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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