Longest subsequence such that difference between adjacents is one

Given an array of n size, the task is to find the longest subsequence such that difference between adjacents is one.

Examples:

Input :  arr[] = {10, 9, 4, 5, 4, 8, 6}
Output :  3
As longest subsequences with difference 1 are, "10, 9, 8", 
"4, 5, 4" and "4, 5, 6"

Input :  arr[] = {1, 2, 3, 2, 3, 7, 2, 1}
Output :  7
As longest consecutive sequence is "1, 2, 3, 2, 3, 2, 1"

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

This problem is based upon the concept of Longest Increasing Subsequence Problem.

Let arr[0..n-1] be the input array and 
dp[i] be the length of the longest subsequence (with
differences one) ending at index i such that arr[i] 
is the last element of the subsequence.

Then, dp[i] can be recursively written as:
dp[i] = 1 + max(dp[j]) where 0 < j < i and 
       [arr[j] = arr[i] -1  or arr[j] = arr[i] + 1]
dp[i] = 1, if no such j exists.

To find the result for a given array, we need 
to return max(dp[i]) where 0 < i < n.

Following is a Dynamic Programming based implementation. It follows the recursive structure discussed above.

C++

// C++ program to find the longest subsequence such
// the difference between adjacent elements of the
// subsequence is one.
#include<bits/stdc++.h>
using namespace std;

// Function to find the length of longest subsequence
int longestSubseqWithDiffOne(int arr[], int n)
{
    // Initialize the dp[] array with 1 as a
    // single element will be of 1 length
    int dp[n];
    for (int i = 0; i< n; i++)
        dp[i] = 1;

    // Start traversing the given array
    for (int i=1; i<n; i++)
    {
        // Compare with all the previous elements
        for (int j=0; j<i; j++)
        {
            // If the element is consecutive then
            // consider this subsequence and update
            // dp[i] if required.
            if ((arr[i] == arr[j]+1) ||
                (arr[i] == arr[j]-1))

                dp[i] = max(dp[i], dp[j]+1);
        }
    }

    // Longest length will be the maximum value
    // of dp array.
    int result = 1;
    for (int i = 0 ; i < n ; i++)
        if (result < dp[i])
            result = dp[i];
    return result;
}

// Driver code
int main()
{
    // Longest subsequence with one difference is
    // {1, 2, 3, 4, 3, 2}
    int arr[] = {1, 2, 3, 4, 5, 3, 2};
    int n = sizeof(arr)/sizeof(arr[0]);
    cout << longestSubseqWithDiffOne(arr, n);
    return 0;
}

Java

// Java program to find the longest subsequence
// such that the difference between adjacent
// elements of the subsequence is one.
import java.io.*;

class GFG {
    
    // Function to find the length of longest 
    // subsequence
    static int longestSubseqWithDiffOne(int arr[], 
                                           int n)
    {
        // Initialize the dp[] array with 1 as a
        // single element will be of 1 length
        int dp[] = new int[n];
        for (int i = 0; i< n; i++)
            dp[i] = 1;

        // Start traversing the given array
        for (int i = 1; i < n; i++)
        {
            // Compare with all the previous
            // elements
            for (int j = 0; j < i; j++)
            {
                // If the element is consecutive 
                // then consider this subsequence
                // and update dp[i] if required.
                if ((arr[i] == arr[j] + 1) ||
                    (arr[i] == arr[j] - 1))

                dp[i] = Math.max(dp[i], dp[j]+1);
            }
        }

        // Longest length will be the maximum 
        // value of dp array.
        int result = 1;
        for (int i = 0 ; i < n ; i++)
            if (result < dp[i])
                result = dp[i];
        return result;
    }

    // Driver code
    public static void main(String[] args)
    {
        // Longest subsequence with one 
        // difference is
        // {1, 2, 3, 4, 3, 2}
        int arr[] = {1, 2, 3, 4, 5, 3, 2};
        int n = arr.length;
        System.out.println(longestSubseqWithDiffOne(
                                           arr, n));
    }
}

