Sort on the basis of number of factors using STL

• Difficulty Level : Easy
• Last Updated : 11 Feb, 2018

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.

We have already discussed a structure based solution to sort according to number of factors.

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

1. Count distinct number of factors of each element. Refer this.
2. Create a vector of pairs which stores elements and their factor counts.
3. Sort this array on the basis of the problem statement using any sorting algorithm.
 // Sort an array of numbers according// to number of factors.#include using namespace std;  // Function that helps to sort elements // in descending orderbool compare(const pair &a,             const pair &b) {  return (a.first > b.first);}  // Prints array elements sorted in descending// order of number of factors.void printSorted(int arr[], int n) {    vector> v;    for (int i = 0; i < n; i++) {      // Count factors of arr[i].    int count = 0;    for (int j = 1; j * j <= arr[i]; j++) {        // To check Given Number is Exactly      // divisible      if (arr[i] % j == 0) {        count++;          // To check Given number is perfect        // square        if (arr[i] / j != j)          count++;      }    }      // Insert factor count and array element    v.push_back(make_pair(count, arr[i]));  }    // Sort the vector  sort(v.begin(), v.end(), compare);    // Print the vector  for (int i = 0; i < v.size(); i++)     cout << v[i].second << " ";}  // Driver's Functionint main() {  int arr[] = {5, 11, 10, 20, 9, 16, 23};  int n = sizeof(arr) / sizeof(arr);    printSorted(arr, n);    return 0;}
Output:
20 16 10 9 5 11 23

Time Complexity: O(n*sqrt(n))
Auxiliary Complexity: O(n)

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