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

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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.

Algorithm:

  1. Initialize a flag variable “isPermutation” to true.
  2. Initialize a variable “N” to the length of the array.
  3. Loop through integers from 1 to N:
    1. Initialize a flag variable “found” to false.
    2. Loop through the array elements and If the integer “i” is found in the array, set “found” to true and break the loop.
    3. If “found” is false, set “isPermutation” to false and break the loop.
  4. If “isPermutation” is true, print “Array represents a permutation”.
  5. Else, print “Array does not represent a permutation”.

Below is the implementation of the approach:

C++

// C++ code for the approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to check if an Array is a
// permutation of numbers from 1 to N
bool isPermutation(int arr[], int n) {
    // Check for each element from 1 to N in the array
    for(int i=1; i<=n; i++) {
        bool found = false;
        for(int j=0; j<n; j++) {
            if(arr[j] == i) {
                found = true;
                break;
            }
        }
       
        // If any element is not found, array is not a permutation
        if(!found) {
            return false;
        }
    }
   
    // All elements found, array is a permutation
    return true;
}
 
// Driver's code
int main() {
      // Input
    int arr[] = { 1, 2, 5, 3, 2 };
    int n = sizeof(arr)/sizeof(arr[0]);
   
      // Function Call
    if(isPermutation(arr, n)) {
        cout << "Yes" << endl;
    }
    else {
        cout << "No" << endl;
    }
   
    return 0;
}

                    

Java

// Java code for the approach
 
import java.util.*;
 
public class GFG {
 
    // Function to check if an Array is a
    // permutation of numbers from 1 to N
    public static boolean isPermutation(int[] arr, int n)
    {
 
        // Check for each element from
        // 1 to N in the array
        for (int i = 1; i <= n; i++) {
            boolean found = false;
            for (int j = 0; j < n; j++) {
                if (arr[j] == i) {
                    found = true;
                    break;
                }
            }
            // If any element is not found,
            // array is not a permutation
            if (!found) {
                return false;
            }
        }
 
        // All elements found, array
        // is a permutation
        return true;
    }
 
    // Driver's code
    public static void main(String[] args)
    {
        int[] arr = { 1, 2, 5, 3, 2 };
        int n = arr.length;
 
        if (isPermutation(arr, n)) {
            System.out.println("Yes");
        }
        else {
            System.out.println("No");
        }
    }
}

                    

Python3

# Function to check if a list is a permutation of numbers from 1 to N
def is_permutation(arr):
    n = len(arr)
     
    # Check for each element from 1 to N in the list
    for i in range(1, n + 1):
        found = False
        for j in range(n):
            if arr[j] == i:
                found = True
                break
         
        # If any element is not found, the list is not a permutation
        if not found:
            return False
     
    # All elements found, the list is a permutation
    return True
 
# Driver's code
arr = [1, 2, 5, 3, 2]
 
# Function Call
if is_permutation(arr):
    print("Yes")
else:
    print("No")

                    

C#

using System;
 
public class GFG {
    // Function to check if an Array is a
    // permutation of numbers from 1 to N
    public static bool IsPermutation(int[] arr, int n)
    {
        // Check for each element from
        // 1 to N in the array
        for (int i = 1; i <= n; i++) {
            bool found = false;
            for (int j = 0; j < n; j++) {
                if (arr[j] == i) {
                    found = true;
                    break;
                }
            }
            // If any element is not found,
            // array is not a permutation
            if (!found) {
                return false;
            }
        }
 
        // All elements found, array
        // is a permutation
        return true;
    }
 
    // Driver's code
    public static void Main(string[] args)
    {
        int[] arr = { 1, 2, 5, 3, 2 };
        int n = arr.Length;
 
        if (IsPermutation(arr, n)) {
            Console.WriteLine("Yes");
        }
        else {
            Console.WriteLine("No");
        }
    }
}
 
// This code is contributed by Samim Hossain Mondal.

                    

Javascript

// Function to check if an Array is a permutation of numbers from 1 to N
function isPermutation(arr, n) {
    // Check for each element from 1 to N in the array
    for (let i = 1; i <= n; i++) {
        let found = false;
        for (let j = 0; j < n; j++) {
            if (arr[j] === i) {
                found = true;
                break;
            }
        }
        // If any element is not found, array is not a permutation
        if (!found) {
            return false;
        }
    }
 
    // All elements found, array is a permutation
    return true;
}
 
// Driver's code
const arr = [1, 2, 5, 3, 2];
const n = arr.length;
 
if (isPermutation(arr, n)) {
    console.log("Yes");
} else {
    console.log("No");
}

                    

Output
No




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)

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 <bits/stdc++.h>
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<n;i++)
    {
          //if i+1 element not present
          //or dublicacy is present
          if(arr[i]!=i+1)return false;
    }
   
      return true;
}
 
// 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;
}
//This code is contributed by shubhamrajput6156

                    

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 contributed 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 <cstring>
#include <iostream>
 
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;
}

                    

Java

import java.util.Arrays;
 
public class Main {
 
    public static boolean permutation(int[] arr, int n) {
        // create a boolean array to keep track of which numbers
        // have been seen before
        boolean[] x = new boolean[n];
 
        // initialize the boolean array with false values
        Arrays.fill(x, false);
 
        // 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;
    }
 
    public static void main(String[] args) {
        // initialize the array to be checked
        int[] arr = { 1, 2, 3, 4, 5 };
        int n = arr.length;
 
        // check if the array represents a permutation
        if (permutation(arr, n)) {
            System.out.println("YES");
        } else {
            System.out.println("NO");
        }
    }
}
// This code is contributed by shiv1043g

                    

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

                    

C#

using System;
 
public class Gfg
{
  public static bool permutation(int[] arr, int n)
  {
     
    // create a boolean array to keep track of which numbers
    // have been seen before
    bool[] x = new bool[n];
 
    // initialize the boolean array with false values
    for (int i = 0; i < n; i++)
    {
      x[i] = false;
    }
 
    // 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;
  }
 
  public static void Main()
  {
    // initialize the array to be checked
    int[] arr = { 1, 2, 3, 4, 5 };
    int n = arr.Length;
 
    // check if the array represents a permutation
    if (permutation(arr, n))
    {
      Console.WriteLine("YES");
    }
    else
    {
      Console.WriteLine("NO");
    }
  }
}

                    

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)



Last Updated : 05 Oct, 2023
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