Open In App
Related Articles

Longest alternating subsequence

Improve
Improve
Improve
Like Article
Like
Save Article
Save
Report issue
Report

A sequence {X1, X2, .. Xn} is an alternating sequence if its elements satisfy one of the following relations : 

  X1 < X2 > X3 < X4 > X5 < …. xn or 
  X1 > X2 < X3 > X4 < X5 > …. xn

Examples:

Input: arr[] = {1, 5, 4}
Output: 3
Explanation: The whole arrays is of the form  x1 < x2 > x3 

Input: arr[] = {10, 22, 9, 33, 49, 50, 31, 60}
Output: 6
Explanation: The subsequences {10, 22, 9, 33, 31, 60} or
{10, 22, 9, 49, 31, 60} or {10, 22, 9, 50, 31, 60}
are longest subsequence of length 6

Note: This problem is an extension of the longest increasing subsequence problem, but requires more thinking for finding optimal substructure property in this

Longest alternating subsequence using dynamic programming:

To solve the problem follow the below idea:

We will solve this problem by dynamic Programming method, as it has optimal substructure and overlapping subproblems

Follow the below steps to solve the problem:

  • Let A is given an array of length N 
  • We define a 2D array las[n][2] such that las[i][0] contains the longest alternating subsequence ending at index i and the last element is greater than its previous element 
  • las[i][1] contains the longest alternating subsequence ending at index i and the last element is smaller than its previous element, then we have the following recurrence relation between them,  

las[i][0] = Length of the longest alternating subsequence 
                  ending at index i and last element is greater
                  than its previous element

las[i][1] = Length of the longest alternating subsequence 
                  ending at index i and last element is smaller
                  than its previous element

Recursive Formulation:

   las[i][0] = max (las[i][0], las[j][1] + 1); 
                  for all j < i and A[j] < A[i] 

   las[i][1] = max (las[i][1], las[j][0] + 1); 
                 for all j < i and A[j] > A[i]

  • The first recurrence relation is based on the fact that, If we are at position i and this element has to be bigger than its previous element then for this sequence (upto i) to be bigger we will try to choose an element j ( < i) such that A[j] < A[i] i.e. A[j] can become A[i]’s previous element and las[j][1] + 1 is bigger than las[i][0] then we will update las[i][0]. 
  • Remember we have chosen las[j][1] + 1 not las[j][0] + 1 to satisfy the alternate property because in las[j][0] the last element is bigger than its previous one and A[i] is greater than A[j] which will break the alternating property if we update. So above fact derives the first recurrence relation, a similar argument can be made for the second recurrence relation also. 

Below is the implementation of the above approach:

C++

// C++ program to find longest alternating
// subsequence in an array
#include <bits/stdc++.h>
using namespace std;
 
// Function to return max of two numbers
int max(int a, int b) { return (a > b) ? a : b; }
 
// Function to return longest alternating
// subsequence length
int zzis(int arr[], int n)
{
 
    /*las[i][0] = Length of the longest
        alternating subsequence ending at
        index i and last element is greater
        than its previous element
    las[i][1] = Length of the longest
        alternating subsequence ending
        at index i and last element is
        smaller than its previous element */
    int las[n][2];
 
    // Initialize all values from 1
    for (int i = 0; i < n; i++)
        las[i][0] = las[i][1] = 1;
 
    // Initialize result
    int res = 1;
 
    // Compute values in bottom up manner
    for (int i = 1; i < n; i++) {
 
        // Consider all elements as
        // previous of arr[i]
        for (int j = 0; j < i; j++) {
 
            // If arr[i] is greater, then
            // check with las[j][1]
            if (arr[j] < arr[i]
                && las[i][0] < las[j][1] + 1)
                las[i][0] = las[j][1] + 1;
 
            // If arr[i] is smaller, then
            // check with las[j][0]
            if (arr[j] > arr[i]
                && las[i][1] < las[j][0] + 1)
                las[i][1] = las[j][0] + 1;
        }
 
        // Pick maximum of both values at index i
        if (res < max(las[i][0], las[i][1]))
            res = max(las[i][0], las[i][1]);
    }
    return res;
}
 
