Leonardo’s Prime Factors- Hackerrank Problem (C++)

https://www.hackerrank.com/challenges/leonardo-and-prime/problem?isFullScreen=true

Leonardo loves primes and created queries where each query takes the form of an integer, . For each , count the maximum number of distinct prime factors of any number in the inclusive range .

Note: Recall that a prime number is only divisible by and itself, and is not a prime number.

Example

The maximum number of distinct prime factors for values less than or equal to is . One value with distinct prime factors is . Another is .

Function Description

Complete the primeCount function in the editor below.

primeCount has the following parameters:

• int n: the inclusive limit of the range to check

Returns

• int: the maximum number of distinct prime factors of any number in the inclusive range .

Input Format

The first line contains an integer, , the number of queries. Each of the next lines contains a single integer, .

Constraints

Sample Input

6
1
2
3
500
5000
10000000000

Sample Output

0
1
1
4
5
10

Explanation

1. is not prime and its only factor is itself.
2. has prime factor, .
3. The number has prime factor, , has and has prime factors.
4. The product of the first four primes is . While higher value primes may be a factor of some numbers, there will never be more than distinct prime factors for a number in this range.

Solution

#include <bits/stdc++.h>

using namespace std;

string ltrim(const string &);
string rtrim(const string &);

/*
 * Complete the 'primeCount' function below.
 *
 * The function is expected to return an INTEGER.
 * The function accepts LONG_INTEGER n as parameter.
 */

int primeCount(long n) {
vector<int> primes = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97};

  unsigned long long product = 1;
  int count = 0;

  if (n == 1)
  {
    return count;
  }

  for (int prime : primes)
  {
// product of primes 
    product *= (unsigned long long)prime;

// exit conditions
    if (product > n)
      break;
    if (product == n)
    {
      count++;
      break;
    }

    count++;
  }
  return count;
}

int main()
{
    ofstream fout(getenv("OUTPUT_PATH"));

    string q_temp;
    getline(cin, q_temp);

    int q = stoi(ltrim(rtrim(q_temp)));

    for (int q_itr = 0; q_itr < q; q_itr++) {
        string n_temp;
        getline(cin, n_temp);

        long n = stol(ltrim(rtrim(n_temp)));

        int result = primeCount(n);

        fout << result << "\n";
    }

    fout.close();

    return 0;
}

string ltrim(const string &str) {
    string s(str);

    s.erase(
        s.begin(),
        find_if(s.begin(), s.end(), not1(ptr_fun<int, int>(isspace)))
    );

    return s;
}

string rtrim(const string &str) {
    string s(str);

    s.erase(
        find_if(s.rbegin(), s.rend(), not1(ptr_fun<int, int>(isspace))).base(),
        s.end()
    );

    return s;
}