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:

CPP

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output:

No

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

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Improved By : princi singh, Yash_R