Maximum length subsequence with difference between adjacent elements as either 0 or 1

20 October 2023

Given an array of n integers. The problem is to find maximum length of the subsequence with difference between adjacent elements as either 0 or 1.
Examples:

Input : arr[] = {2, 5, 6, 3, 7, 6, 5, 8}
Output : 5
The subsequence is {5, 6, 7, 6, 5}.

Input : arr[] = {-2, -1, 5, -1, 4, 0, 3}
Output : 4
The subsequence is {-2, -1, -1, 0}.

Source: Expedia Interview Experience | Set 12

The solution to this problem closely resembles the Longest Increasing Subsequence problem. The only difference is that here we have to check whether the absolute difference between the adjacent elements of the subsequence is either 0 or 1.

C++

// C++ implementation to find maximum length
// subsequence with difference between adjacent 
// elements as either 0 or 1
#include <bits/stdc++.h>
using namespace std;
 
// function to find maximum length subsequence 
// with difference between adjacent elements as
// either 0 or 1
int maxLenSub(int arr[], int n)
{
    int mls[n], max = 0;
     
    // Initialize mls[] values for all indexes
    for (int i=0; i<n; i++)
        mls[i] = 1;
     
    // Compute optimized maximum length subsequence 
    // values in bottom up manner
    for (int i=1; i<n; i++)
        for (int j=0; j<i; j++)
            if (abs(arr[i] - arr[j]) <= 1 &&
                    mls[i] < mls[j] + 1)
                mls[i] = mls[j] + 1;
     
    // Store maximum of all 'mls' values in 'max'    
    for (int i=0; i<n; i++)
        if (max < mls[i])
            max = mls[i];
     
    // required maximum length subsequence
    return max;        
}
 
// Driver program to test above
int main()
{
    int arr[] = {2, 5, 6, 3, 7, 6, 5, 8};
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << "Maximum length subsequence = "
         << maxLenSub(arr, n);
    return 0;
} 

Java

// JAVA Code for Maximum length subsequence 
// with difference between adjacent elements
// as either 0 or 1
import java.util.*;
 
class GFG {
     
    // function to find maximum length subsequence 
    // with difference between adjacent elements as
    // either 0 or 1
     public static int maxLenSub(int arr[], int n)
    {
        int mls[] = new int[n], max = 0;
          
        // Initialize mls[] values for all indexes
        for (int i = 0; i < n; i++)
            mls[i] = 1;
          
        // Compute optimized maximum length 
        // subsequence values in bottom up manner
        for (int i = 1; i < n; i++)
            for (int j = 0; j < i; j++)
                if (Math.abs(arr[i] - arr[j]) <= 1
                      && mls[i] < mls[j] + 1)
                    mls[i] = mls[j] + 1;
          
        // Store maximum of all 'mls' values in 'max'    
        for (int i = 0; i < n; i++)
            if (max < mls[i])
                max = mls[i];
          
        // required maximum length subsequence
        return max;        
    }
     
    /* Driver program to test above function */
    public static void main(String[] args) 
    {
        int arr[] = {2, 5, 6, 3, 7, 6, 5, 8};
        int n = arr.length;
        System.out.println("Maximum length subsequence = "+
                               maxLenSub(arr, n));
           
    }
}
   
// This code is contributed by Arnav Kr. Mandal.

Python3

# Python implementation to find maximum length
# subsequence with difference between adjacent 
# elements as either 0 or 1
 
# function to find maximum length subsequence 
# with difference between adjacent elements as
# either 0 or 1
def maxLenSub( arr, n):
    mls=[]
    max = 0
     
    #Initialize mls[] values for all indexes
    for i in range(n):
        mls.append(1)
     
    #Compute optimized maximum length subsequence
    # values in bottom up manner
    for i in range(n):
        for j in range(i):
            if (abs(arr[i] - arr[j]) <= 1 and mls[i] < mls[j] + 1):
                mls[i] = mls[j] + 1
                 
    # Store maximum of all 'mls' values in 'max'
    for i in range(n):
        if (max < mls[i]):
            max = mls[i]
     
    #required maximum length subsequence
    return max
     
#Driver program to test above
arr = [2, 5, 6, 3, 7, 6, 5, 8]
n = len(arr)
print("Maximum length subsequence = ",maxLenSub(arr, n))
 
#This code is contributed by "Abhishek Sharma 44"