// Driver code
int main()
{
    int arr[] = { 10, 22, 9, 33, 49, 50, 31, 60 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << "Length of Longest alternating "
         << "subsequence is " << zzis(arr, n);
 
    return 0;
}
 
// This code is contributed by shivanisinghss2110

                    

C

// C program to find longest alternating subsequence in
// an array
#include <stdio.h>
#include <stdlib.h>
 
// function to return max of two numbers
int max(int a, int b) { return (a > b) ? a : b; }
 
// Function to return longest alternating subsequence length
int zzis(int arr[], int n)
{
    /*las[i][0] = Length of the longest alternating
     subsequence ending at index i and last element is
     greater than its previous element las[i][1] = Length of
     the longest alternating subsequence ending at index i
     and last element is smaller than its previous element
   */
    int las[n][2];
 
    /* Initialize all values from 1  */
    for (int i = 0; i < n; i++)
        las[i][0] = las[i][1] = 1;
 
    int res = 1; // Initialize result
 
    /* Compute values in bottom up manner */
    for (int i = 1; i < n; i++) {
        // Consider all elements as previous of arr[i]
        for (int j = 0; j < i; j++) {
            // If arr[i] is greater, then check with
            // las[j][1]
            if (arr[j] < arr[i]
                && las[i][0] < las[j][1] + 1)
                las[i][0] = las[j][1] + 1;
 
            // If arr[i] is smaller, then check with
            // las[j][0]
            if (arr[j] > arr[i]
                && las[i][1] < las[j][0] + 1)
                las[i][1] = las[j][0] + 1;
        }
 
        /* Pick maximum of both values at index i  */
        if (res < max(las[i][0], las[i][1]))
            res = max(las[i][0], las[i][1]);
    }
 
    return res;
}
 
/* Driver code */
int main()
{
    int arr[] = { 10, 22, 9, 33, 49, 50, 31, 60 };
    int n = sizeof(arr) / sizeof(arr[0]);
    printf(
        "Length of Longest alternating subsequence is %d\n",
        zzis(arr, n));
    return 0;
}

                    

Java

// Java program to find longest
// alternating subsequence in an array
import java.io.*;
 
class GFG {
 
    // Function to return longest
    // alternating subsequence length
    static int zzis(int arr[], int n)
    {
        /*las[i][0] = Length of the longest
            alternating subsequence ending at
            index i and last element is
            greater than its previous element
        las[i][1] = Length of the longest
            alternating subsequence ending at
            index i and last element is
            smaller than its previous
            element */
        int las[][] = new int[n][2];
 
        /* Initialize all values from 1 */
        for (int i = 0; i < n; i++)
            las[i][0] = las[i][1] = 1;
 
        int res = 1; // Initialize result
 
        /* Compute values in bottom up manner */
        for (int i = 1; i < n; i++) {
            // Consider all elements as
            // previous of arr[i]
            for (int j = 0; j < i; j++) {
                // If arr[i] is greater, then
                // check with las[j][1]
                if (arr[j] < arr[i]
                    && las[i][0] < las[j][1] + 1)
                    las[i][0] = las[j][1] + 1;
 
                // If arr[i] is smaller, then
                // check with las[j][0]
                if (arr[j] > arr[i]
                    && las[i][1] < las[j][0] + 1)
                    las[i][1] = las[j][0] + 1;
            }
 
            /* Pick maximum of both values at
            index i */
            if (res < Math.max(las[i][0], las[i][1]))
                res = Math.max(las[i][0], las[i][1]);
        }
 
        return res;
    }
 
    /* Driver code*/
    public static void main(String[] args)
    {
        int arr[] = { 10, 22, 9, 33, 49, 50, 31, 60 };
        int n = arr.length;
        System.out.println("Length of Longest "
                           + "alternating subsequence is "
                           + zzis(arr, n));
    }
}
// This code is contributed by Prerna Saini

                    

Python3

# Python3 program to find longest
# alternating subsequence in an array
 
# Function to return max of two numbers
 
 
def Max(a, b):
 
    if a > b:
        return a
    else:
        return b
 
# Function to return longest alternating
# subsequence length
 
 
def zzis(arr, n):
    """las[i][0] = Length of the longest
        alternating subsequence ending at
        index i and last element is greater
        than its previous element
    las[i][1] = Length of the longest
        alternating subsequence ending
        at index i and last element is
        smaller than its previous element"""
    las = [[0 for i in range(2)]
           for j in range(n)]
 
