Longest Mountain Subarray
Last Updated :
12 Jul, 2025
Given an array arr[] with N elements, the task is to find out the longest sub-array which has the shape of a mountain.
Note: A mountain sub-array starts with an increasing sequence, reaches a peak, and then follows a decreasing sequence.
Examples:
Input: arr = [2, 2, 2]
Output: 0
Explanation: No sub-array exists that shows the behavior of a mountain sub-array.
Input: arr = [1, 3, 1, 4, 5, 6, 7, 8, 9, 8, 7, 6, 5]
Output: 11
Explanation: Longest sub-array Mountain [1, 4, 5, 6, 7, 8, 9, 8, 7, 6, 5] which have length 11
[Naive Approach] Using the Nested loops - O(N^2) Time and O(1) Space
We can solve this problem using nested loops: the outer loop iterates over the array, while the inner loop finds the mountain length starting at the outer loop's index. The inner loop first checks the increasing slope, then the decreasing slope. If both exist, we update the maximum mountain length accordingly.
C++
#include <bits/stdc++.h>
using namespace std;
int LongestMountain(vector<int> &a)
{
int ans = 0;
int n = a.size();
// iterate over the array
for (int i = 0; i < n; i++)
{
int j = i + 1;
int inc = 0, dec = 0;
// check weather it make it is increase first or not
while (j < n && a[j] > a[j - 1])
{
inc = 1;
j++;
}
// check weather it is decreasing after checking the increaseing part
while (j < n && a[j] < a[j - 1])
{
dec = 1;
j++;
}
// if mountain
if (inc && dec)
{
ans = max(j - i, ans);
inc = 0, dec = 0;
}
}
// return maximum length
return ans;
}
int main()
{
vector<int> d = {1, 3, 1, 4,
5, 6, 7, 8,
9, 8, 7, 6, 5};
cout << LongestMountain(d)
<< endl;
return 0;
}
import java.util.*;
public class GfG {
public static int findLongestMountain(int[] a) {
int ans = 0;
int n = a.length;
// Iterate over the array
for (int i = 0; i < n; i++) {
int j = i + 1;
int inc = 0, dec = 0;
// Check increasing sequence
while (j < n && a[j] > a[j - 1]) {
inc = 1;
j++;
}
// Check decreasing sequence
while (j < n && a[j] < a[j - 1]) {
dec = 1;
j++;
}
// If a mountain is found, update max length
if (inc == 1 && dec == 1) {
ans = Math.max(ans, j - i);
}
}
return ans;
}
public static void main(String[] args) {
int[] d = {1, 3, 1, 4, 5, 6, 7, 8, 9, 8, 7, 6, 5};
System.out.println(findLongestMountain(d));
}
}
def longest_mountain(a):
ans = 0
n = len(a)
# Iterate over the array
for i in range(n):
j = i + 1
inc, dec = 0, 0
# Check increasing sequence
while j < n and a[j] > a[j - 1]:
inc = 1
j += 1
# Check decreasing sequence
while j < n and a[j] < a[j - 1]:
dec = 1
j += 1
# If a mountain is found, update max length
if inc and dec:
ans = max(ans, j - i)
return ans
# Driver code
d = [1, 3, 1, 4, 5, 6, 7, 8, 9, 8, 7, 6, 5]
print(longest_mountain(d))
using System;
using System.Collections.Generic;
class GfG
{
static int LongestMountain(List<int> a)
{
int ans = 0; // Store the answer
int n = a.Count;
// Iterate over the array
for (int i = 0; i < n; i++)
{
int j = i + 1;
int inc = 0, dec = 0;
// Check if it's increasing first
while (j < n && a[j] > a[j - 1])
{
inc = 1;
j++;
