# Longest Consecutive Subsequence

Given an array of integers, find the length of the longest sub-sequence such that elements in the subsequence are consecutive integers, the consecutive numbers can be in any order.

**Examples**:

Input: arr[] = {1, 9, 3, 10, 4, 20, 2}

Output: 4

The subsequence 1, 3, 4, 2 is the longest subsequence

of consecutive elements

Input: arr[] = {36, 41, 56, 35, 44, 33, 34, 92, 43, 32, 42}

Output: 5

The subsequence 36, 35, 33, 34, 32 is the longest subsequence

of consecutive elements.

**One Solution **is to first sort the array and find the longest subarray with consecutive elements. Time complexity of this solution is O(nLogn). Thanks to Hao.W for suggesting this solution.

We can solve this problem in O(n) time using an **Efficient Solution**. The idea is to use Hashing. We first insert all elements in a Set. Then check all the possible starts of consecutive subsequences. Below is the complete algorithm.

- Create an empty hash.
- Insert all array elements to hash.
- Do following for every element arr[i]
- Check if this element is the starting point of a subsequence. To check this, we simply look for arr[i] – 1 in the hash, if not found, then this is the first element a subsequence.
- If this element is the first element, then count number of elements in the consecutive starting with this element. Iterate from arr[i] + 1 till the last element that can be found.
- If the count is more than the previous longest subsequence found, then update this.

Below image is a dry run of the above approach:

Below is the implementation of the above approach:

## C/C++

`// C++ program to find longest contiguous subsequence ` `#include<bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Returns length of the longest contiguous subsequence ` `int` `findLongestConseqSubseq(` `int` `arr[], ` `int` `n) ` `{ ` ` ` `unordered_set<` `int` `> S; ` ` ` `int` `ans = 0; ` ` ` ` ` `// Hash all the array elements ` ` ` `for` `(` `int` `i = 0; i < n; i++) ` ` ` `S.insert(arr[i]); ` ` ` ` ` `// check each possible sequence from the start ` ` ` `// then update optimal length ` ` ` `for` `(` `int` `i=0; i<n; i++) ` ` ` `{ ` ` ` `// if current element is the starting ` ` ` `// element of a sequence ` ` ` `if` `(S.find(arr[i]-1) == S.end()) ` ` ` `{ ` ` ` `// Then check for next elements in the ` ` ` `// sequence ` ` ` `int` `j = arr[i]; ` ` ` `while` `(S.find(j) != S.end()) ` ` ` `j++; ` ` ` ` ` `// update optimal length if this length ` ` ` `// is more ` ` ` `ans = max(ans, j - arr[i]); ` ` ` `} ` ` ` `} ` ` ` `return` `ans; ` `} ` ` ` `// Driver program ` `int` `main() ` `{ ` ` ` `int` `arr[] = {1, 9, 3, 10, 4, 20, 2}; ` ` ` `int` `n = ` `sizeof` `arr/ ` `sizeof` `arr[0]; ` ` ` `cout << ` `"Length of the Longest contiguous subsequence is "` ` ` `<< findLongestConseqSubseq(arr, n); ` ` ` `return` `0; ` `} ` |

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## Java

`// Java program to find longest consecutive subsequence ` `import` `java.io.*; ` `import` `java.util.*; ` ` ` `class` `ArrayElements ` `{ ` ` ` `// Returns length of the longest consecutive subsequence ` ` ` `static` `int` `findLongestConseqSubseq(` `int` `arr[],` `int` `n) ` ` ` `{ ` ` ` `HashSet<Integer> S = ` `new` `HashSet<Integer>(); ` ` ` `int` `ans = ` `0` `; ` ` ` ` ` `// Hash all the array elements ` ` ` `for` `(` `int` `i=` `0` `; i<n; ++i) ` ` ` `S.add(arr[i]); ` ` ` ` ` `// check each possible sequence from the start ` ` ` `// then update optimal length ` ` ` `for` `(` `int` `i=` `0` `; i<n; ++i) ` ` ` `{ ` ` ` `// if current element is the starting ` ` ` `// element of a sequence ` ` ` `if` `(!S.contains(arr[i]-` `1` `)) ` ` ` `{ ` ` ` `// Then check for next elements in the ` ` ` `// sequence ` ` ` `int` `j = arr[i]; ` ` ` `while` `(S.contains(j)) ` ` ` `j++; ` ` ` ` ` `// update optimal length if this length ` ` ` `// is more ` ` ` `if` `(ans<j-arr[i]) ` ` ` `ans = j-arr[i]; ` ` ` `} ` ` ` `} ` ` ` `return` `ans; ` ` ` `} ` ` ` ` ` `// Testing program ` ` ` `public` `static` `void` `main(String args[]) ` ` ` `{ ` ` ` `int` `arr[] = {` `1` `, ` `9` `, ` `3` `, ` `10` `, ` `4` `, ` `20` `, ` `2` `}; ` ` ` `int` `n = arr.length; ` ` ` `System.out.println(` `"Length of the Longest consecutive subsequence is "` `+ ` ` ` `findLongestConseqSubseq(arr,n)); ` ` ` `} ` `} ` `// This code is contributed by Aakash Hasija ` |

