Given an array arr[] of size N, the task is to find the count of distinct prime factors of each element of the given array.
Examples:
Input: arr[] = {6, 9, 12}
Output: 2 1 2
Explanation:
- 6 = 2 × 3. Therefore, count = 2
- 9 = 3 × 3. Therefore, count = 1
- 12 = 2 × 2 × 3. Therefore, count = 2
The count of distinct prime factors of each array element are 2, 1, 2 respectively.
Input: arr[] = {4, 8, 10, 6}
Output: 1 1 2 2
Naive Approach: The simplest approach to solve the problem is to find the distinct prime factors of each array element. Then, print that count for each array element.
steps to implement-
- Run a loop over the given array to find the count of distinct prime factors of each element
- For finding the count of distinct prime factors of each element-
- Initialize a variable count with a value of 0
- First, check if the number can be divided by 2. If it can be then divide it by 2 till it can be divided and after that increment the count by 1.
- Then check for all odd numbers from 3 till its square root that it can divide the number or not
- If any odd number can divide then it should divide till it can and after that increment count by 1 for each odd number
- In last, if still we have number>2 then this number that is remaining is any prime number so increment count by 1
- In the last print/return the count
Code-
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// A function to provide count of//distinct prime factor of a given numberint primeFactors(int n){ //To store count of distinct prime factor of given number int count=0; //Boolean variable to check an element //is included in its prime factor or not bool val=false; // Remove all 2s that can be prime factor of n while (n % 2 == 0) { val=true; n = n/2; } //If 2 is included if(val==true){count++;} // n must be odd at this point. So we can skip // one element (Note i = i +2) for (int i = 3; i <= sqrt(n); i = i + 2) { //To check whether i is included in prime factor val=false; // While i divides n,divide n while (n % i == 0) { val=true; n = n/i; } //If i is included in prime factor if(val==true){count++;} } // This condition is to handle the case when n // is a prime number greater than 2 if (n > 2){ count++; } return count; }// Function to get the distinct// factor count of arr[]void getFactorCount(int arr[], int N){ //Traverse every array element for(int i=0;i<N;i++){ cout<<primeFactors(arr[i])<<" "; }}// Driver Codeint main(){ int arr[] = { 6, 9, 12 }; int N = sizeof(arr) / sizeof(arr[0]); getFactorCount(arr, N); return 0;} |
Java
import java.util.*;public class Main { // A function to provide count of // distinct prime factors of a given number static int primeFactors(int n) { // To store the count of distinct prime factors of the given number int count = 0; // Boolean variable to check if an element is included in its prime factor or not boolean val = false; // Remove all 2s that can be prime factors of n while (n % 2 == 0) { val = true; n = n / 2; } // If 2 is included if (val) { count++; } // n must be odd at this point. So we can skip one element (Note i = i + 2) for (int i = 3; i <= Math.sqrt(n); i = i + 2) { // To check whether i is included in prime factor val = false; // While i divides n, divide n while (n % i == 0) { val = true; n = n / i; } // If i is included in the prime factor if (val) { count++; } } // This condition is to handle the case when n is a prime number greater than 2 if (n > 2) { count++; } return count; } // Function to get the distinct factor count of arr[] static void getFactorCount(int[] arr, int N) { // Traverse every array element for (int i = 0; i < N; i++) { System.out.print(primeFactors(arr[i]) + " "); } } public static void main(String[] args) { int[] arr = { 6, 9, 12 }; int N = arr.length; getFactorCount(arr, N); }} |
Python3
