# Longest permutation subsequence in a given array

Given an array arr containing N elements, find the length of the longest subsequence such that it is a valid permutation of a particular length. If no such permutation sequence exists then print 0.

Examples:

Input: arr[] = {3, 2, 1, 6, 5}
Output: 3
Explanation:
Longest permutation subsequence will be [3, 2, 1].

Input: arr[]= {2, 3, 4, 5}
Output: 0
Explanation:
No valid permutation subsequence present as element 1 is missing.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: The above-mentioned problem is on permutation subsequence so the order of the array elements is irrelevant, only what matter is the frequency of each element. If array is of length N then the maximum possible length for the permutation sequence is N and minimum possible length is 0. If the subsequence of length L is a valid permutation then all elements from 1 to L should be present.

1. Count the frequency of the elements in the range [1, N] from the array
2. Iterate through all elements from 1 to N in the array and count the iterations till a 0 frequency is observed. If the frequency of an element is ‘0’, return the current count of iterations as the required length.

Below is the implementation of the above approach:

## C++

 `// C++ Program to find length of ` `// Longest Permutaion Subsequence ` `// in a given array ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function to find the ` `// longest permutation subsequence ` `int` `longestPermutation(``int` `a[], ``int` `n) ` `{ ` ` `  `    ``// Map data structure to ` `    ``// count the frequency of each element ` `    ``map<``int``, ``int``> freq; ` ` `  `    ``for` `(``int` `i = 0; i < n; i++) { ` ` `  `        ``freq[a[i]]++; ` `    ``} ` ` `  `    ``int` `len = 0; ` ` `  `    ``for` `(``int` `i = 1; i <= n; i++) { ` ` `  `        ``// If frequency of element is 0, ` `        ``// then we can not move forward ` `        ``// as every element should be present ` `        ``if` `(freq[i] == 0) { ` `            ``break``; ` `        ``} ` ` `  `        ``// Increasing the length by one ` `        ``len++; ` `    ``} ` ` `  `    ``return` `len; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` ` `  `    ``int` `arr[] = { 3, 2, 1, 6, 5 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` ` `  `    ``cout << longestPermutation(arr, n) ` `         ``<< ``"\n"``; ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java Program to find length of ` `// Longest Permutaion Subsequence ` `// in a given array ` `import` `java.util.*; ` ` `  `class` `GFG{ ` `  `  `// Function to find the ` `// longest permutation subsequence ` `static` `int` `longestPermutation(``int` `arr[], ``int` `n) ` `{ ` `  `  `    ``// Map data structure to ` `    ``// count the frequency of each element ` `    ``HashMap freq = ``new` `HashMap(); ` `  `  `    ``for` `(``int` `i = ``0``; i < n; i++) { ` `  `  `        ``if``(freq.containsKey(arr[i])){ ` `            ``freq.put(arr[i], freq.get(arr[i])+``1``); ` `        ``}``else``{ ` `            ``freq.put(arr[i], ``1``); ` `        ``} ` `    ``} ` `  `  `    ``int` `len = ``0``; ` `  `  `    ``for` `(``int` `i = ``1``; i <= n; i++) { ` `  `  `        ``// If frequency of element is 0, ` `        ``// then we can not move forward ` `        ``// as every element should be present ` `        ``if` `(!freq.containsKey(i)) { ` `            ``break``; ` `        ``} ` `  `  `        ``// Increasing the length by one ` `        ``len++; ` `    ``} ` `  `  `    ``return` `len; ` `} ` `  `  `// Driver Code ` `public` `static` `void` `main(String[] args) ` `{ ` `  `  `    ``int` `arr[] = { ``3``, ``2``, ``1``, ``6``, ``5` `}; ` `    ``int` `n = arr.length; ` `  `  `    ``System.out.print(longestPermutation(arr, n)); ` `  `  `} ` `} ` ` `  `// This code is contributed by Rajput-Ji `

## C#

 `// C# Program to find length of ` `// longest Permutaion Subsequence ` `// in a given array ` ` `  `using` `System; ` `using` `System.Collections.Generic; ` ` `  `public` `class` `GFG{ ` ` `  `// Function to find the ` `// longest permutation subsequence ` `static` `int` `longestPermutation(``int` `[]arr, ``int` `n) ` `{ ` ` `  `    ``// Map data structure to ` `    ``// count the frequency of each element ` `    ``Dictionary<``int``,``int``> freq = ``new` `Dictionary<``int``,``int``>(); ` ` `  `    ``for` `(``int` `i = 0; i < n; i++) { ` ` `  `        ``if``(freq.ContainsKey(arr[i])){ ` `            ``freq[arr[i]] = freq[arr[i]] + 1; ` `        ``}``else``{ ` `            ``freq.Add(arr[i], 1); ` `        ``} ` `    ``} ` ` `  `    ``int` `len = 0; ` ` `  `    ``for` `(``int` `i = 1; i <= n; i++) { ` ` `  `        ``// If frequency of element is 0, ` `        ``// then we can not move forward ` `        ``// as every element should be present ` `        ``if` `(!freq.ContainsKey(i)) { ` `            ``break``; ` `        ``} ` ` `  `        ``// Increasing the length by one ` `        ``len++; ` `    ``} ` ` `  `    ``return` `len; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main(String[] args) ` `{ ` ` `  `    ``int` `[]arr = { 3, 2, 1, 6, 5 }; ` `    ``int` `n = arr.Length; ` ` `  `    ``Console.Write(longestPermutation(arr, n)); ` ` `  `} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

## Python3

 `# Program to find length of ` `# Longest Permutaion Subsequence ` `# in a given array ` `from` `collections ``import` `defaultdict ` ` `  `# Function to find the ` `# longest permutation subsequence ` `def` `longestPermutation(a, n): ` `  `  `    ``# Map data structure to ` `    ``# count the frequency of each element ` `    ``freq ``=` `defaultdict(``int``) ` `  `  `    ``for` `i ``in` `range``( n ): ` `  `  `        ``freq[a[i]] ``+``=` `1` `  `  `    ``length ``=` `0` `  `  `    ``for` `i ``in` `range``(``1` `, n ``+` `1``): ` `  `  `        ``# If frequency of element is 0, ` `        ``# then we can not move forward ` `        ``# as every element should be present ` `        ``if` `(freq[i] ``=``=` `0``): ` `            ``break` `  `  `        ``# Increasing the length by one ` `        ``length ``+``=` `1`  `         `  `    ``return` `length ` `  `  `# Driver Code ` `if` `__name__ ``=``=` `"__main__"``: ` `  `  `    ``arr ``=` `[ ``3``, ``2``, ``1``, ``6``, ``5` `] ` `    ``n ``=` `len``(arr) ` `  `  `    ``print``(longestPermutation(arr, n)) ` ` `  `# This code is contributed by chitranayal `

Output:

```3
```

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

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