    # Initialize all values from 1
    for i in range(n):
        las[i][0], las[i][1] = 1, 1
 
    # Initialize result
    res = 1
 
    # Compute values in bottom up manner
    for i in range(1, n):
 
        # Consider all elements as
        # previous of arr[i]
        for j in range(0, i):
 
            # If arr[i] is greater, then
            # check with las[j][1]
            if (arr[j] < arr[i] and
                    las[i][0] < las[j][1] + 1):
                las[i][0] = las[j][1] + 1
 
            # If arr[i] is smaller, then
            # check with las[j][0]
            if(arr[j] > arr[i] and
               las[i][1] < las[j][0] + 1):
                las[i][1] = las[j][0] + 1
 
        # Pick maximum of both values at index i
        if (res < max(las[i][0], las[i][1])):
            res = max(las[i][0], las[i][1])
 
    return res
 
 
# Driver Code
arr = [10, 22, 9, 33, 49, 50, 31, 60]
n = len(arr)
 
print("Length of Longest alternating subsequence is",
      zzis(arr, n))
 
# This code is contributed by divyesh072019

                    

C#

// C# program to find longest
// alternating subsequence
// in an array
using System;
 
class GFG {
 
    // Function to return longest
    // alternating subsequence length
    static int zzis(int[] arr, int n)
    {
        /*las[i][0] = Length of the
            longest alternating subsequence
            ending at index i and last
            element is greater than its
            previous element
        las[i][1] = Length of the longest
            alternating subsequence ending at
            index i and last element is
            smaller than its previous
            element */
        int[, ] las = new int[n, 2];
 
        /* Initialize all values from 1 */
        for (int i = 0; i < n; i++)
            las[i, 0] = las[i, 1] = 1;
 
        // Initialize result
        int res = 1;
 
        /* Compute values in
        bottom up manner */
        for (int i = 1; i < n; i++) {
            // Consider all elements as
            // previous of arr[i]
            for (int j = 0; j < i; j++) {
                // If arr[i] is greater, then
                // check with las[j][1]
                if (arr[j] < arr[i]
                    && las[i, 0] < las[j, 1] + 1)
                    las[i, 0] = las[j, 1] + 1;
 
                // If arr[i] is smaller, then
                // check with las[j][0]
                if (arr[j] > arr[i]
                    && las[i, 1] < las[j, 0] + 1)
                    las[i, 1] = las[j, 0] + 1;
            }
 
            /* Pick maximum of both
            values at index i */
            if (res < Math.Max(las[i, 0], las[i, 1]))
                res = Math.Max(las[i, 0], las[i, 1]);
        }
 
        return res;
    }
 
    // Driver Code
    public static void Main()
    {
        int[] arr = { 10, 22, 9, 33, 49, 50, 31, 60 };
        int n = arr.Length;
        Console.WriteLine("Length of Longest "
                          + "alternating subsequence is "
                          + zzis(arr, n));
    }
}
 
// This code is contributed by anuj_67.

                    

PHP

<?php
// PHP program to find longest
// alternating subsequence in
// an array
 
// Function to return longest
// alternating subsequence length
function zzis($arr, $n)
{
    /*las[i][0] = Length of the
        longest alternating subsequence
        ending at index i and last element
        is greater than its previous element
    las[i][1] = Length of the longest
        alternating subsequence ending at
        index i and last element is
        smaller than its previous element */
    $las = array(array());
 
    /* Initialize all values from 1 */
    for ( $i = 0; $i < $n; $i++)
        $las[$i][0] = $las[$i][1] = 1;
 
    $res = 1; // Initialize result
 
    /* Compute values in
    bottom up manner */
    for ( $i = 1; $i < $n; $i++)
    {
        // Consider all elements
        // as previous of arr[i]
        for ($j = 0; $j < $i; $j++)
        {
            // If arr[i] is greater, then
            // check with las[j][1]
            if ($arr[$j] < $arr[$i] and
                $las[$i][0] < $las[$j][1] + 1)
               $las[$i][0] = $las[$j][1] + 1;
 