import math# A function to provide count of distinct prime factor of a given numberdef prime_factors(n): # To store count of distinct prime factors of the given number count = 0 # Boolean variable to check if an element is included in its prime factors or not val = False # Remove all 2s that can be prime factors of n while n % 2 == 0: val = True n = n // 2 # If 2 is included if val == True: count += 1 # n must be odd at this point. So we can skip one element (Note i = i + 2) for i in range(3, int(math.sqrt(n)) + 1, 2): # To check whether i is included in prime factors val = False # While i divides n, divide n while n % i == 0: val = True n = n // i # If i is included in prime factors if val == True: count += 1 # This condition is to handle the case when n is a prime number greater than 2 if n > 2: count += 1 return count# Function to get the distinct factor count of arr[]def get_factor_count(arr): # Traverse every array element for num in arr: print(prime_factors(num), end=" ")# Driver Codeif __name__ == "__main__": arr = [6, 9, 12] get_factor_count(arr) |
Javascript
// JavaScript program for the above approach// A function to provide count of//distinct prime factor of a given numberfunction primeFactors(n){ //To store count of distinct prime factor of given number let count=0; //Boolean variable to check an element //is included in its prime factor or not let val=false; // Remove all 2s that can be prime factor of n while (n % 2 == 0) { val=true; n = n/2; } //If 2 is included if(val==true){count++;} // n must be odd at this point. So we can skip // one element (Note i = i +2) for (let i = 3; i <= Math.sqrt(n); i = i + 2) { //To check whether i is included in prime factor val=false; // While i divides n,divide n while (n % i == 0) { val=true; n = n/i; } //If i is included in prime factor if(val==true){count++;} } // This condition is to handle the case when n // is a prime number greater than 2 if (n > 2){ count++; } return count; }// Function to get the distinct// factor count of arr[]function getFactorCount(arr, N){ //Traverse every array element for(let i=0;i<N;i++){ document.write(primeFactors(arr[i])+ " "); }}// Driver Code let arr = [ 6, 9, 12 ]; let N = arr.length; getFactorCount(arr, N); |
Output-
2 1 2
Time Complexity: O(N * (√Maximum value present in array)), because O(N) in traversing the array and (√Maximum value present in array) is the maximum time to find the count of distinct prime factors of each Number
Auxiliary Space: O(1), because no extra space has been used
Efficient Approach: The above approach can be optimized by precomputing the distinct factors of all the numbers using their Smallest Prime Factors. Follow the steps below to solve the problem
- Initialize a vector, say v, to store the distinct prime factors.
- Store the Smallest Prime Factor(SPF) up to 105 using a Sieve of Eratosthenes.
- Calculate the distinct prime factors of all the numbers by dividing the numbers recursively with their smallest prime factor till it reduces to 1 and store distinct prime factors of X, in v[X].
- Traverse the array arr[] and for each array element, print the count as v[arr[i]].size().
Below is the implementation of the above approach :
C++14
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;#define MAX 100001// Stores smallest prime// factor for every numberint spf[MAX];// Stores distinct prime factorsvector<int> v[MAX];// Function to find the smallest// prime factor of every numbervoid sieve(){ // Mark the smallest prime factor // of every number to itself for (int i = 1; i < MAX; i++) spf[i] = i; // Separately mark all the // smallest prime factor of // every even number to be 2 for (int i = 4; i < MAX; i = i + 2) spf[i] = 2; for (int i = 3; i * i < MAX; i++) // If i is prime if (spf[i] == i) { // Mark spf for all numbers // divisible by i for (int j = i * i; j < MAX; j = j + i) { // Mark spf[j] if it is // not previously marked if (spf[j] == j) spf[j] = i; } }}// Function to find the distinct// prime