Longest Mountain Subarray

Difficulty Level : Medium
Last Updated : 09 Oct, 2020

Given an array arr[] with N elements, the task is to find out the longest sub-array which has the shape of a mountain.

A mountain sub-array consists of elements that are initially in ascending order until a peak element is reached and beyond the peak element all other elements of the sub-array are in decreasing order.

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:
There are two sub-arrays that can be considered as mountain sub-arrays. The first one is from index 0 – 2 (3 elements) and next one is from index 2 – 12 (11 elements). As 11 > 2, our answer is 11.

Naive Approach:
Go through every possible sub-array and check whether it is a mountain sub-array or not. This might take a long time to find the solution and the time complexity for the above approach can be estimated as O(N*N) to go through every possible sub-array and O(N) to check whether it is a mountain sub-array or not. Thus, the overall time complexity for the program is O(N³) which is very high.

Efficient Approach:

If the length of the given array is less than 3, print 0 as it is not possible to have a mountain sub-array in such a case.
Set the maximum length to 0 initially.
Use the two-pointer technique (‘begin’ pointer and ‘end’ pointer) to find out the longest mountain sub-array in the given array.
When an increasing sub-array is encountered, mark the beginning index of that increasing sub-array in the ‘begin’ pointer.
If an index value is found in the ‘end’ pointer then reset the values in both the pointers as it marks the beginning of a new mountain sub-array.
When a decreasing sub-array us encountered, mark the ending index of the mountain sub-array in the ‘end’ pointer.
Calculate the length of the current mountain sub-array, compare it with the current maximum length of all-mountain sub-arrays traversed until now and keep updating the current maximum length.

Below is the implementation of the above described efficient approach:

C++

// C++ codde for Longest Mountain Subarray
 
#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;
 
    // If the size of array is less
    // than 3, the array won't show
    // mountain like behaviour
    if (a.size() < 3) {
        return 0;
    }
 
    for (i = 0; i < a.size() - 1; i++) {
 
        if (a[i + 1] > a[i]) {
 
            // When a new mountain sub-array
            // is found, there is a need to
            // set the variables k, j to -1
            // in order to help calculate the
            // length of new mountain sub-array
            if (k != -1) {
                k = -1;
                j = -1;
            }
 
            // j marks the starting index of a
            // new mountain sub-array. So 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]) {
 
                // Checks if starting element exists
                // or not, 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;
                }
 
                // This condition checks if both
                // starting index and ending index
                // exists or not, if yes, the
                // length is calculated.
                if (k != -1 && j != -1) {
 
                    // d holds the lenght of the
                    // longest mountain sub-array.
                    // If the current length is
                    // greater than the
                    // calculated length, then
                    // value of d is updated.
                    if (d < k - j + 1) {
                        d = k - j + 1;
                    }
                }
            }
 
            // ignore if there is no
            // increase or decrease in
            // the value of the next element
            else {
                k = -1;
                j = -1;
            }
        }
    }
 
    // Checks and calculates
    // the length if last element
    // of the array is the last
    // element of a mountain sub-array
    if (k != -1 && j != -1) {
        if (d < k - j + 1) {
            d = k - j + 1;
        }
    }
    return d;
}
 
// 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;
}

Java

// Java code for Longest Mountain Subarray
import java.io.*;
 
class GFG{
     
// Function to find the
// longest mountain subarray
public static int LongestMountain(int a[])
{
    int i = 0, j = -1, k = -1,
        p = 0, d = 0, n = 0;
 
    // If the size of array is less than 3, 
    // the array won't show mountain like 
    // behaviour
    if (a.length < 3)
        return 0;
 
    for(i = 0; i < a.length - 1; i++)
    {
        if (a[i + 1] > a[i])
        {
             
            // When a new mountain sub-array is 
            // found, there is a need to set the
            // variables k, j to -1 in order to 
            // help calculate the length of new 
            // mountain sub-array
            if (k != -1)
            {
                k = -1;
                j = -1;
            }
 
            // j marks the starting index of a
            // new mountain sub-array. So 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])
            {
                 
                // Checks if starting element exists
                // or not, 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;
 
                // This condition checks if both 
                // starting index and ending index
                // exists or not, if yes,the length 
                // is calculated.
                if (k != -1 && j != -1)
                {
 
