Skip to content
Related Articles

Related Articles

Count distinct prime factors for each element of an array
  • Difficulty Level : Medium
  • Last Updated : 16 Mar, 2021

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 prime factors of each array element. Then, find the distinct prime numbers among them and print that count for each array element. 
Time Complexity: O(N2)
Auxiliary Space: O(1)

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;
 
    // Seperately 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;
 
    // Seperately 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.
Output: 
2 1 2

 

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

Attention reader! Don’t stop learning now. Get hold of all the important mathematical concepts for competitive programming with the Essential Maths for CP Course at a student-friendly price. To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

My Personal Notes arrow_drop_up
Recommended Articles
Page :