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# 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"```

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 ``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`

## Python3

 `# 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

 ``

## Javascript

 ``

Output

`6`

Time Complexity: O(n2
Auxiliary Space: O(n)

Efficient Approach

## C++

 `#include``using` `namespace` `std;``int` `longestSubsequence(``int` `n, ``int` `arr[])``    ``{``        ``if``(n==1)``            ``return` `1;``        ``unordered_map<``int``,``int``> mapp;``        ``int` `res = 1;``        ``for``(``int` `i=0;i0 || mapp.count(arr[i]-1)>0){``                ``mapp[arr[i]]=1+max(mapp[arr[i]+1],mapp[arr[i]-1]);``            ``}``            ``else``                ``mapp[arr[i]]=1;``            ``res = max(res, mapp[arr[i]]);``        ``}``        ``return` `res;``          ``//This code is contributed by Akansha Mittal``    ``}``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 << longestSubsequence(n, arr);``    ``return` `0;``}`

## Java

 `import` `java.lang.Math;``import` `java.util.*;` `class` `GFG {` `    ``static` `int` `longestSubsequence(``int` `n, ``int` `arr[])``    ``{``        ``if` `(n == ``1``)``            ``return` `1``;``        ``Integer dp[] = ``new` `Integer[n];``        ``HashMap mapp = ``new` `HashMap<>();``        ``dp[``0``] = ``1``;``        ``mapp.put(arr[``0``], ``0``);``        ``for` `(``int` `i = ``1``; i < n; i++) {``            ``if` `(Math.abs(arr[i] - arr[i - ``1``]) == ``1``)``                ``dp[i] = dp[i - ``1``] + ``1``;``            ``else` `{``                ``if` `(mapp.containsKey(arr[i] + ``1``)``                    ``|| mapp.containsKey(arr[i] - ``1``)) {``                    ``dp[i] = ``1``                            ``+ Math.max(mapp.getOrDefault(``                                           ``arr[i] + ``1``, ``0``),``                                       ``mapp.getOrDefault(``                                           ``arr[i] - ``1``, ``0``));``                ``}``                ``else``                    ``dp[i] = ``1``;``            ``}``            ``mapp.put(arr[i], dp[i]);``        ``}``        ``return` `Collections.max(Arrays.asList(dp));``    ``}` `    ``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(longestSubsequence(n, arr));``    ``}``}` `// This code is contributed by rajsanghavi9.`

## Python3

 `def` `longestSubsequence(A, N):``    ``L ``=` `[``1``]``*``N``    ``hm ``=` `{}``    ``for` `i ``in` `range``(``1``,N):``        ``if` `abs``(A[i]``-``A[i``-``1``]) ``=``=` `1``:``            ``L[i] ``=` `1` `+` `L[i``-``1``]``        ``elif` `hm.get(A[i]``+``1``,``0``) ``or` `hm.get(A[i]``-``1``,``0``):``            ``L[i] ``=` `1``+``max``(hm.get(A[i]``+``1``,``0``), hm.get(A[i]``-``1``,``0``))``        ``hm[A[i]] ``=` `L[i]``    ``return` `max``(L)``# Driver code``A ``=`  `[``1``, ``2``, ``3``, ``4``, ``5``, ``3``, ``2``]``N ``=` `len``(A)``print``(longestSubsequence(A, N))`

## C#

 `using` `System;``using` `System.Collections.Generic;``using` `System.Linq;` `class` `GFG {` `    ``static` `int` `longestSubsequence(``int` `n, ``int``[] arr)``    ``{``        ``if` `(n == 1)``            ``return` `1;``        ``int``[] dp = ``new` `int``[n];``        ``Dictionary<``int``, ``int``> mapp = ``new` `Dictionary<``int``, ``int``>();``        ``dp[0] = 1;``        ``mapp.Add(arr[0], 0);``        ``for` `(``int` `i = 1; i < n; i++)``        ``{``            ``if` `(Math.Abs(arr[i] - arr[i - 1]) == 1)``                ``dp[i] = dp[i - 1] + 1;``            ``else``            ``{``                ``if` `(mapp.ContainsKey(arr[i] + 1)``                    ``|| mapp.ContainsKey(arr[i] - 1))``                ``{``                    ``dp[i] = 1``                  ``+ Math.Max(mapp.ContainsKey(arr[i] + 1) ? mapp[arr[i] + 1] : 0,``                             ``mapp.ContainsKey(arr[i] - 1) ? mapp[arr[i] - 1] : 0);``                ``}``                ``else``                    ``dp[i] = 1;``            ``}``            ``mapp[arr[i]] = dp[i];``        ``}``        ``return` `dp.Max();``    ``}` `    ``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;``        ``Console.WriteLine(longestSubsequence(n, arr));``    ``}``}`

## Javascript

 `function` `longestSubsequence(n, arr) {``        ``var` `L = Array(n).fill(1);``        ``hm = {};` `        ``for` `(``var` `i = 1; i < n; i++) {``          ``if` `(Math.abs(arr[i] - arr[i - 1]) == 1) L[i] = 1 + L[i - 1];``          ``else` `if` `(get(hm, arr[i] + 1, 0) || get(hm, arr[i] - 1, 0)) {``            ``L[i] = 1 + Math.max(get(hm, arr[i] + 1, 0), get(hm, arr[i] - 1, 0));``          ``}``          ``hm[arr[i]] = L[i];``        ``}` `        ``return` `Math.max(...L);``      ``}` `      ``function` `get(object, key, default_value) {``        ``var` `result = object[key];``        ``return` `typeof` `result !== ``"undefined"` `? result : default_value;``      ``}` `      ``var` `arr = [1, 2, 3, 4, 5, 3, 2];``      ``var` `n = arr.length;``      ``console.log(longestSubsequence(n, arr));``      ` `      ``// This code is contributed by satwiksuman.`

Output

`6`

Time Complexity : O(n)

Space Complexity : O(n)

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