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. 
 

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++ 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 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# 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

# 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

<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>

3

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