Longest permutation subsequence in a given array Last Updated : 25 Aug, 2021 Comments Improve Suggest changes Like Article Like Report 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. 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. Count the frequency of the elements in the range [1, N] from the arrayIterate 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 Permutation Subsequence // in a given array #include <bits/stdc++.h> 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[0]); cout << longestPermutation(arr, n) << "\n"; return 0; } Java // Java Program to find length of // Longest Permutation 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<Integer,Integer> freq = new HashMap<Integer,Integer>(); 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 Permutation 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 Permutation 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 JavaScript <script> // Javascript Program to find length of // Longest Permutation Subsequence // in a given array // Function to find the // longest permutation subsequence function longestPermutation(arr, n) { // Map data structure to // count the frequency of each element let freq = new Map(); for (let i = 0; i < n; i++) { if(freq.has(arr[i])){ freq.set(arr[i], freq.get(arr[i])+1); }else{ freq.set(arr[i], 1); } } let len = 0; for (let 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.has(i)) { break; } // Increasing the length by one len++; } return len; } // Driver code let arr = [ 3, 2, 1, 6, 5 ]; let n = arr.length; document.write(longestPermutation(arr, n)); </script> Output: 3 Time Complexity: O(N) Auxiliary Space Complexity: O(N) Comment More infoAdvertise with us Next Article Longest permutation subsequence in a given array S souradeep Follow Improve Article Tags : Searching Hash Combinatorial DSA Arrays permutation cpp-map frequency-counting Natural Numbers +5 More Practice Tags : ArraysCombinatorialHashpermutationSearching +1 More Similar Reads DSA Tutorial - Learn Data Structures and Algorithms DSA (Data Structures and Algorithms) is the study of organizing data efficiently using data structures like arrays, stacks, and trees, paired with step-by-step procedures (or algorithms) to solve problems effectively. 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