                    // d holds the lenght of the 
                    // longest mountain sub-array.
                    // If the current length is 
                    // greater than the calculated 
                    // length, then value of d is 
                    // updated.
                    if (d < k - j + 1)
                        d = k - j + 1;
                }
            }
             
            // Ignore if there is no increase
            // or decrease in the value of the
            // next element
            else
            {
                k = -1;
                j = -1;
            }
        }
    }
 
    // Checks and calculates the length
    // if last element of the array is
    // the last element of a mountain sub-array
    if (k != -1 && j != -1)
    {
        if (d < k - j + 1) 
            d = k - j + 1;
    }
    return d;
}
 
// Driver code
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));
}
}
 
// This code is contributed by piyush3010

Python3

# Python3 code for longest mountain subarray
 
# Function to find the
# longest mountain subarray
def LongestMountain(a):
     
    i = 0
    j = -1
    k = -1
    p = 0
    d = 0
    n = 0
     
    # If the size of the array is less
    # than 3, the array won't show
    # mountain like behaviour
    if (len(a) < 3):
        return 0
         
    for i in range(len(a) - 1):
        if (a[i + 1] > a[i]):
             
            # When a new mountain sub-array
            # is found, there is a need to
            # set the variables k, j to -1
            # in order to help calculate the
            # length of new mountain sub-array
            if (k != -1):
                k = -1
                j = -1
             
            # j marks the starting index of a
            # new mountain sub-array. So 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]):
                 
                # Checks if starting element exists
                # or not, 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
                     
                # This condition checks if both
                # starting index and ending index
                # exists or not, if yes, the
                # length is calculated.
                if (k != -1 and j != -1):
                     
                    # d holds the lenght of the
                    # longest mountain sub-array.
                    # If the current length is
                    # greater than the
                    # calculated length, then
                    # value of d is updated.
                    if (d < k - j + 1):
                        d = k - j + 1
             
            # Ignore if there is no
            # increase or decrease in
            # the value of the next element
            else:
                k = -1
                j = -1
     
    # Checks and calculates
    # the length if last element
    # of the array is the last
    # element of a mountain sub-array
    if (k != -1 and j != -1):
        if (d < k - j + 1):
            d = k - j + 1
             
    return d
 
# Driver code
d = [ 1, 3, 1, 4, 5, 6, 
      7, 8, 9, 8, 7, 6, 5 ]
 
print(LongestMountain(d))
     
# This code is contributed by shubhamsingh10

C#

// C# code for the 
// longest Mountain Subarray
using System;
class GFG{
     
// Function to find the
// longest mountain subarray
public static int longestMountain(int []a)
{
  int i = 0, j = -1, k = -1,
  p = 0, d = 0;
 
  // If the size of array is less than 3, 
  // the array won't show mountain like 
  // behaviour
  if (a.Length < 3)
    return 0;
 
  for(i = 0; i < a.Length - 1; i++)
  {
    if (a[i + 1] > a[i])
    {
      // When a new mountain sub-array is 
      // found, there is a need to set the
      // variables k, j to -1 in order to 
      // help calculate the length of new 
      // mountain sub-array
      if (k != -1)
      {
        k = -1;
        j = -1;
      }
 
      // j marks the starting index of a
      // new mountain sub-array. So 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])
      {
        // Checks if starting element exists
        // or not, 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;
 
        // This condition checks if both 
        // starting index and ending index
        // exists or not, if yes,the length 
        // is calculated.
        if (k != -1 && j != -1)
        {
          // d holds the lenght of the 
          // longest mountain sub-array.
          // If the current length is 
          // greater than the calculated 
          // length, then value of d is 
          // updated.
          if (d < k - j + 1)
            d = k - j + 1;
        }
      }
 
      // Ignore if there is no increase
      // or decrease in the value of the
      // next element
      else
      {
        k = -1;
        j = -1;
      }
    }
  }
 
  // Checks and calculates the length
  // if last element of the array is
  // the last element of a mountain sub-array
  if (k != -1 && j != -1)
  {
    if (d < k - j + 1) 
      d = k - j + 1;
  }
  return d;
}
 
// Driver code
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));
}
}
 
// This code is contributed by Princi Singh

Output:

Time Complexity: O(N)
Auxiliary Space Complexity: O(1)

My Personal Notes

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Python3

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