Sort elements on the basis of number of factors

Given an array of positive integers. Sort the given array in decreasing order of number of factors of each element, i.e., element having the highest number of factors should be the first to be displayed and the number having least number of factors should be the last one. Two elements with equal number of factors should be in the same order as in the original array.

Examples:

Input : {5, 11, 10, 20, 9, 16, 23}
Output : 20 16 10 9 5 11 23
Number of distinct factors:
For 20 = 6, 16 = 5, 10 = 4, 9 = 3
and for 5, 11, 23 = 2 (same number of factors
therefore sorted in increasing order of index)

Input : {104, 210, 315, 166, 441, 180}
Output : 180 210 315 441 104 166

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

The following steps sort numbers in decreasing order of count of factors.

Count distinct number of factors of each element. Refer this.
You can use a structure for each element to store its original index and count of factors. Create an array of such structures to store such information for all the elements.
Sort this array of structures on the basis of the problem statement using any sorting algorithm.
Traverse this array of structures from the beginning and get the number from the original array with the help of the index stored in the structure of each element of the sorted array of structures.

C++

// C++ implementation to sort numbers on 
// the basis of factors 
#include <bits/stdc++.h> 
  
using namespace std; 
  
// structure of each element having its index 
// in the input array and number of factors 
struct element 
{ 
    int index, no_of_fact; 
}; 
  
// function to count factors for 
// a given number n 
int countFactors(int n) 
{ 
    int count = 0; 
    int sq = sqrt(n); 
      
    // if the number is a perfect square 
    if (sq * sq == n) 
        count++; 
      
    // count all other factors 
    for (int i=1; i<sqrt(n); i++) 
    { 
        // if i is a factor then n/i will be 
        // another factor. So increment by 2 
        if (n % i == 0)     
            count += 2; 
    }         
      
    return count; 
} 
  
// comparsion function for the elements  
// of the structure 
bool compare(struct element e1, struct element e2) 
{ 
    // if two elements have the same number 
    // of factors then sort them in increasing 
    // order of their index in the input array 
    if (e1.no_of_fact == e2.no_of_fact) 
        return e1.index < e2.index; 
      
    // sort in decreasing order of number of factors 
    return e1.no_of_fact > e2.no_of_fact;     
} 
  
// function to print numbers after sorting them in 
// decreasing order of number of factors 
void printOnBasisOfFactors(int arr[], int n) 
{     
    struct element num[n]; 
      
    // for each element of input array create a 
    // structure element to store its index and  
    // factors count  
    for (int i=0; i<n; i++) 
    { 
        num[i].index = i; 
        num[i].no_of_fact = countFactors(arr[i]);         
    } 
      
    // sort the array of structures as defined 
    sort(num, num+n, compare); 
      
    // access index from the structure element and correponding 
    // to that index access the element from arr 
    for (int i=0; i<n; i++) 
        cout << arr[num[i].index] << " "; 
} 
  
// Driver program to test above 
int main() 
{ 
    int arr[] = {5, 11, 10, 20, 9, 16, 23}; 
    int n = sizeof(arr) / sizeof(arr[0]); 
    printOnBasisOfFactors(arr, n); 
    return 0; 
}  

Java

// Java implementation to sort numbers on 
// the basis of factors 
  
import java.util.Arrays; 
import java.util.Comparator; 
  
class Element  
{ 
    //each element having its index 
    // in the input array and number of factors 
    int index, no_of_fact; 
   
    public Element(int i, int countFactors) 
    { 
        index = i; 
        no_of_fact = countFactors; 
    } 
  
    // method to count factors for 
    // a given number n 
    static int countFactors(int n) 
    { 
        int count = 0; 
        int sq = (int)Math.sqrt(n); 
       
        // if the number is a perfect square 
        if (sq * sq == n) 
            count++; 
       
        // count all other factors 
        for (int i=1; i<Math.sqrt(n); i++) 
        { 
            // if i is a factor then n/i will be 
            // another factor. So increment by 2 
            if (n % i == 0)     
                count += 2; 
        }         
       
        return count; 
    } 
   
    // function to print numbers after sorting them in 
    // decreasing order of number of factors 
    static void printOnBasisOfFactors(int arr[], int n) 
    {     
        Element num[] = new Element[n]; 
       
        // for each element of input array create a 
        // structure element to store its index and  
        // factors count  
        for (int i=0; i<n; i++) 
        { 
            num[i] = new Element(i,countFactors(arr[i])); 
        } 
       
        // sort the array of structures as defined 
        Arrays.sort(num,new Comparator<Element>() { 
  
            @Override
            // compare method for the elements  
            // of the structure 
            public int compare(Element e1, Element e2) { 
                // if two elements have the same number 
                // of factors then sort them in increasing 
                // order of their index in the input array 
                if (e1.no_of_fact == e2.no_of_fact) 
                 return e1.index < e2.index ? -1 : 1; 
                
                // sort in decreasing order of number of factors 
                return e1.no_of_fact > e2.no_of_fact ? -1 : 1;   
            } 
              
        }); 
       
        // access index from the structure element and correponding 
        // to that index access the element from arr 
        for (int i=0; i<n; i++) 
            System.out.print(arr[num[i].index]+" "); 
    } 
   
    // Driver program to test above 
    public static void main(String[] args)  
    { 
          
        int arr[] = {5, 11, 10, 20, 9, 16, 23}; 
          
        printOnBasisOfFactors(arr, arr.length); 
  
    } 
} 
// This code is contributed by Gaurav Miglani 

Output:

20 16 10 9 5 11 23

Time Complexity: O(n √n)

This article is contributed by Ayush Jauhari. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

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