}
// Check if it's decreasing after increasing
while (j < n && a[j] < a[j - 1])
{
dec = 1;
j++;
}
// If it's a valid mountain
if (inc == 1 && dec == 1)
{
ans = Math.Max(j - i, ans);
}
}
return ans;
}
static void Main()
{
List<int> d = new List<int> { 1, 3, 1, 4, 5, 6, 7, 8, 9, 8, 7, 6, 5 };
Console.WriteLine(LongestMountain(d));
}
}
function LongestMountain(arr) {
let ans = 0; // Store the answer
let n = arr.length;
// Iterate over the array
for (let i = 0; i < n; i++) {
let j = i + 1;
let inc = 0, dec = 0;
// Check if it's increasing first
while (j < n && arr[j] > arr[j - 1]) {
inc = 1;
j++;
}
// Check if it's decreasing after increasing
while (j < n && arr[j] < arr[j - 1]) {
dec = 1;
j++;
}
// If it's a valid mountain
if (inc && dec) {
ans = Math.max(j - i, ans);
}
}
return ans;
}
// Driver Code
let d = [1, 3, 1, 4, 5, 6, 7, 8, 9, 8, 7, 6, 5];
console.log(LongestMountain(d));
[Expected Approach] Peak Expansion Strategy - O(N) Time and O(1) Space
Use the two-pointer approach to find the longest mountain subarray efficiently. Traverse the array to identify peak elements (arr[i-1] < arr[i] > arr[i+1]). For each peak, expand outward with two pointers: move left while elements increase and right while they decrease. Update the maximum length accordingly and continue traversal. Return the longest recorded mountain length.
C++
#include <bits/stdc++.h>
using namespace std;
// Function to find the
// longest mountain subarray
int LongestMountain(vector<int> &arr)
{
int n = arr.size();
if (n < 3)
return 0;
int ans = 0;
for (int i = 1; i <= n - 2;)
{
if (arr[i] > arr[i - 1] and arr[i] > arr[i + 1])
{
int count = 0;
int j = i;
while (arr[j] > arr[j - 1] and j > 0)
count++, j--;
while (arr[i] > arr[i + 1] and i <= n - 2)
count++, i++;
ans = max(ans, count);
}
else
i++;
}
if (ans > 0)
return ans + 1;
return ans;
}
// Driver code
int main()
{
vector<int> d = {1, 3, 1, 4, 5, 6, 7, 8, 9, 8, 7, 6, 5};
cout << LongestMountain(d) << endl;
return 0;
}
import java.util.ArrayList;
public class GfG {
// Function to find the
// longest mountain subarray
public static int
LongestMountain(ArrayList<Integer> arr)
{
int n = arr.size();
if (n < 3) // base condition for having the mountain
return 0;
int ans = 0;
for (int i = 1; i <= n - 2;) {
if (arr.get(i) > arr.get(i - 1)
&& arr.get(i) > arr.get(i + 1)) {
int count = 0;
// now we find the index where a peak is
// present
int j = i;
while (j > 0
&& arr.get(j) > arr.get(j - 1)) {
count++;
j--;
}
while (i <= n - 2
&& arr.get(i) > arr.get(i + 1)) {
count++;
i++;
}
ans = Math.max(ans, count);
}
else {
i++;
}
}
if (ans > 0)
return ans + 1;
return ans;
}
// Driver code
public static void main(String[] args)
{
ArrayList<Integer> d = new ArrayList<>();
d.add(1);
d.add(3);
d.add(1);
d.add(4);
d.add(5);
d.add(6);
d.add(7);
d.add(8);
d.add(9);
d.add(8);
d.add(7);
d.add(6);
d.add(5);
System.out.println(LongestMountain(d));
}
}
import math
def LongestMountain(arr):
n = len(arr)
if n < 3: # base condition for having the mountain
return 0
ans = 0
i = 1
while i <= n - 2:
