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; ` `} ` |

*chevron_right*

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**Output:**

20 16 10 9 5 11 23

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

Auxiliary Complexity: O(n)

## Recommended Posts:

- Sort elements on the basis of number of factors
- Sort an array according to the increasing count of distinct Prime Factors
- Number which has the maximum number of distinct prime factors in the range M to N
- Queries on sum of odd number digit sums of all the factors of a number
- Super Ugly Number (Number whose prime factors are in given set)
- Number with maximum number of prime factors
- Sum of all the factors of a number
- Product of factors of number
- Find sum of odd factors of a number
- Prime factors of a big number
- Find sum of even factors of a number
- Find minimum sum of factors of number
- Sum of Factors of a Number using Prime Factorization
- Check whether a number has exactly three distinct factors or not
- Queries to find whether a number has exactly four distinct factors or not

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