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"

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 C++ 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

Output:

6

**Time Complexity: ** O(n^{2})

**Auxiliary Space: ** O(n)

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