# 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.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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

 `// C++ program to decide if an array ` `// represents a permutation or not ` `#include ` `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

 `// 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

 `# 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#

 `// 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 `

Output:

```No
```

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

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Improved By : princi singh, Yash_R