Find prime factors of Array elements whose sum of exponents is divisible by K

Given an array arr[] of N positive integers and an integer K., The task is to create a set of prime numbers such that the sum of all the powers of prime numbers in the prime factorization of all the array elements is divisible by K.

Examples:

Input: arr[] = {1, 2, 3}, K = 1 
Output: {2, 3} 
Explanation:
2 = 2¹
3 = 3¹
The power of 2 is 1 which is divisible by K(=1). 
The power of 2 is 1 which is divisible by K(=1).

Input: arr[] = {2, 2, 4, 8}, K = 10 
Output: {} 
Explanation: 
2 = 2¹
2 = 2¹
4 = 2²
8 = 2³
The power of 2 is (1 + 1 + 2 + 3) = 7 which is not divisible by K(=10). 
Thus, the output empty set.

Naive approach: The idea is to find all prime numbers less than or equal to the maximum element of the array arr[]. For each prime number count number of times, it divides the array element. If the value of count is divisible by K, then insert the prime number into the resultant set. At the end print elements of the set. 

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

Efficient Approach: To optimize the above approach the idea is to precompute the count of all the prime factors of all the numbers. Below are the steps:

1. Create the smallest prime factorization array spf[] up to the maximum number in the array. This step is used to precalculate the prime factors of a number.
2. Traverse the given array arr[] and for each element find the sum of all the count of factors stored in spf[] array.
3. For each sum of the power of a prime number in the above steps stored it frequency in a Map.
4. Traverse the map if, for any number, frequency is divisible by K then store that number.
5. Finally, print all the numbers stored in the above step.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <iostream>
#include <unordered_map>
#include <vector>
using namespace std;

// To store the smallest prime
// factor till 10^5
int spf[10001];

// Function to compute smallest
// prime factor array
void spf_array(int spf[])
{
    // Initialize the spf array
    // first element
    spf[1] = 1;

    for (int i = 2; i < 1000; i++)

        // Marking smallest prime
        // factor for every number
        // to be itself
        spf[i] = i;

    // Separately marking smallest
    // prime factor for every even
    // number as 2
    for (int i = 4; i < 1000; i += 2)
        spf[i] = 2;

    for (int i = 3; i * i < 1000; i++) {

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

            // Marking SPF for all
            // numbers divisible by i
            for (int j = i * i;
                j < 1000; j += i)

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

// Function that finds minimum operation
void frequent_prime(int arr[], int N,
                    int K)
{

    // Create a spf[] array
    spf_array(spf);

    // Map created to store the
    // unique prime numbers
    unordered_map<int, int> Hmap;

    // To store the result
    vector<int> result;
    int i = 0;

    // To store minimum operations
    int c = 0;

    // To store every unique
    // prime number
    for (i = 0; i < N; i++) {

        int x = arr[i];
        while (x != 1) {

            Hmap[spf[x]]
                = Hmap[spf[x]] + 1;
            x = x / spf[x];
        }
    }

    // Erase 1 as a key because
    // it is not a prime number
    Hmap.erase(1);

    for (auto x : Hmap) {

        // First Prime Number
        int primeNum = x.first;
        int frequency = x.second;

        // Frequency is divisible
        // by K then insert primeNum
        // in the result[]
        if (frequency % K == 0) {
            result.push_back(primeNum);
        }
    }

    // Print the elements
    // if it exists
    if (result.size() > 0) {

        for (auto& it : result) {
            cout << it << ' ';
        }
    }
    else {
        cout << "{}";
    }
}

// Driver Code
int main()
{
    // Given array arr[]
    int arr[] = { 1, 4, 6 };

    // Given K
    int K = 1;

    int N = sizeof(arr) / sizeof(arr[0]);

    // Function Call
    frequent_prime(arr, N, K);
}

Java

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

class GFG{

// To store the smallest prime
// factor till 10^5
static int[] spf = new int[10001];

// Function to compute smallest
// prime factor array
static void spf_array(int spf[])
{

    // Initialize the spf array
    // first element
    spf[1] = 1;

    for(int i = 2; i < 1000; i++)

        // Marking smallest prime
        // factor for every number
        // to be itself
        spf[i] = i;

