Sort an array according to the increasing count of distinct Prime Factors

Given an array of integers. The task is to sort the given array on the basis increasing count of distinct prime factors.

Examples:

Input : arr[] = {30, 2, 1024, 210, 3, 6}
Output : 2 1024 3 6 30 210 

Input : arr[] = {12, 16, 27, 6}
Output : 16 27 6 12

A naive approach is to find all the prime factors of each elements of the array and pair the count of the prime factors with the element in a vector and sort the array with respect to the count of prime factors.



An Efficient approach is to use a sieve to find the count of distinct prime factors and store them in a vector. Now, traverse through the array and pair the count of distinct prime factors with the element in a vector and sort the array with respect to the count of prime factors using a comparator function.

Below is the implementation of this approach:

C++

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// C++ program to sort array according to the
// count of distinct prime factors
#include <bits/stdc++.h>
using namespace std;
  
// array to store the count of distinct prime
int prime[100001];
  
void SieveOfEratosthenes()
{
    // Create a int array "prime[0..n]" and initialize
    // all entries it as 0. A value in prime[i] will
    // count of distinct prime factors.
  
    memset(prime, 0, sizeof(prime));
  
    // 0 and 1 does not have any prime factors
    prime[0] = 0;
    prime[1] = 0;
  
    for (int p = 2; p * p <= 100001; p++) {
        // If prime[p] is not changed,
        // then it is a prime
        if (prime[p] == 0) {
            prime[p]++;
  
            // Update all multiples of p greater than or
            // equal to the square of it
            // numbers which are multiple of p and are
            // less than p^2 are already been marked.
            for (int i = p * p; i <= 100001; i += p)
                prime[i]++;
        }
    }
}
  
// comparator function to sort the vector in
// ascending order of second element of the pair
bool Compare(pair<int, int> p1, pair<int, int> p2)
{
    return (p1.second < p2.second);
}
  
// Function to sort the array on the
// basis increasing count of distinct
// prime factors
void sortArr(int arr[], int n)
{
    // vector to store the number and
    // count of prime factor
    vector<pair<int, int> > v;
  
    for (int i = 0; i < n; i++) {
        // push_back the element and
        // count of distinct
        // prime factors
        v.push_back(make_pair(arr[i], prime[arr[i]]));
    }
  
    // sort the array on the
    // basis increasing count of
    // distinct prime factors
    sort(v.begin(), v.end(), Compare);
  
    // display the sorted array
    for (int i = 0; i < n; i++)
        cout << v[i].first << " ";
  
    cout << endl;
}
  
// Driver code
int main()
{
    // create the seive
    SieveOfEratosthenes();
  
    int arr[] = { 30, 2, 1024, 210, 3, 6 };
  
    int n = sizeof(arr) / sizeof(int);
  
    sortArr(arr, n);
  
    return 0;
}

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Java

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import java.util.Arrays;
  
class GFG
{
      
    static class Pair implements Comparable<Pair>
    {
        int first, second;
          
        Pair(int f, int s)
        {
            first = f;
            second = s;
        }
  
        @Override
        public int compareTo(Pair o)
        {
            if(this.second > o.second)
                return 1;
            else if(this.second == o.second)
                return 0;
            return -1;
        }
    }
    // array to store the count of distinct prime 
    static int prime[] = new int[100002]; 
      
    static void SieveOfEratosthenes()
    
        // Create a int array "prime[0..n]" and initialize 
        // all entries it as 0. A value in prime[i] will 
        // count of distinct prime factors. 
        Arrays.fill(prime, 0);
      
        // 0 and 1 does not have any prime factors 
        prime[0] = 0
        prime[1] = 0
      
        for (int p = 2; p * p <= 100001; p++) 
        
            // If prime[p] is not changed, 
            // then it is a prime 
            if (prime[p] == 0
            
                prime[p]++; 
      
                // Update all multiples of p greater than or 
                // equal to the square of it 
                // numbers which are multiple of p and are 
                // less than p^2 are already been marked. 
                for (int i = p * p; i <= 100001; i += p) 
                    prime[i]++; 
            
        
    
      
    // Function to sort the array on the 
    // basis increasing count of distinct 
    // prime factors 
    static void sortArr(int arr[], int n) 
    
        // Array to store the number and 
        // count of prime factor
        Pair v[] = new Pair[n];
      
        for (int i = 0; i < n; i++) 
        
            // update the element and 
            // count of distinct 
            // prime factors 
            v[i] = new Pair(arr[i], prime[arr[i]]);
        
      
        // sort the array on the 
        // basis increasing count of 
        // distinct prime factors 
        Arrays.sort(v);
      
        // display the sorted array 
        for (int i = 0; i < n; i++) 
            System.out.print(v[i].first + " "); 
      
        System.out.println();
    
      
    // Driver code 
    public static void main(String args[])
    
        // create the seive 
        SieveOfEratosthenes(); 
      
        int arr[] = { 30, 2, 1024, 210, 3, 6 }; 
      
        int n = arr.length; 
      
        sortArr(arr, n);
    
}
  
// This code is contributed by ghanshyampandey

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Python3

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# Python3 program to sort array according to the
# count of distinct prime factors
import functools as ft
  
# array to store the count of distinct prime
prime = [0 for i in range(100001)]
  
def SieveOfEratosthenes():
  
    # Create a array "prime[0..n]" and initialize
    # all entries it as 0. A value in prime[i]
    # will count of distinct prime factors.
  
    # memset(prime, 0, sizeof(prime))
  
    # 0 and 1 does not have any prime factors
    prime[0] = 0
    prime[1] = 0
  
    for p in range(2, 100002):
  
        if p * p > 100001:
            break
  
        # If prime[p] is not changed,
        # then it is a prime
        if (prime[p] == 0):
            prime[p] += 1
  
            # Update all multiples of p greater than
            # or equal to the square of it
            # numbers which are multiple of p and are
            # less than p^2 are already been marked.
            for i in range(2 * p, 100001, p):
                prime[i] += 1
  
# Function to sort the array on the
# basis increasing count of distinct
# prime factors
def sortArr(arr, n):
  
    # vector to store the number and
    # count of prime factor
    v = []
  
    for i in range(n):
  
        # append the element and
        # count of distinct
        # prime factors
        v.append([arr[i], prime[arr[i]]])
  
    # sort the array on the
    # basis increasing count of
    # distinct prime factors
    v.sort(key= lambda x:x[1])
  
    # display the sorted array
  
    for i in range(n):
        print(v[i][0], end = " ")
  
    print()
  
# Driver code
  
# create the seive
SieveOfEratosthenes()
  
arr = [30, 2, 1024, 210, 3, 6]
  
n = len(arr)
  
sortArr(arr, n)
  
# This code is contributed by Mohit Kumar

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Output:

2 1024 3 6 30 210


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Second year Department of Information Technology Jadavpur University

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