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

• Difficulty Level : Basic
• Last Updated : 24 Mar, 2023

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 ``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 hash = ``new` `HashSet();` `    ``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

 ``

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)

Efficient Approach:-

• As we have to check all elements from 1 to N in the array
• So think that if we just sort the array then if the array element will be from 1 to N then the sequence will be like 1,2,3_____,N.
• So we can just sort the array and can check is all the elements are like 1,2,3,____,N or not.

Implementation:-

## C++

 `// 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``bool` `permutation(``int` `arr[], ``int` `n)``{``      ``//sorting the array``      ``sort(arr,arr+n);``  ` `      ``//traversing the array to find if it is a valid permutation ot not``      ``for``(``int` `i=0;i

## 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)``  ``{``    ``// sorting the array``    ``Arrays.sort(arr);` `    ``// traversing the array to find if it is a valid permutation ot not``    ``for``(``int` `i = ``0``; i < n; i++)``    ``{``      ``//if i+1 element not present``      ``//or dublicacy is present``      ``if``(arr[i]!=i+``1``)``return` `false``;``    ``}` `    ``return` `true``;``  ``}` `  ``// 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"``);` `    ``return` `;``  ``}``}` `// this code is contibuted by bhardwajji`

## 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):``    ``# sorting the array``    ``arr.sort()` `    ``# traversing the array to find if it is a valid permutation or not``    ``for` `i ``in` `range``(n):``        ``# if i+1 element not present``        ``# or dublicacy is present``        ``if` `arr[i] !``=` `i ``+` `1``:``            ``return` `False` `    ``return` `True`  `# Driver code``if` `__name__ ``=``=` `'__main__'``:``    ``arr ``=` `[``1``, ``2``, ``5``, ``3``, ``2``]``    ``n ``=` `len``(arr)` `    ``if` `permutation(arr, n):``        ``print``(``"Yes"``)``    ``else``:``        ``print``(``"No"``)`

## C#

 `// C# Program to decide if an``// array represents a permutation or not``using` `System;` `public` `class` `GFG``{``    ``// Function to check if an``    ``// array represents a permutation or not``    ``public` `static` `bool` `permutation(``int``[] arr, ``int` `n)``    ``{``          ``//sorting the array``          ``Array.Sort(arr);` `          ``//traversing the array to find if it is a valid permutation ot not``          ``for` `(``int` `i = 0;i < n;i++)``          ``{``              ``//if i+1 element not present``              ``//or dublicacy is present``              ``if` `(arr[i] != i + 1)``              ``{``                  ``return` `false``;``              ``}``          ``}` `          ``return` `true``;``    ``}``    ` `    ``internal` `static` `void` `Main()``    ``{``        ``int``[] arr = {1, 2, 5, 3, 2};``        ``int` `n = arr.Length;` `        ``if` `(permutation(arr, n))``        ``{``            ``Console.Write(``"Yes"``);``            ``Console.Write(``"\n"``);``        ``}``        ``else``        ``{``            ``Console.Write(``"No"``);``            ``Console.Write(``"\n"``);``        ``}``    ``}``}` `//This code is contributed by bhardwajji`

## Javascript

 `// Function to check if an``// array represents a permutation or not``function` `permutation(arr, n) {``  ``// sorting the array``  ``arr.sort();` `  ``// traversing the array to find if it is a valid permutation or not``  ``for` `(let i = 0; i < n; i++) {``    ``// if i+1 element not present``    ``// or dublicacy is present``    ``if` `(arr[i] !== i + 1) {``      ``return` `false``;``    ``}``  ``}` `  ``return` `true``;``}` `// Driver code``const arr = [1, 2, 5, 3, 2];``const n = arr.length;` `if` `(permutation(arr, n)) {``  ``console.log(``"Yes"``);``} ``else` `{``  ``console.log(``"No"``);``}``// This code is Contributed by Shushant Kumar`

Output

`No`

Time Complexity:- O(NLogN)

Space Complexity:- O(1)