    // Separately marking smallest
    // prime factor for every even
    // number as 2
    for(int i = 4; i < 1000; i += 2)
        spf[i] = 2;

    for(int i = 3; i * i < 1000; i++)
    {

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

            // Marking SPF for all
            // numbers divisible by i
            for(int j = i * i;
                    j < 1000; j += i)

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

// Function that finds minimum operation
static void frequent_prime(int arr[], int N,
                                      int K)
{

    // Create a spf[] array
    spf_array(spf);

    // Map created to store the
    // unique prime numbers
    Map<Integer, Integer> Hmap = new TreeMap<>();

    // To store the result
    ArrayList<Integer> result = new ArrayList<>();
    int i = 0;

    // To store minimum operations
    int c = 0;

    // To store every unique
    // prime number
    for(i = 0; i < N; i++)
    {
        int x = arr[i];
        while (x != 1)
        {
            Hmap.put(spf[x],
                     Hmap.getOrDefault(spf[x], 0) + 1);
            x = x / spf[x];
        }
    }

    // Erase 1 as a key because
    // it is not a prime number
    Hmap.remove(1);

    for(Map.Entry<Integer,
                  Integer> x : Hmap.entrySet())
    {

        // First prime number
        int primeNum = x.getKey();
        int frequency = x.getValue();

        // Frequency is divisible
        // by K then insert primeNum
        // in the result[]
        if (frequency % K == 0)
        {
            result.add(primeNum);
        }
    }

    // Print the elements
    // if it exists
    if (result.size() > 0)
    {
        for(Integer it : result)
        {
            System.out.print(it + " ");
        }
    }
    else
    {
        System.out.print("{}");
    }
}

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

    // Given array arr[]
    int arr[] = { 1, 4, 6 };

    // Given K
    int K = 1;

    int N = arr.length;

    // Function call
    frequent_prime(arr, N, K);
}
}

// This code is contributed by offbeat

C#

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

  // To store the smallest prime
  // factor till 10^5
  static int[] spf = new int[10001];

  // Function to compute smallest
  // prime factor array
  static void spf_array(int[] spf)
  {

    // Initialize the spf array
    // first element
    spf[1] = 1;

    for (int i = 2; i < 1000; i++)

      // Marking smallest prime
      // factor for every number
      // to be itself
      spf[i] = i;

    // Separately marking smallest
    // prime factor for every even
    // number as 2
    for (int i = 4; i < 1000; i += 2)
      spf[i] = 2;

    for (int i = 3; i * i < 1000; i++)
    {

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

        // Marking SPF for all
        // numbers divisible by i
        for (int j = i * i; j < 1000; j += i)

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

  // Function that finds minimum operation
  static void frequent_prime(int[] arr, int N, int K)
  {

    // Create a spf[] array
    spf_array(spf);

    // Map created to store the
    // unique prime numbers
    SortedDictionary<int,
                     int> Hmap = new SortedDictionary<int,
                                                      int>();

    // To store the result
    List<int> result = new List<int>();
    int i = 0;

    // To store every unique
    // prime number
    for (i = 0; i < N; i++)
    {
      int x = arr[i];
      while (x != 1)
      {
        if (Hmap.ContainsKey(spf[x]))
          Hmap[spf[x]] = spf[x] + 1;
        else
          Hmap.Add(spf[x], 1);
        x = x / spf[x];
      }
    }

    // Erase 1 as a key because
    // it is not a prime number
    Hmap.Remove(1);

    foreach(KeyValuePair<int, int> x in Hmap)
    {

      // First prime number
      int primeNum = x.Key;
      int frequency = x.Value;

      // Frequency is divisible
      // by K then insert primeNum
      // in the result[]
      if (frequency % K == 0)
      {
        result.Add(primeNum);
      }
    }

    // Print the elements
    // if it exists
    if (result.Count > 0)
    {
      foreach(int it in result)
      {
        Console.Write(it + " ");
      }
    }
    else
    {
      Console.Write("{}");
    }
  }

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

    // Given array []arr
    int[] arr = {1, 4, 6};

    // Given K
    int K = 1;

    int N = arr.Length;

    // Function call
    frequent_prime(arr, N, K);
  }
}

// This code is contributed by Rajput-Ji

Python3

# Python3 program for the above approach

# To store the smallest prime
# factor till 10^5
spf  = [0 for i in range(10001)]

# Function to compute smallest
# prime factor array
def spf_array(spf):

    # Initialize the spf array
    # first element
    spf[1] = 1

    for i in range(2, 1000, 1):

