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++

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// C++ Program to find length of
// Longest Permutaion 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;
}

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Java

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// 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<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

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C#

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// 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

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Python3

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# 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

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Output:

3

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

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