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

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(n^{2}) 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 <bits/stdc++.h>` `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<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.find(arr[i]-1) != um.end() &&` ` ` `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.find(arr[i]+1) != um.end() &&` ` ` `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` `int` `main()` `{` ` ` `int` `arr[] = {1, 2, 3, 4, 5, 3, 2};` ` ` `int` `n = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]);` ` ` `cout << ` `"Longest length subsequence = "` ` ` `<< longLenSub(arr, n);` ` ` `return` `0;` `} ` |

## 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<Integer,` ` ` `Integer> um = ` `new` `HashMap<Integer,` ` ` `Integer>();` ` ` ` ` `// 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

`<script>` `// JavaScript implementation to find longest subsequence` `// such that difference between adjacents is one` ` ` ` ` `// function to find longest subsequence such` `// that difference between adjacents is one` ` ` `function` `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` ` ` `let um = ` `new` `Map();` ` ` ` ` `// to store the longest length subsequence` ` ` `let longLen = 0;` ` ` ` ` `// traverse the array elements` ` ` `for` `(let i = 0; i < n; i++)` ` ` `{` ` ` `// initialize current length` ` ` `// for element arr[i] as 0` ` ` `let len = 0;` ` ` ` ` `// if 'arr[i]-1' is in 'um' and its length` ` ` `// of subsequence is greater than 'len'` ` ` `if` `(um.has(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.has(arr[i] + 1) &&` ` ` `len < um.get(arr[i] + 1))` ` ` `len = um.get(arr[i] + 1);` ` ` ` ` `// update arr[i] subsequence length in 'um'` ` ` `um.set(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` ` ` `let arr=[1, 2, 3, 4, 5, 3, 2];` ` ` `let n = arr.length;` ` ` `document.write(` `"Longest length subsequence = "` `+` ` ` `longLenSub(arr, n));` ` ` `// This code is contributed by unknown2108` `</script>` |

**Output:**

Longest length subsequence = 6

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

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