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 array
- 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)