Aquileo | Generate an array having sum of Euler Totient Function of all elements equal to N

Given a positive integer N, the task is to generate an array such that the sum of the Euler Totient Function of each element is equal to N.

Examples:

Input: N = 6
Output: 1 6 2 3

Input: N = 12
Output: 1 12 2 6 3 4

Approach: The given problem can be solved based on the divisor sum property of the Euler Totient Function, i.e.,

The Euler Totient Function of a number N< is the number of integers from 1 to N that gives GCD(i, N) as 1 and a number N can be represented as the summation of the Euler Totient Function values of all the divisors of N.
Therefore, the idea is to find the divisors of the given number N as the resultant array.

Below is the implementation of the above approach:

C++

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to construct the array such
// the sum of values of Euler Totient
// functions of all array elements is N
void constructArray(int N)
{
    // Stores the resultant array
    vector<int> ans;

    // Find divisors in sqrt(N)
    for (int i = 1; i * i <= N; i++) {

        // If N is divisible by i
        if (N % i == 0) {

            // Push the current divisor
            ans.push_back(i);

            // If N is not a
            // perfect square
            if (N != (i * i)) {

                // Push the second divisor
                ans.push_back(N / i);
            }
        }
    }

    // Print the resultant array
    for (auto it : ans) {
        cout << it << " ";
    }
}

// Driver Code
int main()
{
    int N = 12;

    // Function Call
    constructArray(N);

    return 0;
}

Java

// Java program for the above approach
import java.util.*;

class GFG{
    
// Function to construct the array such
// the sum of values of Euler Totient
// functions of all array elements is N
static void constructArray(int N)
{
    
    // Stores the resultant array
    ArrayList<Integer> ans = new ArrayList<Integer>();

    // Find divisors in sqrt(N)
    for(int i = 1; i * i <= N; i++)
    {
        
        // If N is divisible by i
        if (N % i == 0) 
        {
            
            // Push the current divisor
            ans.add(i);

            // If N is not a
            // perfect square
            if (N != (i * i)) 
            {
                
                // Push the second divisor
                ans.add(N / i);
            }
        }
    }

    // Print the resultant array
    for(int it : ans) 
    { 
        System.out.print(it + " ");
    }
}

// Driver Code
public static void main(String[] args)
{
    int N = 12;

    // Function Call
    constructArray(N);
}
}

// This code is contributed by splevel62

Python3

# Python3 program for the above approach
from math import sqrt

# Function to construct the array such
# the sum of values of Euler Totient
# functions of all array elements is N
def constructArray(N):
    
    # Stores the resultant array
    ans = []

    # Find divisors in sqrt(N)
    for i in range(1, int(sqrt(N)) + 1, 1):
        
        # If N is divisible by i
        if (N % i == 0):
            
            # Push the current divisor
            ans.append(i)

            # If N is not a
            # perfect square
            if (N != (i * i)):
                
                # Push the second divisor
                ans.append(N / i)

    # Print the resultant array
    for it in ans:
        print(int(it), end = " ")

# Driver Code
if __name__ == '__main__':
    
    N = 12
    
    # Function Call
    constructArray(N)

# This code is contributed by ipg2016107

// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
    
// Function to construct the array such
// the sum of values of Euler Totient
// functions of all array elements is N
static void constructArray(int N)
{
    
    // Stores the resultant array
    List<int> ans = new List<int>();

    // Find divisors in sqrt(N)
    for(int i = 1; i * i <= N; i++)
    {
        
        // If N is divisible by i
        if (N % i == 0) 
        {
            
            // Push the current divisor
            ans.Add(i);

            // If N is not a
            // perfect square
            if (N != (i * i)) 
            {
                
                // Push the second divisor
                ans.Add(N / i);
            }
        }
    }

    // Print the resultant array
    foreach(int it in ans) 
    { 
        Console.Write(it + " ");
    }
}

// Driver Code
public static void Main()
{
    int N = 12;

    // Function Call
    constructArray(N);
}
}

// This code is contributed by ukasp

JavaScript

<script>
// javascript program for the above approach
   
// Function to construct the array such
// the sum of values of Euler Totient
// functions of all array elements is N
function constructArray(N)
{
    
    // Stores the resultant array
    var ans = [];

    // Find divisors in sqrt(N)
    for(var i = 1; i * i <= N; i++)
    {
        
        // If N is divisible by i
        if (N % i == 0) 
        {
            
            // Push the current divisor
            ans.push(i);

            // If N is not a
            // perfect square
            if (N != (i * i)) 
            {
                
                // Push the second divisor
                ans.push(N / i);
            }
        }
    }

    // Print the resultant array
    document.write(ans);
}

// Driver Code
var N = 12;

// Function Call
constructArray(N);

// This code contributed by shikhasingrajput 
</script>

Output:

1 12 2 6 3 4

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

Generate an array having sum of Euler Totient Function of all elements equal to N

Explore