# Sort an array according to the increasing count of distinct Prime Factors

Given an array of integers. The task is to sort the given array on the basis increasing count of distinct prime factors.

Examples:

```Input : arr[] = {30, 2, 1024, 210, 3, 6}
Output : 2 1024 3 6 30 210

Input : arr[] = {12, 16, 27, 6}
Output : 16 27 6 12
```

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

A naive approach is to find all the prime factors of each elements of the array and pair the count of the prime factors with the element in a vector and sort the array with respect to the count of prime factors.

An Efficient approach is to use a sieve to find the count of distinct prime factors and store them in a vector. Now, traverse through the array and pair the count of distinct prime factors with the element in a vector and sort the array with respect to the count of prime factors using a comparator function.

Below is the implementation of this approach:

## C++

 `// C++ program to sort array according to the ` `// count of distinct prime factors ` `#include ` `using` `namespace` `std; ` ` `  `// array to store the count of distinct prime ` `int` `prime; ` ` `  `void` `SieveOfEratosthenes() ` `{ ` `    ``// Create a int array "prime[0..n]" and initialize ` `    ``// all entries it as 0. A value in prime[i] will ` `    ``// count of distinct prime factors. ` ` `  `    ``memset``(prime, 0, ``sizeof``(prime)); ` ` `  `    ``// 0 and 1 does not have any prime factors ` `    ``prime = 0; ` `    ``prime = 0; ` ` `  `    ``for` `(``int` `p = 2; p * p <= 100001; p++) { ` `        ``// If prime[p] is not changed, ` `        ``// then it is a prime ` `        ``if` `(prime[p] == 0) { ` `            ``prime[p]++; ` ` `  `            ``// Update all multiples of p greater than or ` `            ``// equal to the square of it ` `            ``// numbers which are multiple of p and are ` `            ``// less than p^2 are already been marked. ` `            ``for` `(``int` `i = p * p; i <= 100001; i += p) ` `                ``prime[i]++; ` `        ``} ` `    ``} ` `} ` ` `  `// comparator function to sort the vector in ` `// ascending order of second element of the pair ` `bool` `Compare(pair<``int``, ``int``> p1, pair<``int``, ``int``> p2) ` `{ ` `    ``return` `(p1.second < p2.second); ` `} ` ` `  `// Function to sort the array on the ` `// basis increasing count of distinct ` `// prime factors ` `void` `sortArr(``int` `arr[], ``int` `n) ` `{ ` `    ``// vector to store the number and ` `    ``// count of prime factor ` `    ``vector > v; ` ` `  `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``// push_back the element and ` `        ``// count of distinct ` `        ``// prime factors ` `        ``v.push_back(make_pair(arr[i], prime[arr[i]])); ` `    ``} ` ` `  `    ``// sort the array on the ` `    ``// basis increasing count of ` `    ``// distinct prime factors ` `    ``sort(v.begin(), v.end(), Compare); ` ` `  `    ``// display the sorted array ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``cout << v[i].first << ``" "``; ` ` `  `    ``cout << endl; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``// create the seive ` `    ``SieveOfEratosthenes(); ` ` `  `    ``int` `arr[] = { 30, 2, 1024, 210, 3, 6 }; ` ` `  `    ``int` `n = ``sizeof``(arr) / ``sizeof``(``int``); ` ` `  `    ``sortArr(arr, n); ` ` `  `    ``return` `0; ` `} `