if arr[i] > arr[i - 1] and arr[i] > arr[i + 1]:
count = 0
# now we find the index where a peak is present
j = i
while j > 0 and arr[j] > arr[j - 1]:
count += 1
j -= 1
while i <= n - 2 and arr[i] > arr[i + 1]:
count += 1
i += 1
ans = max(ans, count)
else:
i += 1
if ans > 0:
return ans + 1
return ans
# Driver code
d = [1, 3, 1, 4, 5, 6, 7, 8, 9, 8, 7, 6, 5]
print(LongestMountain(d))
using System;
public class GfG {
public static int LongestMountain(int[] arr)
{
int n = arr.Length;
if (n < 3) { // base condition for having the mountain
return 0;
}
int ans = 0;
int i = 1;
while (i <= n - 2) {
if (arr[i] > arr[i - 1]
&& arr[i] > arr[i + 1]) {
int count = 0;
// now we find the index where a peak is
// present
int j = i;
while (j > 0 && arr[j] > arr[j - 1]) {
count += 1;
j -= 1;
}
while (i <= n - 2 && arr[i] > arr[i + 1]) {
count += 1;
i += 1;
}
ans = Math.Max(ans, count);
}
else {
i += 1;
}
}
if (ans > 0) {
return ans + 1;
}
return ans;
}
public static void Main()
{
int[] d = { 1, 3, 1, 4, 5, 6, 7, 8, 9, 8, 7, 6, 5 };
Console.WriteLine(LongestMountain(d));
}
}
function LongestMountain(arr)
{
// define the length of the array
let n = arr.length;
// base condition for having the mountain
if (n < 3) {
return 0;
}
// set the ans to 0
let ans = 0;
// set i to 1
let i = 1;
// while loop
while (i <= n - 2) {
// if statement to check if element at i is greater
// than element at i-1 and i+1
if (arr[i] > arr[i - 1] && arr[i] > arr[i + 1]) {
// set count to 0
let count = 0;
// now we find the index where a peak is present
let j = i;
while (j > 0 && arr[j] > arr[j - 1]) {
// increment count
count++;
// decrement j
j--;
}
// while loop to check if i is less than or
// equal to n - 2 and element at i is greater
// than element at i+1
while (i <= n - 2 && arr[i] > arr[i + 1]) {
// increment count
count++;
// increment i
i++;
}
// set ans to the maximum value of ans and count
ans = Math.max(ans, count);
}
else {
// increment i
i++;
}
}
// if statement to check if ans is greater than 0
if (ans > 0) {
// return ans + 1
return ans + 1;
}
// return ans
return ans;
}
let output = LongestMountain(
[ 1, 3, 1, 4, 5, 6, 7, 8, 9, 8, 7, 6, 5 ]);
// log the output
console.log(output);
[Alternate Approach] Using two Pointers - O(n) Time and O(1) Space
Use the two-pointer approach to find the longest mountain subarray. Set j (start of increase) and k (end of decrease) as -1. Traverse the array, setting j at the start of an increasing slope and updating k when a decreasing slope follows. If both are valid, update the maximum mountain length. Reset j and k if a flat region appears, and return the longest recorded mountain length.
C++
#include <bits/stdc++.h>
using namespace std;
// Function to find the
// longest mountain subarray
int LongestMountain(vector<int> &a)
{
int i = 0, j = -1, k = -1, p = 0, d = 0, n = 0;
// cann't make the mountain
if (a.size() < 3)
{
return 0;
}
for (i = 0; i < a.size() - 1; i++)
{
if (a[i + 1] > a[i])
{
if (k != -1)
{
k = -1;
j = -1;
}
// Set the value of j to current index i.
