Sort elements on the basis of number of factors

• Difficulty Level : Easy
• Last Updated : 24 Jun, 2021

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

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

1. Count distinct number of factors of each element. Refer this.
2. 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.
3. Sort this array of structures on the basis of the problem statement using any sorting algorithm.
4. 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  using namespace std; // structure of each element having its index// in the input array and number of factorsstruct element{    int index, no_of_fact;}; // function to count factors for// a given number nint 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 e2.no_of_fact;   } // function to print numbers after sorting them in// decreasing order of number of factorsvoid 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

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() {             @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 corresponding        // to that index access the element from arr        for (int i=0; i

Output:

20 16 10 9 5 11 23

Time Complexity: O(n √n)
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