        # Marking smallest prime
        # factor for every number
        # to be itself
        spf[i] = i

    # Separately marking smallest
    # prime factor for every even
    # number as 2
    for i in range(4, 1000, 2):
        spf[i] = 2

    i = 3
    while( i* i < 1000):

        # Checking if i is prime
        if (spf[i] == i):

            # Marking SPF for all
            # numbers divisible by i
            j = i * i
            while(j < 1000):

                # Marking spf[j] if it is
                # not previously marked
                if (spf[j] == j):
                    spf[j] = i

                j += i

        i += 1

# Function that finds minimum operation
def frequent_prime(arr, N, K):

    # Create a spf[] array
    spf_array(spf)

    # Map created to store the
    # unique prime numbers
    Hmap = {}

    # To store the result
    result = []
    i = 0

    # To store minimum operations
    c = 0

    # To store every unique
    # prime number
    for i in range(N):
        x = arr[i]

        while (x != 1):
            Hmap[spf[x]] = Hmap.get(spf[x], 0) + 1
            x = x // spf[x]

    # Erase 1 as a key because
    # it is not a prime number
    if (1 in Hmap):
      Hmap.pop(1)

    for key, value in Hmap.items():

        # First Prime Number
        primeNum = key
        frequency = value

        # Frequency is divisible
        # by K then insert primeNum
        # in the result[]
        if (frequency % K == 0):
            result.append(primeNum)

    # Print the elements
    # if it exists
    result = result[::-1]

    if (len(result) > 0):
        for it in result:
            print(it, end = " ")
    else:
        print("{}")

# Driver Code
if __name__ == '__main__':

    # Given array arr[]
    arr =  [ 1, 4, 6 ]

    # Given K
    K = 1

    N = len(arr)

    # Function Call
    frequent_prime(arr, N, K)

# This code is contributed by ipg2016107

Javascript

<script>

// JavaScript program for the above approach

// To store the smallest prime
// factor till 10^5
var spf = Array(10001);

// Function to compute smallest
// prime factor array
function spf_array(spf)
{
    // Initialize the spf array
    // first element
    spf[1] = 1;

    for (var i = 2; i < 1000; i++)

        // Marking smallest prime
        // factor for every number
        // to be itself
        spf[i] = i;

    // Separately marking smallest
    // prime factor for every even
    // number as 2
    for (var i = 4; i < 1000; i += 2)
        spf[i] = 2;

    for (var i = 3; i * i < 1000; i++) {

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

            // Marking SPF for all
            // numbers divisible by i
            for (var j = i * i;
                j < 1000; j += i)

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

// Function that finds minimum operation
function frequent_prime(arr, N, K)
{

    // Create a spf[] array
    spf_array(spf);

    // Map created to store the
    // unique prime numbers
    var Hmap = new Map();

    // To store the result
    var result = [];
    var i = 0;

    // To store minimum operations
    var c = 0;

    // To store every unique
    // prime number
    for (i = 0; i < N; i++) {

        var x = arr[i];
        while (x != 1) {

            if(Hmap.has(spf[x]))
                Hmap.set(spf[x], Hmap.get(spf[x])+1)
            else
                Hmap.set(spf[x], 1);
            x = parseInt(x / spf[x]);
        }
    }

    // Erase 1 as a key because
    // it is not a prime number
    Hmap.delete(1);

    Hmap.forEach((value, key) => {


        // First Prime Number
        var primeNum = key;
        var frequency = value;

        // Frequency is divisible
        // by K then insert primeNum
        // in the result[]
        if (frequency % K == 0) {
            result.push(primeNum);
        }
    });

    // Print the elements
    // if it exists
    if (result.length > 0) {
        result.forEach(it => {
            document.write(it+" ");
        });

    }
    else {
        document.write( "{}");
    }
}

// Driver Code

// Given array arr[]
var arr = [1, 4, 6];

// Given K
var K = 1;
var N = arr.length;

// Function Call
frequent_prime(arr, N, K);


</script>

Output:

3 2

Time Complexity: O(M*log(M)), where M is the maximum element of the array. 
Auxiliary Space: O(M)