Count distinct prime factors for each element of an array

Given an array arr[] of size N, the task is to find the count of distinct prime factors of each element of the given array.

Examples:

Input: arr[] = {6, 9, 12}
Output: 2 1 2
Explanation:

• 6 = 2 × 3. Therefore, count = 2
• 9 = 3 × 3. Therefore, count = 1
• 12 = 2 × 2 × 3. Therefore, count = 2

The count of distinct prime factors of each array element are 2, 1, 2 respectively.

Input: arr[] = {4, 8, 10, 6}
Output: 1 1 2 2

Naive Approach: The simplest approach to solve the problem is to find the distinct prime factors of each array element. Then, print that count for each array element. 

steps to implement-

• Run a loop over the given array to find the count of distinct prime factors of each element
• For finding the count of distinct prime factors of each element-
  • Initialize a variable count with a value of 0
  • First, check if the number can be divided by 2. If it can be then divide it by 2 till it can be divided and after that increment the count by 1.
  • Then check for all odd numbers from 3 till its square root that it can divide the number or not
  • If any odd number can divide then it should divide till it can and after that increment count by 1 for each odd number
  • In last, if still we have number>2 then this number that is remaining is any prime number so increment count by 1
  • In the last print/return the count

Code-

C++

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

// A function to provide count of
//distinct prime factor of a given number
int primeFactors(int n)
{
    //To store count of distinct prime factor of given number
    int count=0;

    //Boolean variable to check an element
    //is included in its prime factor or not
    bool val=false;

    // Remove all 2s that can be prime factor of n
    while (n % 2 == 0)
    {
        val=true;
        n = n/2;
    }

    //If 2 is included
    if(val==true){count++;}

    // n must be odd at this point. So we can skip
    // one element (Note i = i +2)
    for (int i = 3; i <= sqrt(n); i = i + 2)
    {
        //To check whether i is included in prime factor
        val=false;

        // While i divides n,divide n
        while (n % i == 0)
        {
            val=true;
            n = n/i;
        }
        //If i is included in prime factor
        if(val==true){count++;}
    }


    // This condition is to handle the case when n
    // is a prime number greater than 2
    if (n > 2){
       count++;
    }
      return count;
}

// Function to get the distinct
// factor count of arr[]
void getFactorCount(int arr[],
                    int N)
{
    //Traverse every array element
    for(int i=0;i<N;i++){
        cout<<primeFactors(arr[i])<<" ";
    }
}

// Driver Code
int main()
{
    int arr[] = { 6, 9, 12 };
    int N = sizeof(arr) / sizeof(arr[0]);
    getFactorCount(arr, N);
    return 0;
}

Java

import java.util.*;

public class Main {
    // A function to provide count of
    // distinct prime factors of a given number
    static int primeFactors(int n) {
        // To store the count of distinct prime factors of the given number
        int count = 0;

        // Boolean variable to check if an element is included in its prime factor or not
        boolean val = false;

        // Remove all 2s that can be prime factors of n
        while (n % 2 == 0) {
            val = true;
            n = n / 2;
        }

        // If 2 is included
        if (val) {
            count++;
        }

        // n must be odd at this point. So we can skip one element (Note i = i + 2)
        for (int i = 3; i <= Math.sqrt(n); i = i + 2) {
            // To check whether i is included in prime factor
            val = false;

            // While i divides n, divide n
            while (n % i == 0) {
                val = true;
                n = n / i;
            }

            // If i is included in the prime factor
            if (val) {
                count++;
            }
        }

        // This condition is to handle the case when n is a prime number greater than 2
        if (n > 2) {
            count++;
        }

        return count;
    }

    // Function to get the distinct factor count of arr[]
    static void getFactorCount(int[] arr, int N) {
        // Traverse every array element
        for (int i = 0; i < N; i++) {
            System.out.print(primeFactors(arr[i]) + " ");
        }
    }

    public static void main(String[] args) {
        int[] arr = { 6, 9, 12 };
        int N = arr.length;
        getFactorCount(arr, N);
    }
}

