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

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 is not a permutation of numbers from 1 to N, because it 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:
Given array contains all integers from 1 to 5 exactly once. Hence, it represents a permutation of length 5. 

Naive Approach: Clearly, the given array will represent a permutation of length N only, where N is the length of the array. So we have to search for each element from 1 to N in the given array. If all the elements are found then the array represents a permutation else it does not.
Time Complexity: O(N²) 
Efficient Approach:
The above method can be optimized using a set data structure. 

1. Traverse the given array and insert every element in the set data structure.
2. Also, find the maximum element in the array. This maximum element will be value N which will represent the size of the set.
3. After traversal of the array, check if the size of the set is equal to N.
4. If the size of the set is equal to N then the array represents a permutation else it doesn’t.

Below is the implementation of the above approach:

C++

// 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
bool permutation(int arr[], int n)
{
    // Set to check the count
    // of non-repeating elements
    set<int> hash;

    int maxEle = 0;

    for (int i = 0; i < n; i++) {

        // Insert all elements in the set
        hash.insert(arr[i]);

        // Calculating the max element
        maxEle = max(maxEle, arr[i]);
    }

    if (maxEle != n)
        return false;

    // Check if set size is equal to n
    if (hash.size() == n)
        return true;

    return false;
}

// Driver code
int main()
{
    int arr[] = { 1, 2, 5, 3, 2 };
    int n = sizeof(arr) / sizeof(int);

    if (permutation(arr, n))
        cout << "Yes" << endl;
    else
        cout << "No" << endl;

    return 0;
}

Java

// Java Program to decide if an
// array represents a permutation or not
import java.util.*;

class GFG{

// Function to check if an
// array represents a permutation or not
static boolean permutation(int []arr, int n)
{
    // Set to check the count
    // of non-repeating elements
    Set<Integer> hash = new HashSet<Integer>();

    int maxEle = 0;

    for (int i = 0; i < n; i++) {

        // Insert all elements in the set
        hash.add(arr[i]);

        // Calculating the max element
        maxEle = Math.max(maxEle, arr[i]);
    }

    if (maxEle != n)
        return false;

    // Check if set size is equal to n
    if (hash.size() == n)
        return true;

    return false;
}

// Driver code
public static void main(String args[])
{
    int arr[] = { 1, 2, 5, 3, 2 };
    int n = arr.length;

    if (permutation(arr, n))
        System.out.println("Yes");
    else
        System.out.println("No");
}
}

// This code is contributed by Surendra_Gangwar

Python3

# 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):

        # Set to check the count
    # of non-repeating elements
    s = set()

    maxEle = 0;

    for i in range(n):

        # Insert all elements in the set
        s.add(arr[i]);

        # Calculating the max element
        maxEle = max(maxEle, arr[i]);

    if (maxEle != n):
        return False

    # Check if set size is equal to n
    if (len(s) == n):
        return True;

    return False;

# Driver code
if __name__=='__main__':

    arr = [ 1, 2, 5, 3, 2 ]
    n = len(arr)

    if (permutation(arr, n)):
        print("Yes")
    else:
        print("No")

# This code is contributed by Princi Singh

C#

// C# Program to decide if an
// array represents a permutation or not
using System;
using System.Collections.Generic;

class GFG{

// Function to check if an
// array represents a permutation or not
static bool permutation(int []arr, int n)
{
    // Set to check the count
    // of non-repeating elements
    HashSet<int> hash = new HashSet<int>();

    int maxEle = 0;

    for (int i = 0; i < n; i++) {

        // Insert all elements in the set
        hash.Add(arr[i]);

        // Calculating the max element
        maxEle = Math.Max(maxEle, arr[i]);
    }

    if (maxEle != n)
        return false;

    // Check if set size is equal to n
    if (hash.Count == n)
        return true;

    return false;
}

// Driver code
public static void Main(String []args)
{
    int []arr = { 1, 2, 5, 3, 2 };
    int n = arr.Length;

    if (permutation(arr, n))
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
}
}

// This code is contributed by Princi Singh

Javascript

<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)
{
    // Set to check the count
    // of non-repeating elements
    let hash =  new Set();

    let maxEle = 0;

    for (let i = 0; i < n; i++) {

        // Insert all elements in the set
        hash.add(arr[i]);

        // Calculating the max element
        maxEle = Math.max(maxEle, arr[i]);
    }

    if (maxEle != n)
        return false;

    // Check if set size is equal to n
    if (hash.length == n)
        return true;

    return false;
}

// Driver Code

    let arr = [ 1, 2, 5, 3, 2 ];
    let n = arr.length;

    if (permutation(arr, n))
        document.write("Yes");
    else
        document.write("No");

</script>

Output:

No

Time Complexity: O(N log N)

Since every insert operation in the set is an O(log N) operation. There will be N such operations hence O(N log N).

Auxiliary Space: O(N)