            // If arr[i] is smaller, then
            // check with las[j][0]
            if($arr[$j] > $arr[$i] and
               $las[$i][1] < $las[$j][0] + 1)
                $las[$i][1] = $las[$j][0] + 1;
        }
 
        /* Pick maximum of both
        values at index i */
        if ($res < max($las[$i][0], $las[$i][1]))
            $res = max($las[$i][0], $las[$i][1]);
    }
 
    return $res;
}
 
// Driver Code
$arr = array(10, 22, 9, 33,
             49, 50, 31, 60 );
$n = count($arr);
echo "Length of Longest alternating " .
    "subsequence is ", zzis($arr, $n) ;
 
// This code is contributed by anuj_67.
?>

                    

Javascript

<script>
    // Javascript program to find longest
    // alternating subsequence in an array
     
    // Function to return longest
    // alternating subsequence length
    function zzis(arr, n)
    {
        /*las[i][0] = Length of the longest
            alternating subsequence ending at
            index i and last element is
            greater than its previous element
        las[i][1] = Length of the longest
            alternating subsequence ending at
            index i and last element is
            smaller than its previous
            element */
        let las = new Array(n);
        for (let i = 0; i < n; i++)
        {
            las[i] = new Array(2);
            for (let j = 0; j < 2; j++)
            {
                las[i][j] = 0;
            }
        }
 
        /* Initialize all values from 1 */
        for (let i = 0; i < n; i++)
            las[i][0] = las[i][1] = 1;
 
        let res = 1; // Initialize result
 
        /* Compute values in bottom up manner */
        for (let i = 1; i < n; i++)
        {
            // Consider all elements as
            // previous of arr[i]
            for (let j = 0; j < i; j++)
            {
                // If arr[i] is greater, then
                // check with las[j][1]
                if (arr[j] < arr[i] &&
                    las[i][0] < las[j][1] + 1)
                    las[i][0] = las[j][1] + 1;
 
                // If arr[i] is smaller, then
                // check with las[j][0]
                if( arr[j] > arr[i] &&
                  las[i][1] < las[j][0] + 1)
                    las[i][1] = las[j][0] + 1;
            }
 
            /* Pick maximum of both values at
            index i */
            if (res < Math.max(las[i][0], las[i][1]))
                res = Math.max(las[i][0], las[i][1]);
        }
 
        return res;
    }
     
    let arr = [ 10, 22, 9, 33, 49, 50, 31, 60 ];
    let n = arr.length;
    document.write("Length of Longest "+
                    "alternating subsequence is " +
                    zzis(arr, n));
     
    // This code is contributed by rameshtravel07.
</script>

                    

Output
Length of Longest alternating subsequence is 6

Time Complexity: O(N2
Auxiliary Space: O(N), since N extra space has been taken

Efficient Approach: To solve the problem follow the below idea: 

In the above approach, at any moment we are keeping track of two values (The length of the longest alternating subsequence ending at index i, and the last element is smaller than or greater than the previous element), for every element on the array. To optimize space, we only need to store two variables for element at any index i

inc = Length of longest alternative subsequence so far with current value being greater than it’s previous value.
dec = Length of longest alternative subsequence so far with current value being smaller than it’s previous value.
The tricky part of this approach is to update these two values. 

“inc” should be increased, if and only if the last element in the alternative sequence was smaller than it’s previous element.
“dec” should be increased, if and only if the last element in the alternative sequence was greater than it’s previous element.

Follow the below steps to solve the problem:

  • Declare two integers inc and dec equal to one
  • Run a loop for i [1, N-1]
    • If arr[i] is greater than the previous element then set inc equal to dec + 1
    • Else if arr[i] is smaller than the previous element then set dec equal to inc + 1
  • Return maximum of inc and dec

Below is the implementation of the above approach:

C++

// C++ program for above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function for finding
// longest alternating
// subsequence
int LAS(int arr[], int n)
{
 
    // "inc" and "dec" initialized as 1
    // as single element is still LAS
    int inc = 1;
    int dec = 1;
 
    // Iterate from second element
    for (int i = 1; i < n; i++) {
 
        if (arr[i] > arr[i - 1]) {
 
            // "inc" changes if "dec"
            // changes
            inc = dec + 1;
        }
 
        else if (arr[i] < arr[i - 1]) {
 
            // "dec" changes if "inc"
            // changes
            dec = inc + 1;
        }
    }
 