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## Python

`# Python program to find longest contiguous subsequence ` ` ` `from` `sets ` `import` `Set` `def` `findLongestConseqSubseq(arr, n): ` ` ` ` ` `s ` `=` `Set` `() ` ` ` `ans` `=` `0` ` ` ` ` `# Hash all the array elements ` ` ` `for` `ele ` `in` `arr: ` ` ` `s.add(ele) ` ` ` ` ` `# check each possible sequence from the start ` ` ` `# then update optimal length ` ` ` `for` `i ` `in` `range` `(n): ` ` ` ` ` `# if current element is the starting ` ` ` `# element of a sequence ` ` ` `if` `(arr[i]` `-` `1` `) ` `not` `in` `s: ` ` ` ` ` `# Then check for next elements in the ` ` ` `# sequence ` ` ` `j` `=` `arr[i] ` ` ` `while` `(j ` `in` `s): ` ` ` `j` `+` `=` `1` ` ` ` ` `# update optimal length if this length ` ` ` `# is more ` ` ` `ans` `=` `max` `(ans, j` `-` `arr[i]) ` ` ` `return` `ans ` ` ` `# Driver function ` `if` `__name__` `=` `=` `'__main__'` `: ` ` ` `n ` `=` `7` ` ` `arr ` `=` `[` `1` `, ` `9` `, ` `3` `, ` `10` `, ` `4` `, ` `20` `, ` `2` `] ` ` ` `print` `"Length of the Longest contiguous subsequence is "` `, ` ` ` `print` `findLongestConseqSubseq(arr, n) ` ` ` `# Contributed by: Harshit Sidhwa ` |

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## C#

`using` `System; ` `using` `System.Collections.Generic; ` ` ` `// C# program to find longest consecutive subsequence ` ` ` `public` `class` `ArrayElements ` `{ ` ` ` `// Returns length of the longest consecutive subsequence ` ` ` `public` `static` `int` `findLongestConseqSubseq(` `int` `[] arr, ` `int` `n) ` ` ` `{ ` ` ` `HashSet<` `int` `> S = ` `new` `HashSet<` `int` `>(); ` ` ` `int` `ans = 0; ` ` ` ` ` `// Hash all the array elements ` ` ` `for` `(` `int` `i = 0; i < n; ++i) ` ` ` `{ ` ` ` `S.Add(arr[i]); ` ` ` `} ` ` ` ` ` `// check each possible sequence from the start ` ` ` `// then update optimal length ` ` ` `for` `(` `int` `i = 0; i < n; ++i) ` ` ` `{ ` ` ` `// if current element is the starting ` ` ` `// element of a sequence ` ` ` `if` `(!S.Contains(arr[i] - 1)) ` ` ` `{ ` ` ` `// Then check for next elements in the ` ` ` `// sequence ` ` ` `int` `j = arr[i]; ` ` ` `while` `(S.Contains(j)) ` ` ` `{ ` ` ` `j++; ` ` ` `} ` ` ` ` ` `// update optimal length if this length ` ` ` `// is more ` ` ` `if` `(ans < j - arr[i]) ` ` ` `{ ` ` ` `ans = j - arr[i]; ` ` ` `} ` ` ` `} ` ` ` `} ` ` ` `return` `ans; ` ` ` `} ` ` ` ` ` `// Testing program ` ` ` `public` `static` `void` `Main(` `string` `[] args) ` ` ` `{ ` ` ` `int` `[] arr = ` `new` `int` `[] {1, 9, 3, 10, 4, 20, 2}; ` ` ` `int` `n = arr.Length; ` ` ` `Console.WriteLine(` `"Length of the Longest consecutive subsequence is "` `+ findLongestConseqSubseq(arr,n)); ` ` ` `} ` `} ` ` ` `// This code is contributed by Shrikant13 ` |

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**Output:**

Length of the Longest contiguous subsequence is 4

**Time Complexity:** At first look, time complexity looks more than O(n). If we take a closer look, we can notice that it is O(n) under the assumption that hash insert and search take O(1) time. The function S.find() inside the while loop is called at most twice for every element. For example, consider the case when all array elements are consecutive. In this case, the outer find is called for every element, but we go inside the if condition only for the smallest element. Once we are inside the if condition, we call find() one more time for every other element.

Thanks to Gaurav Ahirwar for above solution.

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