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# Longest subsequence such that difference between adjacents is one | Set 2

• Difficulty Level : Medium
• Last Updated : 19 May, 2021

Given an array of size n. The task is to find the longest subsequence such that difference between adjacents is one. Time Complexity of O(n) is required.
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".```

Method 1: Previously an approach having time complexity of O(n2) have been discussed in this post.
Method 2 (Efficient Approach): The idea is to create a hash map having tuples in the form (ele, len), where len denotes the length of the longest subsequence ending with the element ele. Now, for each element arr[i] we can find the length of the values arr[i]-1 and arr[i]+1 in the hash table and consider the maximum among them. Let this maximum value be max. Now, the length of longest subsequence ending with arr[i] would be max+1. Update this length along with the element arr[i] in the hash table. Finally, the element having the maximum length in the hash table gives the longest length subsequence.

## C++

 `// C++ implementation to find longest subsequence``// such that difference between adjacents is one``#include ``using` `namespace` `std;``  ` `// function to find longest subsequence such``// that difference between adjacents is one``int` `longLenSub(``int` `arr[], ``int` `n)``{``    ``// hash table to map the array element with the``    ``// length of the longest subsequence of which``    ``// it is a part of and is the last element of``    ``// that subsequence``    ``unordered_map<``int``, ``int``> um;``     ` `    ``// to store the longest length subsequence``    ``int` `longLen = 0;``     ` `    ``// traverse the array elements``    ``for` `(``int` `i=0; i

## Java

 `// Java implementation to find longest subsequence``// such that difference between adjacents is one``import` `java.util.*;` `class` `GFG``{``    ` `// function to find longest subsequence such``// that difference between adjacents is one``static` `int` `longLenSub(``int` `[]arr, ``int` `n)``{``    ``// hash table to map the array element with the``    ``// length of the longest subsequence of which``    ``// it is a part of and is the last element of``    ``// that subsequence``    ``HashMap um = ``new` `HashMap();``    ` `    ``// to store the longest length subsequence``    ``int` `longLen = ``0``;``    ` `    ``// traverse the array elements``    ``for` `(``int` `i = ``0``; i < n; i++)``    ``{``        ``// initialize current length``        ``// for element arr[i] as 0``        ``int` `len = ``0``;``        ` `        ``// if 'arr[i]-1' is in 'um' and its length``        ``// of subsequence is greater than 'len'``        ``if` `(um.containsKey(arr[i] - ``1``) &&``              ``len < um.get(arr[i] - ``1``))``              ``len = um.get(arr[i] - ``1``);``        ` `        ``// if 'arr[i]+1' is in 'um' and its length``        ``// of subsequence is greater than 'len'    ``        ``if` `(um.containsKey(arr[i] + ``1``) &&``              ``len < um.get(arr[i] + ``1``))``              ``len = um.get(arr[i] + ``1``);``        ` `        ``// update arr[i] subsequence length in 'um'``        ``um. put(arr[i], len + ``1``);``        ` `        ``// update longest length``        ``if` `(longLen < um.get(arr[i]))``            ``longLen = um.get(arr[i]);``    ``}``        ` `    ``// required longest length subsequence``    ``return` `longLen;    ``}``    ` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``int``[] arr = {``1``, ``2``, ``3``, ``4``, ``5``, ``3``, ``2``};``    ``int` `n = arr.length;``    ``System.out.println(``"Longest length subsequence = "` `+``                                    ``longLenSub(arr, n));``}``}` `// This code is contributed by Princi Singh`

## Python3

 `# Python3 implementation to find longest``# subsequence such that difference between``# adjacents is one``from` `collections ``import` `defaultdict` `# function to find longest subsequence such``# that difference between adjacents is one``def` `longLenSub(arr, n):` `    ``# hash table to map the array element``    ``# with the length of the longest``    ``# subsequence of which it is a part of``    ``# and is the last element of that subsequence``    ``um ``=` `defaultdict(``lambda``:``0``)``    ``longLen ``=` `0``    ``for` `i ``in` `range``(n):` `        ``# / initialize current length``        ``# for element arr[i] as 0``        ``len1 ``=` `0` `        ``# if 'arr[i]-1' is in 'um' and its length``        ``# of subsequence is greater than 'len'``        ``if` `(arr[i ``-` `1``] ``in` `um ``and``            ``len1 < um[arr[i] ``-` `1``]):``            ``len1 ``=` `um[arr[i] ``-` `1``]` `        ``# f 'arr[i]+1' is in 'um' and its length``        ``# of subsequence is greater than 'len'    ``        ``if` `(arr[i] ``+` `1` `in` `um ``and``            ``len1 < um[arr[i] ``+` `1``]):``            ``len1 ``=` `um[arr[i] ``+` `1``]` `        ``# update arr[i] subsequence``        ``# length in 'um'    ``        ``um[arr[i]] ``=` `len1 ``+` `1` `        ``# update longest length``        ``if` `longLen < um[arr[i]]:``            ``longLen ``=` `um[arr[i]]` `    ``# required longest length``    ``# subsequence``    ``return` `longLen` `# Driver code``arr ``=` `[``1``, ``2``, ``3``, ``4``, ``5``, ``3``, ``2``]``n ``=` `len``(arr)``print``(``"Longest length subsequence ="``,``                  ``longLenSub(arr, n))` `# This code is contributed by Shrikant13`

## C#

 `// C# implementation to find longest subsequence``// such that difference between adjacents is one``using` `System;``using` `System.Collections.Generic;` `class` `GFG``{``    ` `// function to find longest subsequence such``// that difference between adjacents is one``static` `int` `longLenSub(``int` `[]arr, ``int` `n)``{``    ``// hash table to map the array element with the``    ``// length of the longest subsequence of which``    ``// it is a part of and is the last element of``    ``// that subsequence``    ``Dictionary<``int``,``               ``int``> um = ``new` `Dictionary<``int``,``                                        ``int``>();``    ` `    ``// to store the longest length subsequence``    ``int` `longLen = 0;``    ` `    ``// traverse the array elements``    ``for` `(``int` `i = 0; i < n; i++)``    ``{``        ``// initialize current length``        ``// for element arr[i] as 0``        ``int` `len = 0;``        ` `        ``// if 'arr[i]-1' is in 'um' and its length``        ``// of subsequence is greater than 'len'``        ``if` `(um.ContainsKey(arr[i] - 1) &&``            ``len < um[arr[i] - 1])``            ``len = um[arr[i] - 1];``        ` `        ``// if 'arr[i]+1' is in 'um' and its length``        ``// of subsequence is greater than 'len'    ``        ``if` `(um.ContainsKey(arr[i] + 1) &&``            ``len < um[arr[i] + 1])``            ``len = um[arr[i] + 1];``        ` `        ``// update arr[i] subsequence length in 'um'``        ``um[arr[i]] = len + 1;``        ` `        ``// update longest length``        ``if` `(longLen < um[arr[i]])``            ``longLen = um[arr[i]];``    ``}``        ` `    ``// required longest length subsequence``    ``return` `longLen;    ``}``    ` `// Driver program to test above``static` `void` `Main()``{``    ``int``[] arr = {1, 2, 3, 4, 5, 3, 2};``    ``int` `n = arr.Length;``    ``Console.Write(``"Longest length subsequence = "` `+``                               ``longLenSub(arr, n));``}``}` `// This code is contributed by Mohit Kumar`

## Javascript

 ``

Output:

`Longest length subsequence = 6`

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