Python3

import math

# A function to provide count of distinct prime factor of a given number
def prime_factors(n):
    # To store count of distinct prime factors of the given number
    count = 0

    # Boolean variable to check if an element is included in its prime factors or not
    val = False

    # Remove all 2s that can be prime factors of n
    while n % 2 == 0:
        val = True
        n = n // 2

    # If 2 is included
    if val == True:
        count += 1

    # n must be odd at this point. So we can skip one element (Note i = i + 2)
    for i in range(3, int(math.sqrt(n)) + 1, 2):
        # To check whether i is included in prime factors
        val = False

        # While i divides n, divide n
        while n % i == 0:
            val = True
            n = n // i

        # If i is included in prime factors
        if val == True:
            count += 1

    # This condition is to handle the case when n is a prime number greater than 2
    if n > 2:
        count += 1

    return count

# Function to get the distinct factor count of arr[]
def get_factor_count(arr):
    # Traverse every array element
    for num in arr:
        print(prime_factors(num), end=" ")

# Driver Code
if __name__ == "__main__":
    arr = [6, 9, 12]
    get_factor_count(arr)

C#

using System;

class Program
{
    // A function to provide the count of distinct prime factors of a given number
    static int PrimeFactors(int n)
    {
        // To store the count of distinct prime factors of the given number
        int count = 0;

        // Boolean variable to check if an element is included in its prime factors or not
        bool val = false;

        // Remove all 2s that can be prime factors of n
        while (n % 2 == 0)
        {
            val = true;
            n = n / 2;
        }

        // If 2 is included
        if (val)
        {
            count++;
        }

        // n must be odd at this point, so we can skip one element (Note i = i + 2)
        for (int i = 3; i <= Math.Sqrt(n); i += 2)
        {
            // To check whether i is included in prime factors
            val = false;

            // While i divides n, divide n
            while (n % i == 0)
            {
                val = true;
                n = n / i;
            }

            // If i is included in prime factors
            if (val)
            {
                count++;
            }
        }

        // This condition is to handle the case when n is a prime number greater than 2
        if (n > 2)
        {
            count++;
        }

        return count;
    }

    // Function to get the distinct factor count of an array
    static void GetFactorCount(int[] arr)
    {
        // Traverse every array element
        foreach (int num in arr)
        {
            Console.Write(PrimeFactors(num) + " ");
        }
    }

    static void Main()
    {
        int[] arr = { 6, 9, 12 };
        GetFactorCount(arr);
    }
}

Javascript

// JavaScript program for the above approach

// A function to provide count of
//distinct prime factor of a given number
function primeFactors(n)
{
    //To store count of distinct prime factor of given number
    let count=0;

    //Boolean variable to check an element
    //is included in its prime factor or not
    let val=false;

    // Remove all 2s that can be prime factor of n
    while (n % 2 == 0)
    {
        val=true;
        n = n/2;
    }

    //If 2 is included
    if(val==true){count++;}

    // n must be odd at this point. So we can skip
    // one element (Note i = i +2)
    for (let i = 3; i <= Math.sqrt(n); i = i + 2)
    {
        //To check whether i is included in prime factor
        val=false;

        // While i divides n,divide n
        while (n % i == 0)
        {
            val=true;
            n = n/i;
        }
        //If i is included in prime factor
        if(val==true){count++;}
    }


    // This condition is to handle the case when n
    // is a prime number greater than 2
    if (n > 2){
       count++;
    }
      return count;
}

// Function to get the distinct
// factor count of arr[]
function getFactorCount(arr, N)
{
    //Traverse every array element
    for(let i=0;i<N;i++){
        document.write(primeFactors(arr[i])+ " ");
    }
}

// Driver Code
    let arr = [ 6, 9, 12 ];
    let N = arr.length;
    getFactorCount(arr, N);