factorsvoid DistPrime(){ for (int i = 1; i < MAX; i++) { int idx = 1; int x = i; // Push all distinct of x // prime factor in v[x] if (x != 1) v[i].push_back(spf[x]); x = x / spf[x]; while (x != 1) { if (v[i][idx - 1] != spf[x]) { // Pushback into v[i] v[i].push_back(spf[x]); // Increment the idx idx += 1; } // Update x = (x / spf[x]) x = x / spf[x]; } }}// Function to get the distinct// factor count of arr[]void getFactorCount(int arr[], int N){ // Precompute the smallest // Prime Factors sieve(); // For distinct prime factors // Fill the v[] vector DistPrime(); // Count of Distinct Prime // Factors of each array element for (int i = 0; i < N; i++) { cout << (int)v[arr[i]].size() << " "; }}// Driver Codeint main(){ int arr[] = { 6, 9, 12 }; int N = sizeof(arr) / sizeof(arr[0]); getFactorCount(arr, N); return 0;} |
Java
// Java program for the above approachimport java.io.*;import java.lang.*;import java.util.*;class GFG { static int MAX = 100001; // Stores smallest prime // factor for every number static int spf[]; // Stores distinct prime factors static ArrayList<Integer> v[]; // Function to find the smallest // prime factor of every number static void sieve() { // Mark the smallest prime factor // of every number to itself for (int i = 1; i < MAX; i++) spf[i] = i; // Separately mark all the // smallest prime factor of // every even number to be 2 for (int i = 4; i < MAX; i = i + 2) spf[i] = 2; for (int i = 3; i * i < MAX; i++) // If i is prime if (spf[i] == i) { // Mark spf for all numbers // divisible by i for (int j = i * i; j < MAX; j = j + i) { // Mark spf[j] if it is // not previously marked if (spf[j] == j) spf[j] = i; } } } // Function to find the distinct // prime factors static void DistPrime() { for (int i = 1; i < MAX; i++) { int idx = 1; int x = i; // Push all distinct of x // prime factor in v[x] if (x != 1) v[i].add(spf[x]); x = x / spf[x]; while (x != 1) { if (v[i].get(idx - 1) != spf[x]) { // Pushback into v[i] v[i].add(spf[x]); // Increment the idx idx += 1; } // Update x = (x / spf[x]) x = x / spf[x]; } } } // Function to get the distinct // factor count of arr[] static void getFactorCount(int arr[], int N) { // initialization spf = new int[MAX]; v = new ArrayList[MAX]; for (int i = 0; i < MAX; i++) v[i] = new ArrayList<>(); // Precompute the smallest // Prime Factors sieve(); // For distinct prime factors // Fill the v[] vector DistPrime(); // Count of Distinct Prime // Factors of each array element for (int i = 0; i < N; i++) { System.out.print((int)v[arr[i]].size() + " "); } } // Driver Code public static void main(String[] args) { int arr[] = { 6, 9, 12 }; int N = arr.length; getFactorCount(arr, N); }}// This code is contributed by Kingash. |
Python3
MAX = 100001# Stores smallest prime# factor for every numberspf = [0] * MAX# Stores distinct prime factorsv = [[] for _ in range(MAX)]# Function to find the smallest# prime factor of every numberdef sieve(): # Mark the smallest prime factor # of every number to itself for i in range(1, MAX): spf[i] = i # Separately mark all the # smallest prime factor of # every even number to be 2 for i in range(4, MAX, 2): spf[i] = 2 for i in range(3, int(MAX**0.5)+1): # If i is prime if spf[i] == i: # Mark spf for all numbers # divisible by i for j in range(i*i, MAX, i): # Mark spf[j] if it is # not previously marked if spf[j] == j: spf[j] = i# Function to find the distinct# prime factorsdef DistPrime(): for i in range(1, MAX): idx = 1 x = i # Push all distinct of x # prime factor in v[x] if x != 1: v[i].append(spf[x]) x = x // spf[x] while x != 1: if v[i][idx-1] != spf[x]: # Pushback into v[i] v[i].append(spf[x]) # Increment the idx idx += 1 # Update x = (x / spf[x]) x = x // spf[x]# Function to get the distinct# factor count of arr[]def getFactorCount(arr, N): # Precompute the smallest # Prime Factors sieve() # For distinct prime factors # Fill the v[] vector DistPrime() # Count of Distinct Prime # Factors of each array element for i in range(N): print(len(v[arr[i]]), end=' ')arr = [6, 9, 12]N = len(arr)getFactorCount(arr, N)# This code is contributed by phasing17. |