C#

// C# Code for Maximum length subsequence
// with difference between adjacent elements
// as either 0 or 1
using System;
 
class GFG {
     
    // function to find maximum length subsequence
    // with difference between adjacent elements as
    // either 0 or 1
    public static int maxLenSub(int[] arr, int n)
    {
        int[] mls = new int[n];
        int max = 0;
 
        // Initialize mls[] values for all indexes
        for (int i = 0; i < n; i++)
            mls[i] = 1;
 
        // Compute optimized maximum length
        // subsequence values in bottom up manner
        for (int i = 1; i < n; i++)
            for (int j = 0; j < i; j++)
                if (Math.Abs(arr[i] - arr[j]) <= 1
                    && mls[i] < mls[j] + 1)
                    mls[i] = mls[j] + 1;
 
        // Store maximum of all 'mls' values in 'max'
        for (int i = 0; i < n; i++)
            if (max < mls[i])
                max = mls[i];
 
        // required maximum length subsequence
        return max;
    }
 
    /* Driver program to test above function */
    public static void Main()
    {
        int[] arr = { 2, 5, 6, 3, 7, 6, 5, 8 };
        int n = arr.Length;
        Console.Write("Maximum length subsequence = " + 
                                    maxLenSub(arr, n));
    }
}
 
// This code is contributed by Sam007

PHP

<?php
// PHP implementation to find maximum length
// subsequence with difference between adjacent 
// elements as either 0 or 1
 
// function to find maximum length subsequence 
// with difference between adjacent elements as
// either 0 or 1
function maxLenSub($arr, $n)
{
    $mls = array(); $max = 0;
     
    // Initialize mls[] values
    // for all indexes
    for($i = 0; $i < $n; $i++)
        $mls[$i] = 1;
     
    // Compute optimized maximum
    // length subsequence 
    // values in bottom up manner
    for ($i = 1; $i < $n; $i++)
        for ( $j = 0; $j < $i; $j++)
            if (abs($arr[$i] - $arr[$j]) <= 1 and
                         $mls[$i] < $mls[$j] + 1)
                $mls[$i] = $mls[$j] + 1;
     
    // Store maximum of all
    // 'mls' values in 'max' 
    for($i = 0; $i < $n; $i++)
        if ($max < $mls[$i])
            $max = $mls[$i];
     
    // required maximum 
    // length subsequence
    return $max;     
}
 
    // Driver Code
    $arr = array(2, 5, 6, 3, 7, 6, 5, 8);
    $n = count($arr);
    echo "Maximum length subsequence = "
        , maxLenSub($arr, $n);
         
// This code is contributed by anuj_67.
?>

Javascript

<script>
 
// Javascript Code for 
// Maximum length subsequence 
// with difference between 
// adjacent elements
// as either 0 or 1
 
    // function to find maximum 
    // length subsequence 
    // with difference between
    // adjacent elements as
    // either 0 or 1
     function  maxLenSub(arr, n)
    {
        let mls = new Array(n).fill(1), max = 0;
          
        // Initialize mls[] values for all indexes
        for (let i = 0; i < n; i++)
            mls[i] = 1;
          
        // Compute optimized maximum length 
        // subsequence values in bottom up manner
        for (let i = 1; i < n; i++)
            for (let j = 0; j < i; j++)
                if (Math.abs(arr[i] - arr[j]) <= 1 
                      && mls[i] < mls[j] + 1)
                    mls[i] = mls[j] + 1;
          
        // Store maximum of all 'mls' values in 'max'    
        for (let i = 0; i < n; i++)
            if (max < mls[i])
                max = mls[i];
          
        // required maximum length subsequence
        return max;        
    }
     
     
 
// driver program
 
        let arr = [2, 5, 6, 3, 7, 6, 5, 8];
        let n = arr.length;
        document.write("Maximum length subsequence = "+
                               maxLenSub(arr, n));
         
</script>

Output:

Maximum length subsequence = 5

Time Complexity: O(n²)
Auxiliary Space: O(n)
Maximum length subsequence with difference between adjacent elements as either 0 or 1 | Set 2
This article is contributed by Ayush Jauhari. If you like neveropen and would like to contribute, you can also write an article using write.neveropen.co.za or mail your article to review-team@neveropen.co.za. See your article appearing on the neveropen main page and help other Geeks.
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