Skip to content
Related Articles

Related Articles

Longest increasing sequence by the boundary elements of an Array
  • Difficulty Level : Medium
  • Last Updated : 05 May, 2021

Given an array arr[] of length N with unique elements, the task is to find the length of the longest increasing subsequence that can be formed by the elements from either end of the array.
Examples: 
 

Input: arr[] = {3, 5, 1, 4, 2} 
Output:
Explanation: 
The longest sequence is: {2, 3, 4, 5} 
Pick 2, Sequence is {2}, Array is {3, 5, 1, 4} 
Pick 3, Sequence is {2, 3}, Array is {5, 1, 4} 
Pick 4, Sequence is {2, 3, 4}, Array is {5, 1} 
Pick 5, Sequence is {2, 3, 4, 5}, Array is {1}
Input: arr[] = {3, 1, 5, 2, 4} 
Output:
The longest sequence is {3, 4} 
 

 

Approach: This problem can be solved by Two Pointer Approach. Set two pointers at the first and last indices of the array. Select the minimum of the two values currently pointed and check if it is greater than the previously selected value. If so, update the pointer and increase the length of the LIS and repeat the process. Otherwise, print the length of the LIS obtained. 
Below is the implementation of the above approach: 
 

C++




// C++ Program to print the
// longest increasing
// subsequence from the
// boundary elements of an array
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the length of
// Longest Increasing subsequence
int longestSequence(int n, int arr[])
{
 
    // Set pointers at
    // both ends
    int l = 0, r = n - 1;
 
    // Stores the recent
    // value added to the
    // subsequence
    int prev = INT_MIN;
 
    // Stores the length of
    // the subsequence
    int ans = 0;
 
    while (l <= r) {
 
        // Check if both elements
        // can be added to the
        // subsequence
        if (arr[l] > prev
            && arr[r] > prev) {
 
            if (arr[l] < arr[r]) {
                ans += 1;
                prev = arr[l];
                l += 1;
            }
            else {
                ans += 1;
                prev = arr[r];
                r -= 1;
            }
        }
 
        // Check if the element
        // on the left can be
        // added to the
        // subsequence only
        else if (arr[l] > prev) {
            ans += 1;
            prev = arr[l];
            l += 1;
        }
 
        // Check if the element
        // on the right can be
        // added to the
        // subsequence only
        else if (arr[r] > prev) {
            ans += 1;
            prev = arr[r];
            r -= 1;
        }
 
        // If none of the values
        // can be added to the
        // subsequence
        else {
            break;
        }
    }
    return ans;
}
 
// Driver Code
int main()
{
    int arr[] = { 3, 5, 1, 4, 2 };
 
    // Length of array
    int n = sizeof(arr)
            / sizeof(arr[0]);
 
    cout << longestSequence(n, arr);
 
    return 0;
}

Java




// Java program to print the longest
// increasing subsequence from the
// boundary elements of an array
import java.util.*;
 
class GFG{
 
// Function to return the length of
// Longest Increasing subsequence
static int longestSequence(int n, int arr[])
{
     
    // Set pointers at
    // both ends
    int l = 0, r = n - 1;
 
    // Stores the recent
    // value added to the
    // subsequence
    int prev = Integer.MIN_VALUE;
 
    // Stores the length of
    // the subsequence
    int ans = 0;
 
    while (l <= r)
    {
         
        // Check if both elements
        // can be added to the
        // subsequence
        if (arr[l] > prev &&
            arr[r] > prev)
        {
            if (arr[l] < arr[r])
            {
                ans += 1;
                prev = arr[l];
                l += 1;
            }
            else
            {
                ans += 1;
                prev = arr[r];
                r -= 1;
            }
        }
 
        // Check if the element on the
        // left can be added to the
        // subsequence only
        else if (arr[l] > prev)
        {
            ans += 1;
            prev = arr[l];
            l += 1;
        }
 
        // Check if the element on the
        // right can be added to the
        // subsequence only
        else if (arr[r] > prev)
        {
            ans += 1;
            prev = arr[r];
            r -= 1;
        }
 
        // If none of the values
        // can be added to the
        // subsequence
        else
        {
            break;
        }
    }
    return ans;
}
 
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 3, 5, 1, 4, 2 };
 
    // Length of array
    int n = arr.length;
 
    System.out.print(longestSequence(n, arr));
}
}
 
// This code is contributed by Amit Katiyar

Python3




# Python3 program to print the
# longest increasing subsequence
# from the boundary elements
# of an array
import sys
 
# Function to return the length of
# Longest Increasing subsequence
def longestSequence(n, arr):
     
    # Set pointers at
    # both ends
    l = 0
    r = n - 1
     
    # Stores the recent value
    # added to the subsequence
    prev = -sys.maxsize - 1
     
    # Stores the length of
    # the subsequence
    ans = 0
     
    while (l <= r):
         
