Check if an Array is a permutation of numbers from 1 to N : Set 2

Given an array arr containing N positive integers, the task is to check if the given array arr represents a permutation or not.
 

A sequence of N integers is called a permutation if it contains all integers from 1 to N exactly once.

Examples: 
 

Input: arr[] = {1, 2, 5, 3, 2} 
Output: No 
Explanation: 
The given array contains 2 twice, and 4 is missing for the array to represent a permutation of length 5. 
Input: arr[] = {1, 2, 5, 3, 4} 
Output: Yes 
Explanation: 
The given array contains all integers from 1 to 5 exactly once. Hence, it represents a permutation of length 5. 
 

Naive Approach: in O(N²) Time 
This approach is mentioned here
Another Approach: in O(N) Time and O(N) Space 
This approach is mentioned here.
Efficient Approach: Using HashTable 
 

1. Create a HashTable of N size to store the frequency count of each number from 1 to N
2. Traverse through the given array and store the frequency of each number in the HashTable.
3. Then traverse the HashTable and check if all the numbers from 1 to N have a frequency of 1 or not. 
4. Print "Yes" if the above condition is True, Else "No".

Below is the implementation of the above approach: 
 

// C++ program to decide if an array
// represents a permutation or not
#include <bits/stdc++.h>
using namespace std;

// Function to check if an
// array represents a permutation or not
string permutation(int arr[], int N)
{

    int hash[N + 1] = { 0 };

    // Counting the frequency
    for (int i = 0; i < N; i++) {
        hash[arr[i]]++;
    }

    // Check if each frequency is 1 only
    for (int i = 1; i <= N; i++) {
        if (hash[i] != 1)
            return "No";
    }

    return "Yes";
}

// Driver code
int main()
{
    int arr[] = { 1, 1, 5, 5, 3 };
    int n = sizeof(arr) / sizeof(int);
    cout << permutation(arr, n) << endl;

    return 0;
}

// Java program to decide if an array
// represents a permutation or not
class GFG{
 
// Function to check if an
// array represents a permutation or not
static String permutation(int arr[], int N)
{
 
    int []hash = new int[N + 1];
 
    // Counting the frequency
    for (int i = 0; i < N; i++) {
        hash[arr[i]]++;
    }
 
    // Check if each frequency is 1 only
    for (int i = 1; i <= N; i++) {
        if (hash[i] != 1)
            return "No";
    }
 
    return "Yes";
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 1, 1, 5, 5, 3 };
    int n = arr.length;
    System.out.print(permutation(arr, n) +"\n");
}
}

// This code is contributed by Princi Singh

# Python3 program to decide if an array
# represents a permutation or not

# Function to check if an
# array represents a permutation or not
def permutation(arr,  N) :

    hash = [0]*(N + 1);

    # Counting the frequency
    for i in range(N) :
        hash[arr[i]] += 1;

    # Check if each frequency is 1 only
    for i in range(1, N + 1) :
        if (hash[i] != 1) :
            return "No";

    return "Yes";

# Driver code
if __name__ == "__main__" :

    arr = [ 1, 1, 5, 5, 3 ];
    n = len(arr);
    print(permutation(arr, n));

    # This code is contributed by Yash_R

// C# program to decide if an array
// represents a permutation or not
using System;

class GFG{
 
    // Function to check if an
    // array represents a permutation or not
    static string permutation(int []arr, int N)
    {
     
        int []hash = new int[N + 1];
     
        // Counting the frequency
        for (int i = 0; i < N; i++) {
            hash[arr[i]]++;
        }
     
        // Check if each frequency is 1 only
        for (int i = 1; i <= N; i++) {
            if (hash[i] != 1)
                return "No";
        }
     
        return "Yes";
    }
     
    // Driver code
    public static void Main(string[] args)
    {
        int []arr = { 1, 1, 5, 5, 3 };
        int n = arr.Length;
        Console.Write(permutation(arr, n) +"\n");
    }
}

// This code is contributed by Yash_R

<script>

// JavaScript program to decide if an array
// represents a permutation or not

// Function to check if an
// array represents a permutation or not
function permutation(arr, N)
{

    var hash = Array(N+1).fill(0);

    // Counting the frequency
    for (var i = 0; i < N; i++) {
        hash[arr[i]]++;
    }

    // Check if each frequency is 1 only
    for (var i = 1; i <= N; i++) {
        if (hash[i] != 1)
            return "No";
    }

    return "Yes";
}

// Driver code
var arr = [1, 1, 5, 5, 3];
var n = arr.length;
document.write( permutation(arr, n));

</script>  

No

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