if (j == -1)
{
j = i;
}
}
else
{
// Checks if next element is
// less than current element
if (a[i + 1] < a[i])
{
// if the starting element
// of the mountain sub-array exists
// then the index of ending element
// is stored in k
if (j != -1)
{
k = i + 1;
}
// if mountain take maximum
if (k != -1 && j != -1)
{
if (d < k - j + 1)
{
d = k - j + 1;
}
}
}
else
{
k = -1;
j = -1;
}
}
}
// if mountain take maximum
if (k != -1 && j != -1)
{
if (d < k - j + 1)
{
d = k - j + 1;
}
}
return d;
}
int main()
{
vector<int> d = {1, 3, 1, 4, 5, 6, 7, 8, 9, 8, 7, 6, 5};
cout << LongestMountain(d) << endl;
return 0;
}
import java.io.*;
class GFG {
public static int LongestMountain(int a[])
{
int i = 0, j = -1, k = -1, d = 0;
// cann't mountain
if (a.length < 3)
return 0;
for (i = 0; i < a.length - 1; i++) {
if (a[i + 1] > a[i]) {
if (k != -1) {
k = -1;
j = -1;
}
if (j == -1)
j = i;
}
else {
// Checks if next element is
// less than current element
if (a[i + 1] < a[i]) {
if (j != -1)
k = i + 1;
// if mountain make update d
if (k != -1 && j != -1) {
if (d < k - j + 1)
d = k - j + 1;
}
}
else {
k = -1;
j = -1;
}
}
}
if (k != -1 && j != -1) {
if (d < k - j + 1)
d = k - j + 1;
}
return d;
}
public static void main(String[] args)
{
int a[] = { 1, 3, 1, 4, 5, 6, 7, 8, 9, 8, 7, 6, 5 };
System.out.println(LongestMountain(a));
}
}
# Function to find the
# longest mountain subarray
def LongestMountain(a):
i = 0
j = -1
k = -1
p = 0
d = 0
n = 0
if (len(a) < 3):
return 0
for i in range(len(a) - 1):
if (a[i + 1] > a[i]):
# reset it
if (k != -1):
k = -1
j = -1
# value of j to current index i.
if (j == -1):
j = i
else:
# Checks if next element is
# less than current element
if (a[i + 1] < a[i]):
if (j != -1):
k = i + 1
# if mountain update answer
if (k != -1 and j != -1):
if (d < k - j + 1):
d = k - j + 1
else:
k = -1
j = -1
if (k != -1 and j != -1):
if (d < k - j + 1):
d = k - j + 1
return d
d = [1, 3, 1, 4, 5, 6,
7, 8, 9, 8, 7, 6, 5]
print(LongestMountain(d))
using System;
class GfG {
public static int LongestMountain(int[] a)
{
int i = 0, j = -1, k = -1, d = 0;
// cann't make the mountain
if (a.Length < 3)
return 0;
for (i = 0; i < a.Length - 1; i++) {
if (a[i + 1] > a[i]) {
if (k != -1) {
k = -1;
j = -1;
}
if (j == -1)
j = i;
}
else {
if (a[i + 1] < a[i]) {
if (j != -1)
k = i + 1;
// update the answer if mountain
if (k != -1 && j != -1) {
if (d < k - j + 1)
d = k - j + 1;
}
}
// reset the value of the k and j
else {
k = -1;
j = -1;
}
}
}
// update the value of d if mountain
if (k != -1 && j != -1) {
if (d < k - j + 1)
d = k - j + 1;
}
return d;
}
public static void Main(String[] args)
{
int[] a = { 1, 3, 1, 4, 5, 6, 7, 8, 9, 8, 7, 6, 5 };
Console.WriteLine(LongestMountain(a));
}
}
function LongestMountain(a)
{
let i = 0, j = -1, k = -1, p = 0, d = 0, n = 0;
// cann't about the make the mountain
if (a.length < 3)
return 0;
for (i = 0; i < a.length - 1; i++) {
if (a[i + 1] > a[i]) {
// reset the value of k and j
if (k != -1) {
k = -1;
j = -1;
}
// value of j to current index i.
if (j == -1)
j = i;
}
else {
// Checks if next element is
// less than current element
if (a[i + 1] < a[i]) {
if (j != -1)
k = i + 1;
// if mountain update answer
if (k != -1 && j != -1) {
if (d < k - j + 1)
d = k - j + 1;
}
}
else {
k = -1;
j = -1;
}
}
}
// check the mountain accordingly update the answer
if (k != -1 && j != -1) {
if (d < k - j + 1)
d = k - j + 1;
}
return d;
}
let a = [ 1, 3, 1, 4, 5, 6, 7, 8, 9, 8, 7, 6, 5 ];
console.log(LongestMountain(a));
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