// This code is contributed by Prerna Saini

Python

# Function to find the length of longest subsequence
def longestSubseqWithDiffOne(arr, n):
    # Initialize the dp[] array with 1 as a
    # single element will be of 1 length
    dp = [1 for i in range(n)]

    # Start traversing the given array
    for i in range(n):
        # Compare with all the previous elements
        for j in range(i):
            # If the element is consecutive then
            # consider this subsequence and update
            # dp[i] if required.
            if ((arr[i] == arr[j]+1) or (arr[i] == arr[j]-1)):
                dp[i] = max(dp[i], dp[j]+1)

    # Longest length will be the maximum value
    # of dp array.
    result = 1   
    for i in range(n):
        if (result < dp[i]):
            result = dp[i]
           
    return result

# Driver code
arr = [1, 2, 3, 4, 5, 3, 2]
# Longest subsequence with one difference is
# {1, 2, 3, 4, 3, 2}
n = len(arr)
print longestSubseqWithDiffOne(arr, n)

# This code is contributed by Afzal Ansari

C#

// C# program to find the longest subsequence
// such that the difference between adjacent
// elements of the subsequence is one.
using System;

class GFG {
    
    // Function to find the length of longest 
    // subsequence
    static int longestSubseqWithDiffOne(int []arr, 
                                           int n)
    {
        
        // Initialize the dp[] array with 1 as a
        // single element will be of 1 length
        int []dp = new int[n];
        
        for (int i = 0; i< n; i++)
            dp[i] = 1;

        // Start traversing the given array
        for (int i = 1; i < n; i++)
        {
            
            // Compare with all the previous
            // elements
            for (int j = 0; j < i; j++)
            {
                // If the element is consecutive 
                // then consider this subsequence
                // and update dp[i] if required.
                if ((arr[i] == arr[j] + 1) ||
                         (arr[i] == arr[j] - 1))

                dp[i] = Math.Max(dp[i], dp[j]+1);
            }
        }

        // Longest length will be the maximum 
        // value of dp array.
        int result = 1;
        for (int i = 0 ; i < n ; i++)
            if (result < dp[i])
                result = dp[i];
                
        return result;
    }

    // Driver code
    public static void Main()
    {
        
        // Longest subsequence with one 
        // difference is
        // {1, 2, 3, 4, 3, 2}
        int []arr = {1, 2, 3, 4, 5, 3, 2};
        int n = arr.Length;
        
        Console.Write(
            longestSubseqWithDiffOne(arr, n));
    }
}

// This code is contributed by nitin mittal.

PHP

<?php
// PHP program to find the longest
// subsequence such the difference 
// between adjacent elements of the
// subsequence is one.

// Function to find the length of
// longest subsequence
function longestSubseqWithDiffOne($arr, $n)
{
    
    // Initialize the dp[] 
    // array with 1 as a
    // single element will 
    // be of 1 length
    $dp[$n] = 0;
    
    for($i = 0; $i< $n; $i++)
        $dp[$i] = 1;

    // Start traversing the
    // given array
    for($i = 1; $i < $n; $i++)
    {
        
        // Compare with all the 
        // previous elements
        for($j = 0; $j < $i; $j++)
        {
            
            // If the element is
            // consecutive then
            // consider this 
            // subsequence and 
            // update dp[i] if 
            // required.
            if (($arr[$i] == $arr[$j] + 1) ||
                ($arr[$i] == $arr[$j] - 1))

                $dp[$i] = max($dp[$i],
                         $dp[$j] + 1);
        }
    }

    // Longest length will be 
    // the maximum value
    // of dp array.
    $result = 1;
    for($i = 0 ; $i < $n ; $i++)
        if ($result < $dp[$i])
            $result = $dp[$i];
    return $result;
}

    // Driver code
    // Longest subsequence with
    // one difference is
    // {1, 2, 3, 4, 3, 2}
    $arr = array(1, 2, 3, 4, 5, 3, 2);
    $n = sizeof($arr);
    echo longestSubseqWithDiffOne($arr, $n);
    
// This code is contributed by nitin mittal. 
?>

Output:

Time Complexity: O(n²)
Auxiliary Space: O(n)

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My Personal Notes

Improved By : nitin mittal

Practice Tags :

Dynamic Programming