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: For20= 6,16= 5,10= 4,9= 3 and for5, 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

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.

- Count distinct number of factors of each element. Refer this.
- Create a vector of pairs which stores elements and their factor counts.
- 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 <bits/stdc++.h> using namespace std; // Function that helps to sort elements // in descending order bool compare(const pair<int, int> &a, const pair<int, int> &b) { return (a.first > b.first); } // Prints array elements sorted in descending // order of number of factors. void printSorted(int arr[], int n) { vector<pair<int, int>> 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 Function int main() { int arr[] = {5, 11, 10, 20, 9, 16, 23}; int n = sizeof(arr) / sizeof(arr[0]); 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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