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

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  • Last Updated : 09 Sep, 2022
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Given an array of integers. The task is to sort the given array on the basis of 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++




// 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 sieve
    SieveOfEratosthenes();
 
    int arr[] = { 30, 2, 1024, 210, 3, 6 };
 
    int n = sizeof(arr) / sizeof(int);
 
    sortArr(arr, n);
 
    return 0;
}


Java




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 sieve
        SieveOfEratosthenes();
     
        int arr[] = { 30, 2, 1024, 210, 3, 6 };
     
        int n = arr.length;
     
        sortArr(arr, n);
    }
}
 
// This code is contributed by ghanshyampandey


Python3




# 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 sieve
SieveOfEratosthenes()
 
arr = [30, 2, 1024, 210, 3, 6]
 
n = len(arr)
 
sortArr(arr, n)
 
# This code is contributed by Mohit Kumar


C#




// C# program to sort array according to the
// count of distinct prime factors
using System;
using System.Collections.Generic;
 
class GFG{
 
public class Pair : IComparable<Pair>
{
    public int first, second;
 
    public 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.
    Array.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
    Array.Sort(v);
 
    // Display the sorted array
    for(int i = 0; i < n; i++)
        Console.Write(v[i].first + " ");
 
    Console.WriteLine();
}
 
// Driver code
public static void Main(string[] args)
{
     
    // Create the sieve
    SieveOfEratosthenes();
 
    int[] arr = { 30, 2, 1024, 210, 3, 6 };
    int n = arr.Length;
 
    sortArr(arr, n);
}
}
 
// This code is contributed by grand_master


Javascript




<script>
 
   // JavaScript program to sort array according to the
  // count of distinct prime factors
     
    // array to store the count of distinct prime
    let prime = new Array(100002);
     
    function 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.
        for(let i=0;i<prime.length;i++)
        {
            prime[i]=0;
        }
      
        // 0 and 1 does not have any prime factors
        prime[0] = 0;
        prime[1] = 0;
      
        for (let 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 (let i = p * p; i <= 100001; i += p)
                    prime[i]++;
            }
        }
    }
     
    // Function to sort the array on the
    // basis increasing count of distinct
    // prime factors
    function sortArr(arr,n)
    {
        // Array to store the number and
        // count of prime factor
        let v=[];
      
        for (let i = 0; i < n; i++)
        {
            // update the element and
            // count of distinct
            // prime factors
            v.push([arr[i], prime[arr[i]]]);
             
        }
         
         
      
        // sort the array on the
        // basis increasing count of
        // distinct prime factors
        v.sort(function(a,b){return a[1]-b[1];});
      
        // display the sorted array
        for (let i = 0; i < n; i++)
            document.write(v[i][0] + " ");
      
        document.write("<br>");
    }
     
    // Driver code
     
    // create the sieve
    SieveOfEratosthenes();
     
    let arr=[30, 2, 1024, 210, 3, 6];
    let n = arr.length;
     
    sortArr(arr, n);
 
     
 
// This code is contributed by patel2127
 
</script>


Output

2 1024 3 6 30 210 

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