Longest subsequence such that difference between adjacents is one

Difficulty Level : Easy
Last Updated : 24 May, 2018

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)

This article is contributed by Sahil Chhabra (KILLER). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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.

My Personal Notes

Related Articles

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

C++

Java

Python

C#

PHP