Output-

2 1 2 

Time Complexity: O(N * (√Maximum value present in array)), because O(N) in traversing the array and (√Maximum value present in array) is the maximum time to find the count of distinct prime factors of each Number
Auxiliary Space: O(1), because no extra space has been used

Efficient Approach: The above approach can be optimized by precomputing the distinct factors of all the numbers using their Smallest Prime Factors. Follow the steps below to solve the problem

• Initialize a vector, say v, to store the distinct prime factors.
• Store the Smallest Prime Factor(SPF) up to 10⁵ using a Sieve of Eratosthenes.
• Calculate the distinct prime factors of all the numbers by dividing the numbers recursively with their smallest prime factor till it reduces to 1 and store distinct prime factors of X, in v[X].
• Traverse the array arr[] and for each array element, print the count as v[arr[i]].size().

Below is the implementation of the above approach :

C++14

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;
#define MAX 100001

// Stores smallest prime
// factor for every number
int spf[MAX];

// Stores distinct prime factors
vector<int> v[MAX];

// Function to find the smallest
// prime factor of every number
void sieve()
{
    // Mark the smallest prime factor
    // of every number to itself
    for (int i = 1; i < MAX; i++)
        spf[i] = i;

    // Separately mark all the
    // smallest prime factor of
    // every even number to be 2
    for (int i = 4; i < MAX; i = i + 2)
        spf[i] = 2;

    for (int i = 3; i * i < MAX; i++)

        // If i is prime
        if (spf[i] == i) {

            // Mark spf for all numbers
            // divisible by i
            for (int j = i * i; j < MAX;
                 j = j + i) {

                // Mark spf[j] if it is
                // not previously marked
                if (spf[j] == j)
                    spf[j] = i;
            }
        }
}

// Function to find the distinct
// prime factors
void DistPrime()
{
    for (int i = 1; i < MAX; i++) {

        int idx = 1;
        int x = i;

        // Push all distinct of x
        // prime factor in v[x]
        if (x != 1)
            v[i].push_back(spf[x]);

        x = x / spf[x];

        while (x != 1) {

            if (v[i][idx - 1]
                != spf[x]) {

                // Pushback into v[i]
                v[i].push_back(spf[x]);

                // Increment the idx
                idx += 1;
            }

            // Update x = (x / spf[x])
            x = x / spf[x];
        }
    }
}

// Function to get the distinct
// factor count of arr[]
void getFactorCount(int arr[],
                    int N)
{
    // Precompute the smallest
    // Prime Factors
    sieve();

    // For distinct prime factors
    // Fill the v[] vector
    DistPrime();

    // Count of Distinct Prime
    // Factors of each array element
    for (int i = 0; i < N; i++) {
        cout << (int)v[arr[i]].size()
             << " ";
    }
}

// Driver Code
int main()
{
    int arr[] = { 6, 9, 12 };
    int N = sizeof(arr) / sizeof(arr[0]);

    getFactorCount(arr, N);

    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
class GFG
{

  static int MAX = 100001;

  // Stores smallest prime
  // factor for every number
  static int spf[];

  // Stores distinct prime factors
  static ArrayList<Integer> v[];

  // Function to find the smallest
  // prime factor of every number
  static void sieve()
  {

    // Mark the smallest prime factor
    // of every number to itself
    for (int i = 1; i < MAX; i++)
      spf[i] = i;

    // Separately mark all the
    // smallest prime factor of
    // every even number to be 2
    for (int i = 4; i < MAX; i = i + 2)
      spf[i] = 2;
    for (int i = 3; i * i < MAX; i++)

      // If i is prime
      if (spf[i] == i) {

        // Mark spf for all numbers
        // divisible by i
        for (int j = i * i; j < MAX; j = j + i) {

          // Mark spf[j] if it is
          // not previously marked
          if (spf[j] == j)
            spf[j] = i;
        }
      }
  }

