title: Find Largest Prime Factor Number
signature: pub fn find_largest_prime_factor_number(n: i32) -> i32 {
test_code: use std::collections::HashMap;
  use std::collections::HashSet;
  use std::cmp::*;
  use std::collections::*;
  use std::fmt::*; 
  
  
  // SOLUTION CODE
  // ============================================
  PLACEHOLDER_CODE_BODY
  
  // TESTING CODE 
  // ============================================
  pub fn validate_solution(actual: i32, expected: i32)->bool{
  return actual==expected
  }
  pub fn driver(n: i32, expected: i32)->String{
      if validate_solution(PLACEHOLDER_FN_NAME(n), expected) {
        return "PASSED".to_string();
      };
      return "FAILED".to_string();
  
  }
  
  fn main() {
      println!("TEST-0...{}", driver(
        98, 
        7
      ));
      
      println!("TEST-1...{}", driver(
        8, 
        2
      ));
      
      println!("TEST-2...{}", driver(
        78, 
        13
      ));
      
      println!("TEST-3...{}", driver(
        65, 
        13
      ));
      
      println!("TEST-4...{}", driver(
        55, 
        11
      ));
      
      println!("TEST-5...{}", driver(
        10, 
        5
      ));
      
      println!("TEST-6...{}", driver(
        10, 
        5
      ));
      
      println!("TEST-7...{}", driver(
        37, 
        37
      ));
      
      println!("TEST-8...{}", driver(
        39, 
        13
      ));
      
      println!("TEST-9...{}", driver(
        15, 
        5
      ));
      
  }
source_py: def find_largest_prime_factor_number(n):
      maxPrime = (- 1)
      while ((n % 2) == 0):
          maxPrime = 2
          n >>= 1
      for i in range(3, (int(math.sqrt(n)) + 1), 2):
          while ((n % i) == 0):
              maxPrime = i
              n = (n / i)
      if (n > 2):
          maxPrime = n
      return int(maxPrime)
source_cpp: long long find_largest_prime_factor_number(long long n) {
    long long maxPrime = - 1;
    while ( n % 2 == 0 ) {
      maxPrime = 2;
      n >>= 1;
    }
    for ( int i = 3;
    i <= sqrt ( n );
    i += 2 ) {
      while ( n % i == 0 ) {
        maxPrime = i;
        n = n / i;
      }
    }
    if ( n > 2 ) maxPrime = n;
    return maxPrime;
  }