C#
// C# program for the above approachusing System;using System.Collections.Generic;class GFG { static int MAX = 100001; // Stores smallest prime // factor for every number static int[] spf; // Stores distinct prime factors static List<List<int>> v; // Function to find the smallest // prime factor of every number static void sieve() { // Mark the smallest prime factor // of every number to itself for (int i = 1; i < MAX; i++) spf[i] = i; // Separately mark all the // smallest prime factor of // every even number to be 2 for (int i = 4; i < MAX; i = i + 2) spf[i] = 2; for (int i = 3; i * i < MAX; i++) // If i is prime if (spf[i] == i) { // Mark spf for all numbers // divisible by i for (int j = i * i; j < MAX; j = j + i) { // Mark spf[j] if it is // not previously marked if (spf[j] == j) spf[j] = i; } } } // Function to find the distinct // prime factors static void DistPrime() { for (int i = 1; i < MAX; i++) { int idx = 1; int x = i; // Push all distinct of x // prime factor in v[x] if (x != 1) v[i].Add(spf[x]); x = x / spf[x]; while (x != 1) { if (v[i][idx - 1] != spf[x]) { // Pushback into v[i] v[i].Add(spf[x]); // Increment the idx idx += 1; } // Update x = (x / spf[x]) x = x / spf[x]; } } } // Function to get the distinct // factor count of arr[] static void getFactorCount(int[] arr, int N) { // initialization spf = new int[MAX]; v = new List<List<int>>() ; for (int i = 0; i < MAX; i++) v.Add(new List<int>()); // Precompute the smallest // Prime Factors sieve(); // For distinct prime factors // Fill the v[] vector DistPrime(); // Count of Distinct Prime // Factors of each array element for (int i = 0; i < N; i++) { Console.Write((int)v[arr[i]].Count + " "); } } // Driver Code public static void Main(string[] args) { int[] arr = { 6, 9, 12 }; int N = arr.Length; getFactorCount(arr, N); }}// This code is contributed by phasing17. |
Javascript
<script> // JavaScript program for the above approach let MAX = 100001; // Stores smallest prime // factor for every number let spf; // Stores distinct prime factors let v; // Function to find the smallest // prime factor of every number function sieve() { // Mark the smallest prime factor // of every number to itself for (let i = 1; i < MAX; i++) spf[i] = i; // Separately mark all the // smallest prime factor of // every even number to be 2 for (let i = 4; i < MAX; i = i + 2) spf[i] = 2; for (let i = 3; i * i < MAX; i++) // If i is prime if (spf[i] == i) { // Mark spf for all numbers // divisible by i for (let j = i * i; j < MAX; j = j + i) { // Mark spf[j] if it is // not previously marked if (spf[j] == j) spf[j] = i; } } } // Function to find the distinct // prime factors function DistPrime() { for (let i = 1; i < MAX; i++) { let idx = 1; let x = i; // Push all distinct of x // prime factor in v[x] if (x != 1) v[i].push(spf[x]); x = parseInt(x / spf[x], 10); while (x != 1) { if (v[i][idx - 1] != spf[x]) { // Pushback into v[i] v[i].push(spf[x]); // Increment the idx idx += 1; } // Update x = (x / spf[x]) x = parseInt(x / spf[x], 10); } } } // Function to get the distinct // factor count of arr[] function getFactorCount(arr, N) { // initialization spf = new Array(MAX); v = new Array(MAX); for (let i = 0; i < MAX; i++) v[i] = []; // Precompute the smallest // Prime Factors sieve(); // For distinct prime factors // Fill the v[] vector DistPrime(); // Count of Distinct Prime // Factors of each array element for (let i = 0; i < N; i++) { document.write(v[arr[i]].length + " "); } } let arr = [ 6, 9, 12 ]; let N = arr.length; getFactorCount(arr, N); </script> |
Output
2 1 2
Time Complexity: O(N * log N)
Auxiliary Space: O(N)
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