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 prime factors of each array element. Then, find the distinct prime numbers among them and print that count for each array element.
Time Complexity: O(N2)
Auxiliary Space: O(1)
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 number int spf[MAX]; // Stores distinct prime factors vector< int > v[MAX]; // Function to find the smallest // prime factor of every number void sieve() { // Mark the smallest prime factor // of every number to itself for ( int i = 1; i < MAX; i++) spf[i] = i; // Seperately 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 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].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 Code int main() { int arr[] = { 6, 9, 12 }; int N = sizeof (arr) / sizeof (arr[0]); getFactorCount(arr, N); return 0; } |
Java
// Java program for the above approach import 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; // Seperately 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. |
2 1 2
Time Complexity: O(N * log N)
Auxiliary Space: O(N)
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