        # Check if both elements can be
        # added to the subsequence
        if (arr[l] > prev and
            arr[r] > prev):
         
            if (arr[l] < arr[r]):
                ans += 1
                prev = arr[l]
                l += 1
                 
            else:
                ans += 1
                prev = arr[r]
                r -= 1
                 
        # Check if the element
        # on the left can be
        # added to the
        # subsequence only
        elif (arr[l] > prev):
            ans += 1
            prev = arr[l]
            l += 1
         
        # Check if the element
        # on the right can be
        # added to the
        # subsequence only
        elif (arr[r] > prev):
            ans += 1
            prev = arr[r]
            r -= 1
         
        # If none of the values
        # can be added to the
        # subsequence
        else:
            break
         
    return ans
             
# Driver code
arr = [ 3, 5, 1, 4, 2 ]
 
# Length of array
n = len(arr)
 
print(longestSequence(n, arr))
 
# This code is contributed by sanjoy_62

C#




// C# program to print the longest
// increasing subsequence from the
// boundary elements of an array
using System;
 
class GFG{
 
// Function to return the length of
// longest Increasing subsequence
static int longestSequence(int n, int []arr)
{
     
    // Set pointers at
    // both ends
    int l = 0, r = n - 1;
 
    // Stores the recent value
    // added to the subsequence
    int prev = int.MinValue;
 
    // Stores the length of
    // the subsequence
    int ans = 0;
 
    while (l <= r)
    {
         
        // Check if both elements
        // can be added to the
        // subsequence
        if (arr[l] > prev &&
            arr[r] > prev)
        {
            if (arr[l] < arr[r])
            {
                ans += 1;
                prev = arr[l];
                l += 1;
            }
            else
            {
                ans += 1;
                prev = arr[r];
                r -= 1;
            }
        }
 
        // Check if the element on the
        // left can be added to the
        // subsequence only
        else if (arr[l] > prev)
        {
            ans += 1;
            prev = arr[l];
            l += 1;
        }
 
        // Check if the element on the
        // right can be added to the
        // subsequence only
        else if (arr[r] > prev)
        {
            ans += 1;
            prev = arr[r];
            r -= 1;
        }
 
        // If none of the values
        // can be added to the
        // subsequence
        else
        {
            break;
        }
    }
    return ans;
}
 
// Driver Code
public static void Main(String[] args)
{
    int []arr = { 3, 5, 1, 4, 2 };
 
    // Length of array
    int n = arr.Length;
 
    Console.Write(longestSequence(n, arr));
}
}
 
// This code is contributed by Amit Katiyar

Javascript




<script>
// Javascript Program to print the
// longest increasing
// subsequence from the
// boundary elements of an array
 
// Function to return the length of
// Longest Increasing subsequence
function longestSequence(n, arr)
{
 
    // Set pointers at
    // both ends
    var l = 0, r = n - 1;
 
    // Stores the recent
    // value added to the
    // subsequence
    var prev = -1000000000;
 
    // Stores the length of
    // the subsequence
    var ans = 0;
 
    while (l <= r) {
 
        // Check if both elements
        // can be added to the
        // subsequence
        if (arr[l] > prev
            && arr[r] > prev) {
 
            if (arr[l] < arr[r]) {
                ans += 1;
                prev = arr[l];
                l += 1;
            }
            else {
                ans += 1;
                prev = arr[r];
                r -= 1;
            }
        }
 
        // Check if the element
        // on the left can be
        // added to the
        // subsequence only
        else if (arr[l] > prev) {
            ans += 1;
            prev = arr[l];
            l += 1;
        }
 
        // Check if the element
        // on the right can be
        // added to the
        // subsequence only
        else if (arr[r] > prev) {
            ans += 1;
            prev = arr[r];
            r -= 1;
        }
 
        // If none of the values
        // can be added to the
        // subsequence
        else {
            break;
        }
    }
    return ans;
}
 
// Driver Code
var arr = [ 3, 5, 1, 4, 2 ];
 
// Length of array
var n = arr.length;
document.write( longestSequence(n, arr));
 
// This code is contributed by itsok.
</script>   
Output: 



4

 

Time Complexity: O(N)
 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  Get hold of all the important mathematical concepts for competitive programming with the Essential Maths for CP Course at a student-friendly price.

In case you wish to attend live classes with industry experts, please refer Geeks Classes Live and Geeks Classes Live USA

My Personal Notes arrow_drop_up
Recommended Articles
Page :