    // Return the maximum length
    return max(inc, dec);
}
 
// Driver Code
int main()
{
    int arr[] = { 10, 22, 9, 33, 49, 50, 31, 60 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    cout << LAS(arr, n) << endl;
    return 0;
}

                    

Java

// Java Program for above approach
public class GFG {
 
    // Function for finding
    // longest alternating
    // subsequence
    static int LAS(int[] arr, int n)
    {
 
        // "inc" and "dec" initialized as 1,
        // as single element is still LAS
        int inc = 1;
        int dec = 1;
 
        // Iterate from second element
        for (int i = 1; i < n; i++) {
 
            if (arr[i] > arr[i - 1]) {
                // "inc" changes if "dec"
                // changes
                inc = dec + 1;
            }
            else if (arr[i] < arr[i - 1]) {
 
                // "dec" changes if "inc"
                // changes
                dec = inc + 1;
            }
        }
 
        // Return the maximum length
        return Math.max(inc, dec);
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int[] arr = { 10, 22, 9, 33, 49, 50, 31, 60 };
        int n = arr.length;
 
        // Function Call
        System.out.println(LAS(arr, n));
    }
}

                    

Python3

# Python3 program for above approach
def LAS(arr, n):
 
    # "inc" and "dec" initialized as 1
    # as single element is still LAS
    inc = 1
    dec = 1
 
    # Iterate from second element
    for i in range(1, n):
 
        if (arr[i] > arr[i-1]):
 
            # "inc" changes if "dec"
            # changes
            inc = dec + 1
        elif (arr[i] < arr[i-1]):
 
            # "dec" changes if "inc"
            # changes
            dec = inc + 1
 
    # Return the maximum length
    return max(inc, dec)
 
 
# Driver Code
if __name__ == "__main__":
    arr = [10, 22, 9, 33, 49, 50, 31, 60]
    n = len(arr)
 
    # Function Call
    print(LAS(arr, n))

                    

C#

// C# program for above approach
using System;
 
class GFG {
 
    // Function for finding
    // longest alternating
    // subsequence
    static int LAS(int[] arr, int n)
    {
 
        // "inc" and "dec" initialized as 1,
        // as single element is still LAS
        int inc = 1;
        int dec = 1;
 
        // Iterate from second element
        for (int i = 1; i < n; i++) {
            if (arr[i] > arr[i - 1]) {
 
                // "inc" changes if "dec"
                // changes
                inc = dec + 1;
            }
            else if (arr[i] < arr[i - 1]) {
 
                // "dec" changes if "inc"
                // changes
                dec = inc + 1;
            }
        }
 
        // Return the maximum length
        return Math.Max(inc, dec);
    }
 
    // Driver code
    static void Main()
    {
        int[] arr = { 10, 22, 9, 33, 49, 50, 31, 60 };
        int n = arr.Length;
 
        // Function Call
        Console.WriteLine(LAS(arr, n));
    }
}
 
// This code is contributed by divyeshrabadiya07

                    

Javascript

<script>
    // Javascript program for above approach
     
    // Function for finding
    // longest alternating
    // subsequence
    function LAS(arr, n)
    {
 
        // "inc" and "dec" initialized as 1
        // as single element is still LAS
        let inc = 1;
        let dec = 1;
 
        // Iterate from second element
        for (let i = 1; i < n; i++)
        {
 
            if (arr[i] > arr[i - 1])
            {
 
                // "inc" changes if "dec"
                // changes
                inc = dec + 1;
            }
 
            else if (arr[i] < arr[i - 1])
            {
 
                // "dec" changes if "inc"
                // changes
                dec = inc + 1;
            }
        }
 
        // Return the maximum length
        return Math.max(inc, dec);
    }
 
    let arr = [ 10, 22, 9, 33, 49, 50, 31, 60 ];
    let n = arr.length;
  
    // Function Call
    document.write(LAS(arr, n));
     
     // This code is contributed by mukesh07.
</script>

                    

Output:

6

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



Last Updated : 21 Dec, 2022
Like Article
Save Article
Previous
Next
Share your thoughts in the comments
Similar Reads