Another Efficient Approach: create a boolean array that help in if we already visited that element return False

else  Traverse the Whole array

Below is the implementation of above approach

## C++

 `#include ``#include ` `using` `namespace` `std;` `bool` `permutation(``int` `arr[], ``int` `n)``{``    ``// create a boolean array to keep track of which numbers``    ``// have been seen before``    ``bool` `x[n];` `    ``// initialize the boolean array with false values``    ``memset``(x, ``false``, ``sizeof``(x));` `    ``// check each number in the array``    ``for` `(``int` `i = 0; i < n; i++) {``        ``// if the number has not been seen before, mark it``        ``// as seen``        ``if` `(x[arr[i] - 1] == ``false``) {``            ``x[arr[i] - 1] = ``true``;``        ``}``        ``// if the number has been seen before, the array``        ``// does not represent a permutation``        ``else` `{``            ``return` `false``;``        ``}``    ``}` `    ``// check if all numbers from 1 to n have been seen in``    ``// the array``    ``for` `(``int` `i = 0; i < n; i++) {``        ``// if a number has not been seen in the array, the``        ``// array does not represent a permutation``        ``if` `(x[i] == ``false``) {``            ``return` `false``;``        ``}``    ``}` `    ``// if the array has passed all checks, it represents a``    ``// permutation``    ``return` `true``;``}` `int` `main()``{``    ``// initialize the array to be checked``    ``int` `arr[] = { 1, 2, 3, 4, 5 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);` `    ``// check if the array represents a permutation``    ``if` `(permutation(arr, n)) {``        ``cout << ``"YES"` `<< endl;``    ``}``    ``else` `{``        ``cout << ``"NO"` `<< endl;``    ``}` `    ``return` `0;``}`

## Python3

 `# Python code for the above approach` `# Function to check if an``# array represents a permutation or not` `# time complexity O(N)``# space O(N)`  `def` `permutation(arr, n):``     ``# crete a bool array that check if the element``    ``# we traversing are already exist in array or not``    ``x ``=` `[``0``] ``*` `n``    ``# checking for every element in array``    ``for` `i ``in` `range``(n):``        ``if` `x[arr[i] ``-` `1``] ``=``=` `0``:``            ``x[arr[i] ``-` `1``] ``=` `1``        ``else``:``            ``return` `False``    ``# for corner cases``    ``for` `i ``in` `range``(n):``        ``if` `x[i] ``=``=` `0``:``            ``return` `False``    ``return` `True`  `# Drive code``if` `__name__ ``=``=` `"__main__"``:``    ``arr ``=` `[``1``, ``2``, ``3``, ``4``, ``5``]``    ``n ``=` `len``(arr)``    ``if` `(permutation(arr, n)):``        ``print``(``"YES"``)``    ``else``:``        ``print``(``"NO"``)`  `# This code is contributed by Shushant Kumar`

## Javascript

 `// Function to check if an array represents a permutation or not` `// time complexity O(N)``// space O(N)``function` `permutation(arr, n) {``  ``// create a boolean array to check if the element we're``  ``// traversing already exists in the array or not``  ``let x = ``new` `Array(n).fill(``false``);``  ` `  ``// check for every element in array``  ``for` `(let i = 0; i < n; i++) {``    ``if` `(x[arr[i] - 1] == ``false``) {``      ``x[arr[i] - 1] = ``true``;``    ``} ``else` `{``      ``return` `false``;``    ``}``  ``}``  ` `  ``// for corner cases``  ``for` `(let i = 0; i < n; i++) {``    ``if` `(x[i] == ``false``) {``      ``return` `false``;``    ``}``  ``}``  ` `  ``return` `true``;``}` `// Drive code``let arr = [1, 2, 3, 4, 5];``let n = arr.length;` `if` `(permutation(arr, n)) {``  ``console.log(``"YES"``);``} ``else` `{``  ``console.log(``"NO"``);``}` `// This code is contributed by shushant kumar`

Output

`YES`

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

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