Count of elements having Euler’s Totient value one less than itself

Given an array arr[] of N integers, and a range L to R, the task is to find the total number of elements in the array from index L to R that satisfies the following condition:

where F(x) is Euler’s Totient Function.

Examples:

Input: arr[] = {2, 4, 5, 8}, L = 1, R = 3 
Output: 2 
Explanation:
Here in the given range, F(2) = 1 and (2 – 1) = 1. Similarly, F(5) = 4 and (5 – 1) = 4. 
So the total count of indices which satisfy the condition is 2. 
Input: arr[] = {9, 3, 4, 6, 8}, L = 3, R = 5 
Output: 0 
Explanation :
In the given range there is no element that satisfies the given condition. 
So the total count is 0.

Naive Approach: The naive approach to solve this problem is to iterate over all the elements of the array and check if the Euler’s Totient value of the current element is one less than itself. If yes then increment the count.
Time Complexity: O(N * sqrt(N)) 
Auxiliary Space: O(1)
Efficient Approach:

Euler’s Totient function F(n) for an input n is the count of numbers in {1, 2, 3, …, n} that are relatively prime to n, i.e., the numbers whose Greatest Common Divisor with n is 1.

1. If we observe, we can notice that the above given condition is only satisfied by the Prime numbers.
2. So, all we need to do is to compute the total numbers of prime numbers in the given range.
3. We will use Sieve of Eratosthenes to compute prime numbers efficiently.
4. Also, we will pre-compute the numbers of primes numbers in a count array.

Below is the implementation of the above approach.

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

long long prime[1000001] = { 0 };

// Seiev of Erotosthenes method to compute all primes
void seiveOfEratosthenes()
{
    for (int i = 2; i < 1000001; i++)
        prime[i] = 1;

    for (int i = 2; i * i < 1000001; i++) {
        // If current number is marked prime then mark its
        // multiple as non-prime
        if (prime[i] == 1)
            for (int j = i * i; j < 1000001; j += i)
                prime[j] = 0;
    }
}

// Function to count the number
// of element satisfying the condition
void CountElements(int arr[], int n, int L, int R)
{
    seiveOfEratosthenes();

    long long countPrime[n + 1] = { 0 };

    // Compute the number of primes in count prime array
    for (int i = 1; i <= n; i++)
        countPrime[i] = countPrime[i - 1] + prime[arr[i - 1]];

    // Print the number of elements satisfying the condition
    printf("%lld", (countPrime[R] - countPrime[L - 1]));
    return;
}

// Driver Code
int main()
{
    // Given array
    int arr[] = { 2, 4, 5, 8 };
    // Size of the array
    int N = sizeof(arr) / sizeof(int);
    int L = 1, R = 3;
    // Function Call
    CountElements(arr, N, L, R);
    return 0;
}

// This code is contributed by Sania Kumari Gupta

C

// C program for the above approach
#include <stdio.h>
#include <string.h>

long long prime[1000001];

// Seiev of Erotosthenes method to compute all primes
void seiveOfEratosthenes()
{
    memset(prime, 0, sizeof(prime));
    for (int i = 2; i < 1000001; i++)
        prime[i] = 1;

    for (int i = 2; i * i < 1000001; i++) {
        // If current number is marked prime then mark its
        // multiple as non-prime
        if (prime[i] == 1)
            for (int j = i * i; j < 1000001; j += i)
                prime[j] = 0;
    }
}

// Function to count the number
// of element satisfying the condition
void CountElements(int arr[], int n, int L, int R)
{
    seiveOfEratosthenes();
    long long countPrime[n + 1];
    memset(countPrime, 0, sizeof(countPrime));

    // Compute the number of primes in count prime array
    for (int i = 1; i <= n; i++)
        countPrime[i]
            = countPrime[i - 1] + prime[arr[i - 1]];

    // Print the number of elements satisfying the condition
    printf("%lld", (countPrime[R] - countPrime[L - 1]));
    return;
}

// Driver Code
int main()
{
    // Given array
    int arr[] = { 2, 4, 5, 8 };
    // Size of the array
    int N = sizeof(arr) / sizeof(int);
    int L = 1, R = 3;
    // Function Call
    CountElements(arr, N, L, R);
    return 0;
}

// This code is contributed by Sania Kumari Gupta

Java

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

class GFG{

static int prime[] = new int[1000001];

// Seiev of Erotosthenes method to
// compute all primes
static void seiveOfEratosthenes()
{
    for(int i = 2; i < 1000001; i++)
    {
       prime[i] = 1;
    }

    for(int i = 2; i * i < 1000001; i++)
    {

       // If current number is
       // marked prime then mark
       // its multiple as non-prime
       if (prime[i] == 1)
       {
           for(int j = i * i;
                   j < 1000001; j += i)
           {
              prime[j] = 0;
           }
       }
    }
}

