# Length of the longest subsequence consisting of distinct elements

Given an array **arr[]** of size **N**, the task is to find the length of the longest subsequence consisting of distinct elements only.

**Examples:**

Input:arr[] = {1, 1, 2, 2, 2, 3, 3}Output:3Explanation:

The longest subsequence with distinct elements is {1, 2, 3}Input:arr[] = { 1, 2, 3, 3, 4, 5, 5, 5 }Output:5

**Naive Approach:** The simplest approach is to generate all the subsequences of the array and check if it consists of only distinct elements or not. Keep updating the maximum length of such subsequences obtained. Finally, print the maximum length obtained.

**Time Complexity: **O(2^{N}) **Auxiliary Space: **O(1)

**Efficient Approach: **The length of the longest subsequence containing only distinct elements will be equal to the count of distinct elements in the array. Follow the steps below to solve the problem:

- Traverse the given array keep inserting encountered elements in a Hashset.
- Since HashSet consists of only unique elements, print the size of the HashSet as the required answer after completing the traversal of the array.

Below is the implementation of the above approach:

## C++

`// C++ program for the above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to find length of` `// the longest subsequence` `// consisting of distinct elements` `int` `longestSubseq(` `int` `arr[], ` `int` `n)` `{` ` ` `// Stores the distinct` ` ` `// array elements` ` ` `unordered_set<` `int` `> s;` ` ` `// Traverse the input array` ` ` `for` `(` `int` `i = 0; i < n; i++) {` ` ` `// If current element has not` ` ` `// occurred previously` ` ` `if` `(s.find(arr[i]) == s.end()) {` ` ` `// Insert it into set` ` ` `s.insert(arr[i]);` ` ` `}` ` ` `}` ` ` `return` `s.size();` `}` `// Driver Code` `int` `main()` `{` ` ` `// Given array` ` ` `int` `arr[] = { 1, 2, 3, 3, 4, 5, 5, 5 };` ` ` `int` `n = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]);` ` ` `// Function Call` ` ` `cout << longestSubseq(arr, n);` ` ` `return` `0;` `}` |

## Java

`// Java program for the above approach` `import` `java.util.*;` `class` `GFG{` ` ` `// Function to find length of` `// the longest subsequence` `// consisting of distinct elements` `static` `int` `longestSubseq(` `int` `arr[], ` `int` `n)` `{` ` ` `// Stores the distinct` ` ` `// array elements` ` ` `Set<Integer> s = ` `new` `HashSet<>();` ` ` ` ` `// Traverse the input array` ` ` `for` `(` `int` `i = ` `0` `; i < n; i++)` ` ` `{` ` ` ` ` `// If current element has not` ` ` `// occurred previously` ` ` `if` `(!s.contains(arr[i]))` ` ` `{` ` ` ` ` `// Insert it into set` ` ` `s.add(arr[i]);` ` ` `}` ` ` `}` ` ` `return` `s.size();` `}` `// Driver code` `public` `static` `void` `main (String[] args)` `{` ` ` ` ` `// Given array` ` ` `int` `arr[] = { ` `1` `, ` `2` `, ` `3` `, ` `3` `, ` `4` `, ` `5` `, ` `5` `, ` `5` `};` ` ` `int` `n = arr.length;` ` ` ` ` `// Function call` ` ` `System.out.println(longestSubseq(arr, n)); ` `}` `}` `// This code is contributed by offbeat` |

## Python3

`# Python3 program for` `# the above approach` `# Function to find length of` `# the longest subsequence` `# consisting of distinct elements` `def` `longestSubseq(arr, n):` ` ` `# Stores the distinct` ` ` `# array elements` ` ` `s ` `=` `set` `()` ` ` `# Traverse the input array` ` ` `for` `i ` `in` `range` `(n):` ` ` ` ` `# If current element has not` ` ` `# occurred previously` ` ` `if` `(arr[i] ` `not` `in` `s):` ` ` `# Insert it into set` ` ` `s.add(arr[i])` ` ` `return` `len` `(s)` `# Given array` `arr ` `=` `[` `1` `, ` `2` `, ` `3` `, ` `3` `,` ` ` `4` `, ` `5` `, ` `5` `, ` `5` `]` `n ` `=` `len` `(arr)` `# Function Call` `print` `(longestSubseq(arr, n))` `# This code is contributed by divyeshrabadiya07` |

## C#

`// C# program for the above approach` `using` `System;` `using` `System.Collections.Generic;` `class` `GFG{` ` ` `// Function to find length of` `// the longest subsequence` `// consisting of distinct elements` `static` `int` `longestSubseq(` `int` `[]arr, ` `int` `n)` `{` ` ` ` ` `// Stores the distinct` ` ` `// array elements` ` ` `HashSet<` `int` `> s = ` `new` `HashSet<` `int` `>();` ` ` ` ` `// Traverse the input array` ` ` `for` `(` `int` `i = 0; i < n; i++)` ` ` `{` ` ` ` ` `// If current element has not` ` ` `// occurred previously` ` ` `if` `(!s.Contains(arr[i]))` ` ` `{` ` ` ` ` `// Insert it into set` ` ` `s.Add(arr[i]);` ` ` `}` ` ` `}` ` ` `return` `s.Count;` `}` ` ` `// Driver code` `public` `static` `void` `Main(` `string` `[] args)` `{` ` ` ` ` `// Given array` ` ` `int` `[]arr = { 1, 2, 3, 3, 4, 5, 5, 5 };` ` ` `int` `n = arr.Length;` ` ` ` ` `// Function call` ` ` `Console.Write(longestSubseq(arr, n)); ` `}` `}` `// This code is contributed by rutvik_56` |

## C++

`#include <iostream>` `using` `namespace` `std;` `int` `main() {` ` ` `cout<<` `"GFG!"` `;` ` ` `return` `0;` `}` |

## Javascript

`<script>` ` ` `// Javascript program for the above approach` `// Function to find length of` `// the longest subsequence` `// consisting of distinct elements` `function` `longestSubseq(arr, n)` `{` ` ` `// Stores the distinct` ` ` `// array elements` ` ` `var` `s = ` `new` `Set();` ` ` ` ` `// Traverse the input array` ` ` `for` `(` `var` `i = 0; i < n; i++) {` ` ` ` ` `// If current element has not` ` ` `// occurred previously` ` ` `if` `(s.has(arr[i]) == ` `false` `) {` ` ` ` ` `// Insert it into set` ` ` `s.add(arr[i]);` ` ` `}` ` ` `}` ` ` ` ` `return` `s.size;` `}` ` ` `// Driver Code` ` ` `// Given array` `var` `arr = [ 1, 2, 3, 3, 4, 5, 5, 5 ];` `var` `n = arr.length;` `// Function Call` `document.write(longestSubseq(arr, n));` `// This code is contributed by ShubhamSingh10` `</script>` |

**Output:**

5

**Time Complexity: **O(N) **Auxiliary Space: **O(N)

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