## Java

 `import` `java.util.Arrays; ` ` `  `class` `GFG ` `{ ` `     `  `    ``static` `class` `Pair ``implements` `Comparable ` `    ``{ ` `        ``int` `first, second; ` `         `  `        ``Pair(``int` `f, ``int` `s) ` `        ``{ ` `            ``first = f; ` `            ``second = s; ` `        ``} ` ` `  `        ``@Override` `        ``public` `int` `compareTo(Pair o) ` `        ``{ ` `            ``if``(``this``.second > o.second) ` `                ``return` `1``; ` `            ``else` `if``(``this``.second == o.second) ` `                ``return` `0``; ` `            ``return` `-``1``; ` `        ``} ` `    ``} ` `    ``// array to store the count of distinct prime  ` `    ``static` `int` `prime[] = ``new` `int``[``100002``];  ` `     `  `    ``static` `void` `SieveOfEratosthenes() ` `    ``{  ` `        ``// Create a int array "prime[0..n]" and initialize  ` `        ``// all entries it as 0. A value in prime[i] will  ` `        ``// count of distinct prime factors.  ` `        ``Arrays.fill(prime, ``0``); ` `     `  `        ``// 0 and 1 does not have any prime factors  ` `        ``prime[``0``] = ``0``;  ` `        ``prime[``1``] = ``0``;  ` `     `  `        ``for` `(``int` `p = ``2``; p * p <= ``100001``; p++)  ` `        ``{  ` `            ``// If prime[p] is not changed,  ` `            ``// then it is a prime  ` `            ``if` `(prime[p] == ``0``)  ` `            ``{  ` `                ``prime[p]++;  ` `     `  `                ``// Update all multiples of p greater than or  ` `                ``// equal to the square of it  ` `                ``// numbers which are multiple of p and are  ` `                ``// less than p^2 are already been marked.  ` `                ``for` `(``int` `i = p * p; i <= ``100001``; i += p)  ` `                    ``prime[i]++;  ` `            ``}  ` `        ``}  ` `    ``}  ` `     `  `    ``// Function to sort the array on the  ` `    ``// basis increasing count of distinct  ` `    ``// prime factors  ` `    ``static` `void` `sortArr(``int` `arr[], ``int` `n)  ` `    ``{  ` `        ``// Array to store the number and  ` `        ``// count of prime factor ` `        ``Pair v[] = ``new` `Pair[n]; ` `     `  `        ``for` `(``int` `i = ``0``; i < n; i++)  ` `        ``{  ` `            ``// update the element and  ` `            ``// count of distinct  ` `            ``// prime factors  ` `            ``v[i] = ``new` `Pair(arr[i], prime[arr[i]]); ` `        ``}  ` `     `  `        ``// sort the array on the  ` `        ``// basis increasing count of  ` `        ``// distinct prime factors  ` `        ``Arrays.sort(v); ` `     `  `        ``// display the sorted array  ` `        ``for` `(``int` `i = ``0``; i < n; i++)  ` `            ``System.out.print(v[i].first + ``" "``);  ` `     `  `        ``System.out.println(); ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `main(String args[]) ` `    ``{  ` `        ``// create the seive  ` `        ``SieveOfEratosthenes();  ` `     `  `        ``int` `arr[] = { ``30``, ``2``, ``1024``, ``210``, ``3``, ``6` `};  ` `     `  `        ``int` `n = arr.length;  ` `     `  `        ``sortArr(arr, n); ` `    ``}  ` `} ` ` `  `// This code is contributed by ghanshyampandey `

## Python3

 `# Python3 program to sort array according to the ` `# count of distinct prime factors ` `import` `functools as ft ` ` `  `# array to store the count of distinct prime ` `prime ``=` `[``0` `for` `i ``in` `range``(``100001``)] ` ` `  `def` `SieveOfEratosthenes(): ` ` `  `    ``# Create a array "prime[0..n]" and initialize ` `    ``# all entries it as 0. A value in prime[i] ` `    ``# will count of distinct prime factors. ` ` `  `    ``# memset(prime, 0, sizeof(prime)) ` ` `  `    ``# 0 and 1 does not have any prime factors ` `    ``prime[``0``] ``=` `0` `    ``prime[``1``] ``=` `0` ` `  `    ``for` `p ``in` `range``(``2``, ``100002``): ` ` `  `        ``if` `p ``*` `p > ``100001``: ` `            ``break` ` `  `        ``# If prime[p] is not changed, ` `        ``# then it is a prime ` `        ``if` `(prime[p] ``=``=` `0``): ` `            ``prime[p] ``+``=` `1` ` `  `            ``# Update all multiples of p greater than ` `            ``# or equal to the square of it ` `            ``# numbers which are multiple of p and are ` `            ``# less than p^2 are already been marked. ` `            ``for` `i ``in` `range``(``2` `*` `p, ``100001``, p): ` `                ``prime[i] ``+``=` `1` ` `  `# Function to sort the array on the ` `# basis increasing count of distinct ` `# prime factors ` `def` `sortArr(arr, n): ` ` `  `    ``# vector to store the number and ` `    ``# count of prime factor ` `    ``v ``=` `[] ` ` `  `    ``for` `i ``in` `range``(n): ` ` `  `        ``# append the element and ` `        ``# count of distinct ` `        ``# prime factors ` `        ``v.append([arr[i], prime[arr[i]]]) ` ` `  `    ``# sort the array on the ` `    ``# basis increasing count of ` `    ``# distinct prime factors ` `    ``v.sort(key``=` `lambda` `x:x[``1``]) ` ` `  `    ``# display the sorted array ` ` `  `    ``for` `i ``in` `range``(n): ` `        ``print``(v[i][``0``], end ``=` `" "``) ` ` `  `    ``print``() ` ` `  `# Driver code ` ` `  `# create the seive ` `SieveOfEratosthenes() ` ` `  `arr ``=` `[``30``, ``2``, ``1024``, ``210``, ``3``, ``6``] ` ` `  `n ``=` `len``(arr) ` ` `  `sortArr(arr, n) ` ` `  `# This code is contributed by Mohit Kumar `

Output:

```2 1024 3 6 30 210
```

My Personal Notes arrow_drop_up Second year Department of Information Technology Jadavpur University

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