmpty(s) ? i : i - peek(s) - 1;
res = (res > arr[tp] * width) ? res : arr[tp] * width;
}
push(s, i);
}
// Process remaining elements in the stack
while (!isEmpty(s)) {
tp = pop(s);
curr = arr[tp] * (isEmpty(s) ? n : n - peek(s) - 1);
res = (res > curr) ? res : curr;
}
return res;
}
int main() {
int arr[] = {60, 20, 50, 40, 10, 50, 60};
int n = sizeof(arr) / sizeof(arr[0]);
printf("%d\n", getMaxArea(arr, n));
return 0;
}
Java
// Java program to find the largest rectangular area possible
// in a given histogram
import java.util.Stack;
class GfG {
// Function to calculate the maximum rectangular area
static int getMaxArea(int[] arr) {
int n = arr.length;
Stack<Integer> s = new Stack<>();
int res = 0, tp, curr;
for (int i = 0; i < n; i++) {
// Process the stack while the current element
// is smaller than the element corresponding to
// the top of the stack
while (!s.isEmpty() && arr[s.peek()] >= arr[i]) {
// The popped item is to be considered as the
// smallest element of the Histogram
tp = s.pop();
// For the popped item, previous smaller element
// is just below it in the stack (or current stack
// top) and next smaller element is i
int width = s.isEmpty() ? i : i - s.peek() - 1;
// Update the result if needed
res = Math.max(res, arr[tp] * width);
}
s.push(i);
}
// For the remaining items in the stack, next smaller does
// not exist. Previous smaller is the item just below in
// the stack.
while (!s.isEmpty()) {
tp = s.pop();
curr = arr[tp] * (s.isEmpty() ? n : n - s.peek() - 1);
res = Math.max(res, curr);
}
return res;
}
public static void main(String[] args) {
int[] arr = {60, 20, 50, 40, 10, 50, 60};
System.out.println(getMaxArea(arr));
}
}
Python
# Python program to find the largest rectangular area possible
# in a given histogram
# Function to calculate the maximum rectangular area
def getMaxArea(arr):
n = len(arr)
s = []
res = 0
for i in range(n):
# Process the stack while the current element
# is smaller than the element corresponding to
# the top of the stack
while s and arr[s[-1]] >= arr[i]:
# The popped item is to be considered as the
# smallest element of the Histogram
tp = s.pop()
# For the popped item, the previous smaller
# element is just below it in the stack (or
# the current stack top) and the next smaller
# element is i
width = i if not s else i - s[-1] - 1
# Update the result if needed
res = max(res, arr[tp] * width)
s.append(i)
# For the remaining items in the stack, next smaller does
# not exist. Previous smaller is the item just below in
# the stack.
while s:
tp = s.pop()
width = n if not s else n - s[-1] - 1
res = max(res, arr[tp] * width)
return res
if __name__ == "__main__":
arr = [60, 20, 50, 40, 10, 50, 60]
print(getMaxArea(arr))
C#
// C# program to find the largest rectangular area possible
// in a given histogram
using System;
using System.Collections.Generic;
class GfG {
// Function to calculate the maximum rectangular area
static int getMaxArea(int[] arr) {
int n = arr.Length;
Stack<int> s = new Stack<int>();
int res = 0, tp, curr;
// Traverse all bars of the arrogram
for (int i = 0; i < n; i++) {
// Process the stack while the current element
// is smaller than the element corresponding to
// the top of the stack
while (s.Count > 0 && arr[s.Peek()] >= arr[i]) {
tp = s.Pop();
// Calculate width and update result
int width = s.Count == 0 ? i : i - s.Peek() - 1;
res = Math.Max(res, arr[tp] * width);
}
s.Push(i);
}
// Process remaining elements in the stack
while (s.Count > 0) {
tp = s.Pop();
curr = arr[tp] * (s.Count == 0 ? n : n - s.Peek() - 1);
res = Math.Max(res, curr);
}
return res;
}
public static void Main() {
int[] arr = {60, 20, 50, 40, 10, 50, 60};
Console.WriteLine(getMaxArea(arr));
}
}
JavaScript
// JavaScript program to find the largest rectangular area possible
// in a given histogram
// Function to calculate the maximum rectangular area
function getMaxArea(arr) {
let n = arr.length;
let stack = [];
let res = 0;
// Traverse all bars of the arrogram
for (let i = 0; i < n; i++) {
// Process the stack while the current element
// is smaller than the element corresponding to
// the top of the stack
while (stack.length
&& arr[stack[stack.length - 1]] >= arr[i]) {
let tp = stack.pop();
// Calculate width and update result
let width
= stack.length === 0 ? i: i
- stack[stack.length - 1] - 1;
res = Math.max(res, arr[tp] * width);
}
stack.push(i);
}
// Process remaining elements in the stack
while (stack.length) {
let tp = stack.pop();
let curr = arr[tp] * (stack.length === 0 ? n
: n - stack[stack.length - 1] - 1);
res = Math.max(res, curr);
}
return res;
}
// Driver code
let arr = [ 60, 20, 50, 40, 10, 50, 60 ];
console.log(getMaxArea(arr));
[Alternate Approach] Using Divide and Conquer - O(n Log n) Time
The idea is to find the minimum value in the given array. Once we have index of the minimum value, the max area is maximum of following three values.
- Maximum area in left side of minimum value (Not including the min value)
- Maximum area in right side of minimum value (Not including the min value)
- Number of bars multiplied by minimum value.
Please refer Largest Rectangular Area in a histogram Using Divide and Conquer for detailed implementation.
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