Distinct Prime Factors of Array Product (original) (raw)
Last Updated : 24 Oct, 2022
Given an array of integers. Let us say P is the product of elements of the array. Find the number of distinct prime factors of product P.
Examples:
Input : 1 2 3 4 5
Output : 3
Explanation: Here P = 1 * 2 * 3 * 4 * 5 = 120. Distinct prime divisors of 120 are 2, 3 and 5. So, the output is 3.Input : 21 30 15 24 16
Output : 4
Explanation: Here P = 21 * 30 * 15 * 24 * 16 = 3628800. Distinct prime divisors of 3628800 are 2, 3, 5 and 7. So, the output is 4.
Naive Approach: The simple solution for the problem would be to multiply every number in the array and then find the number of distinct prime factors of the product.
But this method can lead to integer overflow.
Better Approach: To avoid the overflow instead of multiplying the numbers we can find the prime factors of each element separately and store the prime factors in a set or a map for unique factors.
Implementation:
C++ `
// C++ program to count distinct prime // factors of a number. #include <bits/stdc++.h> using namespace std;
// Function to count the number of distinct prime // factors of product of array int Distinct_Prime_factors(vector a) { // use set to store distinct factors unordered_set m;
// iterate over every element of array
for (int i = 0; i < a.size(); i++) {
int sq = sqrt(a[i]);
// from 2 to square root of number run
// a loop and check the numbers which
// are factors.
for (int j = 2; j <= sq; j++) {
if (a[i] % j == 0) {
// if j is a factor store it in the set
m.insert(j);
// divide the number with j till it
// is divisible so that only prime factors
// are stored
while (a[i] % j == 0) {
a[i] /= j;
}
}
}
// if the number is still greater than 1 then
// it is a prime factor, insert in set
if (a[i] > 1) {
m.insert(a[i]);
}
}
// the number of unique prime factors will
// the size of the set
return m.size();}
// Driver Function int main() { vector a = { 1, 2, 3, 4, 5 }; cout << Distinct_Prime_factors(a) << '\n'; return 0; }
Java
// Java program to count distinct // prime factors of a number. import java.util.*;
class GFG {
// Function to count the number
// of distinct prime factors of
// product of array
static int Distinct_Prime_factors(Vector<Integer> a)
{
// use set to store distinct factors
HashSet<Integer> m = new HashSet<Integer>();
// iterate over every element of array
for (int i = 0; i < a.size(); i++) {
int sq = (int)Math.sqrt(a.get(i));
// from 2 to square root of number
// run a loop and check the numbers
// which are factors.
for (int j = 2; j <= sq; j++) {
if (a.get(i) % j == 0) {
// if j is a factor store
// it in the set
m.add(j);
// divide the number with j
// till it is divisible so
// that only prime factors
// are stored
while (a.get(i) % j == 0) {
a.set(i, a.get(i) / j);
}
}
}
// if the number is still greater
// than 1 then it is a prime factor,
// insert in set
if (a.get(i) > 1) {
m.add(a.get(i));
}
}
// the number of unique prime
// factors will the size of the set
return m.size();
}
// Driver Code
public static void main(String args[])
{
Vector<Integer> a = new Vector<Integer>();
a.add(1);
a.add(2);
a.add(3);
a.add(4);
a.add(5);
System.out.println(Distinct_Prime_factors(a));
}}
// This code is contributed by Arnab Kundu
Python3
Python3 program to count distinct
prime factors of a number
import math
Function to count the number of distinct
prime factors of product of array
def Distinct_Prime_factors( a):
# use set to store distinct factors
m = []
# iterate over every element of array
for i in range (len(a)) :
sq = int(math.sqrt(a[i]))
# from 2 to square root of number run
# a loop and check the numbers which
# are factors.
for j in range(2, sq + 1) :
if (a[i] % j == 0) :
# if j is a factor store
# it in the set
m.append(j)
# divide the number with j till it
# is divisible so that only prime
# factors are stored
while (a[i] % j == 0) :
a[i] //= j
# if the number is still greater
# than 1 then it is a prime factor,
# insert in set
if (a[i] > 2) :
m.append(a[i])
# the number of unique prime factors
# will the size of the set
return len(m)Driver Code
if name == "main":
a = [ 1, 2, 3, 4, 5 ]
print (Distinct_Prime_factors(a))This code is contributed by ita_c
C#
// C# program to count distinct // prime factors of a number. using System; using System.Collections.Generic;
class GFG {
// Function to count the number
// of distinct prime factors of
// product of array
static int Distinct_Prime_factors(List<int> a)
{
// use set to store distinct factors
HashSet<int> m = new HashSet<int>();
// iterate over every element of array
for (int i = 0; i < a.Count; i++) {
int sq = (int)Math.Sqrt(a[i]);
// from 2 to square root of number
// run a loop and check the numbers
// which are factors.
for (int j = 2; j <= sq; j++) {
if (a[i] % j == 0) {
// if j is a factor store
// it in the set
m.Add(j);
// divide the number with j
// till it is divisible so
// that only prime factors
// are stored
while (a[i] % j == 0) {
a[i] = a[i] / j;
}
}
}
// if the number is still greater
// than 1 then it is a prime factor,
// insert in set
if (a[i] > 1) {
m.Add(a[i]);
}
}
// the number of unique prime
// factors will the size of the set
return m.Count;
}
// Driver Code
public static void Main()
{
List<int> a = new List<int>();
a.Add(1);
a.Add(2);
a.Add(3);
a.Add(4);
a.Add(5);
Console.WriteLine(Distinct_Prime_factors(a));
}}
// This code is contributed by ihritik
JavaScript
`
Time Complexity: O(n*sqrt(n))
Auxiliary Space: O(n)