# 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 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(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.

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