  // Function to find the distinct
  // prime factors
  static void DistPrime()
  {
    for (int i = 1; i < MAX; i++) {

      int idx = 1;
      int x = i;

      // Push all distinct of x
      // prime factor in v[x]
      if (x != 1)
        v[i].add(spf[x]);

      x = x / spf[x];

      while (x != 1) {

        if (v[i].get(idx - 1) != spf[x]) {

          // Pushback into v[i]
          v[i].add(spf[x]);

          // Increment the idx
          idx += 1;
        }

        // Update x = (x / spf[x])
        x = x / spf[x];
      }
    }
  }

  // Function to get the distinct
  // factor count of arr[]
  static void getFactorCount(int arr[], int N)
  {

    // initialization
    spf = new int[MAX];
    v = new ArrayList[MAX];
    for (int i = 0; i < MAX; i++)
      v[i] = new ArrayList<>();

    // Precompute the smallest
    // Prime Factors
    sieve();

    // For distinct prime factors
    // Fill the v[] vector
    DistPrime();

    // Count of Distinct Prime
    // Factors of each array element
    for (int i = 0; i < N; i++) {
      System.out.print((int)v[arr[i]].size() + " ");
    }
  }

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

    int arr[] = { 6, 9, 12 };
    int N = arr.length;

    getFactorCount(arr, N);
  }
}

// This code is contributed by Kingash.

Python3

MAX = 100001

# Stores smallest prime
# factor for every number
spf = [0] * MAX

# Stores distinct prime factors
v = [[] for _ in range(MAX)]

# Function to find the smallest
# prime factor of every number
def sieve():

    # Mark the smallest prime factor
    # of every number to itself
    for i in range(1, MAX):
        spf[i] = i

    # Separately mark all the
    # smallest prime factor of
    # every even number to be 2
    for i in range(4, MAX, 2):
        spf[i] = 2
    for i in range(3, int(MAX**0.5)+1):

        # If i is prime
        if spf[i] == i:
            # Mark spf for all numbers
            # divisible by i
            for j in range(i*i, MAX, i):
                # Mark spf[j] if it is
                # not previously marked
                if spf[j] == j:
                    spf[j] = i

# Function to find the distinct
# prime factors
def DistPrime():
    for i in range(1, MAX):
        idx = 1
        x = i

        # Push all distinct of x
        # prime factor in v[x]
        if x != 1:
            v[i].append(spf[x])

        x = x // spf[x]

        while x != 1:
            if v[i][idx-1] != spf[x]:
                # Pushback into v[i]
                v[i].append(spf[x])

                # Increment the idx
                idx += 1

            # Update x = (x / spf[x])
            x = x // spf[x]

# Function to get the distinct
# factor count of arr[]
def getFactorCount(arr, N):
    # Precompute the smallest
    # Prime Factors
    sieve()

    # For distinct prime factors
    # Fill the v[] vector
    DistPrime()

    # Count of Distinct Prime
    # Factors of each array element
    for i in range(N):
        print(len(v[arr[i]]), end=' ')

arr = [6, 9, 12]
N = len(arr)

getFactorCount(arr, N)

# This code is contributed by phasing17.

C#

// C# program for the above approach
using System;
using System.Collections.Generic;

class GFG
{

  static int MAX = 100001;

  // Stores smallest prime
  // factor for every number
  static int[] spf;

  // Stores distinct prime factors
  static List<List<int>> v;

  // Function to find the smallest
  // prime factor of every number
  static void sieve()
  {

    // Mark the smallest prime factor
    // of every number to itself
    for (int i = 1; i < MAX; i++)
      spf[i] = i;

    // Separately mark all the
    // smallest prime factor of
    // every even number to be 2
    for (int i = 4; i < MAX; i = i + 2)
      spf[i] = 2;
    for (int i = 3; i * i < MAX; i++)

