Maximum number of unique prime factors in C++

Given the task is to find the maximum number of unique prime factors a number can have in the range of [1, N] where N is given.

Let’s now understand what we have to do using an example −

Input − N=100

Output − 3

Explanation − Let us take 30 in the range of [1, 100]

30 = 3 * 2 * 5 = unique prime factors. Therefore, in the range of [1, 100] a maximum of 3 unique factors can be found.

Input − N=300

Output − 4

Approach used in the below program as follows

  • In function MaxPrime() we will first check if (N < 2). If so then simply return zero, otherwise proceed.

  • Then we will use the sieve of Eratosthenes to find out all the prime numbers before the given number N.

  • Initialize two variables pro=1 and max=0 of type int to store the product and the final answer respectively.

  • Inside the sieve of Eratosthenes we will multiply the first set of prime numbers until the product remains smaller than N by writing − pro=*p;

  • Check if (pro > N). If so then return max and add 1 to max.

  • Outside the sieve, also return max.


 Live Demo

#include <bits/stdc++.h>
using namespace std;
int MaxPrime(int N){
   if (N < 2)
      return 0;
   // Using Sieve of Eratosthenes
   bool Arr[N+1];
   memset(Arr, true, sizeof(Arr));
   int pro = 1, max = 0;
   for (int p=2; p*p<=N; p++){
      if (Arr[p] == true){
         for (int i=p*2; i<=N; i += p)
            Arr[i] = false;
         /*Multiply first set of prime numbers while product remains smaller than N*/
         pro *= p;
         if (pro > N)
            return max;
   return max;
//Main function
int main(){
   int N = 300;
   cout << MaxPrime(N);
   return 0;


If we run the above code we will get the following output −


Updated on: 17-Aug-2020


Kickstart Your Career

Get certified by completing the course

Get Started