// Function to count the number
// of element satisfying the condition
static void CountElements(int arr[], int n,
                          int L, int R)
{
    seiveOfEratosthenes();

    int countPrime[] = new int[n + 1];

    // Compute the number of primes
    // in count prime array
    for(int i = 1; i <= n; i++)
    {
       countPrime[i] = countPrime[i - 1] +
                        prime[arr[i - 1]];
    }

    // Print the number of elements
    // satisfying the condition
    System.out.print(countPrime[R] -
                     countPrime[L - 1] + "\n");

    return;
}

// Driver Code
public static void main(String[] args)
{

    // Given array
    int arr[] = { 2, 4, 5, 8 };

    // Size of the array
    int N = arr.length;
    int L = 1, R = 3;

    // Function Call
    CountElements(arr, N, L, R);
}
}

// This code is contributed by amal kumar choubey

Python3

# Python3 program for the above approach
prime = [0] * (1000001)

# Seiev of Erotosthenes method to
# compute all primes
def seiveOfEratosthenes():

    for i in range(2, 1000001):
        prime[i] = 1

    i = 2
    while(i * i < 1000001):

        # If current number is
        # marked prime then mark
        # its multiple as non-prime
        if (prime[i] == 1):
            for j in range(i * i, 1000001, i):
                prime[j] = 0

        i += 1

# Function to count the number
# of element satisfying the condition
def CountElements(arr, n, L, R):

    seiveOfEratosthenes()

    countPrime = [0] * (n + 1)

    # Compute the number of primes
    # in count prime array
    for i in range(1, n + 1):
        countPrime[i] = (countPrime[i - 1] +
                          prime[arr[i - 1]])

    # Print the number of elements
    # satisfying the condition
    print(countPrime[R] -
          countPrime[L - 1])

    return

# Driver Code

# Given array
arr = [ 2, 4, 5, 8 ]

# Size of the array
N = len(arr)
L = 1
R = 3

# Function call
CountElements(arr, N, L, R)

# This code is contributed by sanjoy_62

C#

// C# program for the above approach
using System;
class GFG{

static int []prime = new int[1000001];

// Seiev of Erotosthenes method to
// compute all primes
static void seiveOfEratosthenes()
{
    for(int i = 2; i < 1000001; i++)
    {
        prime[i] = 1;
    }

    for(int i = 2; i * i < 1000001; i++)
    {

        // If current number is
        // marked prime then mark
        // its multiple as non-prime
        if (prime[i] == 1)
        {
            for(int j = i * i;
                    j < 1000001; j += i)
            {
                prime[j] = 0;
            }
        }
    }
}

// Function to count the number
// of element satisfying the condition
static void CountElements(int []arr, int n,
                          int L, int R)
{
    seiveOfEratosthenes();

    int []countPrime = new int[n + 1];

    // Compute the number of primes
    // in count prime array
    for(int i = 1; i <= n; i++)
    {
        countPrime[i] = countPrime[i - 1] +
                         prime[arr[i - 1]];
    }

    // Print the number of elements
    // satisfying the condition
    Console.Write(countPrime[R] -
                  countPrime[L - 1] + "\n");

    return;
}

// Driver Code
public static void Main(String[] args)
{

    // Given array
    int []arr = { 2, 4, 5, 8 };

    // Size of the array
    int N = arr.Length;
    int L = 1, R = 3;

    // Function Call
    CountElements(arr, N, L, R);
}
}

// This code is contributed by sapnasingh4991

Javascript

<script>

// JavaScript program for the above approach

let prime = new Uint8Array(1000001);

// Seiev of Erotosthenes method to
// compute all primes
function seiveOfEratosthenes()
{

    for (let i = 2; i < 1000001;
        i++) {
        prime[i] = 1;
    }

    for (let i = 2; i * i < 1000001;
        i++) {

        // If current number is
        // marked prime then mark
        // its multiple as non-prime
        if (prime[i] == 1) {
            for (let j = i * i;
                j < 1000001; j += i) {
                prime[j] = 0;
            }
        }
    }
}

// Function to count the number
// of element satisfying the condition
function CountElements(arr, n, L, R)
{
    seiveOfEratosthenes();

    countPrime = new Uint8Array(n + 1);

    // Compute the number of primes
    // in count prime array
    for (let i = 1; i <= n; i++) {
        countPrime[i] = countPrime[i - 1]
                        + prime[arr[i - 1]];
    }

    // Print the number of elements
    // satisfying the condition
    document.write((countPrime[R]
                - countPrime[L - 1])
        + "<br>");

    return;
}

// Driver Code

    // Given array
    let arr = [ 2, 4, 5, 8 ];

    // Size of the array
    let N = arr.length;
    let L = 1, R = 3;

    // Function Call
    CountElements(arr, N, L, R);


// This code is contributed by Surbhi Tyagi.

</script>

Output:

2

Time Complexity: O(N * log(logN))
Auxiliary Space: O(N)