      // If i is prime
      if (spf[i] == i) {

        // Mark spf for all numbers
        // divisible by i
        for (int j = i * i; j < MAX; j = j + i) {

          // Mark spf[j] if it is
          // not previously marked
          if (spf[j] == j)
            spf[j] = i;
        }
      }
  }

  // Function to find the distinct
  // prime factors
  static void DistPrime()
  {
    for (int i = 1; i < MAX; i++) {

      int idx = 1;
      int x = i;

      // Push all distinct of x
      // prime factor in v[x]
      if (x != 1)
        v[i].Add(spf[x]);

      x = x / spf[x];

      while (x != 1) {

        if (v[i][idx - 1] != spf[x]) {

          // Pushback into v[i]
          v[i].Add(spf[x]);

          // Increment the idx
          idx += 1;
        }

        // Update x = (x / spf[x])
        x = x / spf[x];
      }
    }
  }

  // Function to get the distinct
  // factor count of arr[]
  static void getFactorCount(int[] arr, int N)
  {

    // initialization
    spf = new int[MAX];
    v = new List<List<int>>() ;
    for (int i = 0; i < MAX; i++)
      v.Add(new List<int>());

    // Precompute the smallest
    // Prime Factors
    sieve();

    // For distinct prime factors
    // Fill the v[] vector
    DistPrime();

    // Count of Distinct Prime
    // Factors of each array element
    for (int i = 0; i < N; i++) {
      Console.Write((int)v[arr[i]].Count + " ");
    }
  }

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

    int[] arr = { 6, 9, 12 };
    int N = arr.Length;

    getFactorCount(arr, N);
  }
}

// This code is contributed by phasing17.

Javascript

<script>

    // JavaScript program for the above approach

    let MAX = 100001;

    // Stores smallest prime
    // factor for every number
    let spf;

    // Stores distinct prime factors
    let v;

    // Function to find the smallest
    // prime factor of every number
    function sieve()
    {

      // Mark the smallest prime factor
      // of every number to itself
      for (let i = 1; i < MAX; i++)
        spf[i] = i;

      // Separately mark all the
      // smallest prime factor of
      // every even number to be 2
      for (let i = 4; i < MAX; i = i + 2)
        spf[i] = 2;
      for (let i = 3; i * i < MAX; i++)

        // If i is prime
        if (spf[i] == i) {

          // Mark spf for all numbers
          // divisible by i
          for (let j = i * i; j < MAX; j = j + i) {

            // Mark spf[j] if it is
            // not previously marked
            if (spf[j] == j)
              spf[j] = i;
          }
        }
    }

    // Function to find the distinct
    // prime factors
    function DistPrime()
    {
      for (let i = 1; i < MAX; i++) {

        let idx = 1;
        let x = i;

        // Push all distinct of x
        // prime factor in v[x]
        if (x != 1)
          v[i].push(spf[x]);

        x = parseInt(x / spf[x], 10);

        while (x != 1) {

          if (v[i][idx - 1] != spf[x]) {

            // Pushback into v[i]
            v[i].push(spf[x]);

            // Increment the idx
            idx += 1;
          }

          // Update x = (x / spf[x])
          x = parseInt(x / spf[x], 10);
        }
      }
    }

    // Function to get the distinct
    // factor count of arr[]
    function getFactorCount(arr, N)
    {

      // initialization
      spf = new Array(MAX);
      v = new Array(MAX);
      for (let i = 0; i < MAX; i++)
        v[i] = [];

      // Precompute the smallest
      // Prime Factors
      sieve();

      // For distinct prime factors
      // Fill the v[] vector
      DistPrime();

      // Count of Distinct Prime
      // Factors of each array element
      for (let i = 0; i < N; i++) {
        document.write(v[arr[i]].length + " ");
      }
    }

    let arr = [ 6, 9, 12 ];
    let N = arr.length;

    getFactorCount(arr, N);

</script>

Output

2 1 2

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