title: Ugly Numbers
signature: function ugly_numbers(n)
test_code: using Printf
  
  # SOLUTION CODE
  # ============================================
  PLACEHOLDER_CODE_BODY
  
  # TESTING CODE 
  # ============================================
  function validate_solution(actual, expected)
      return actual == expected
  end
  
  function driver(n, expected)
      try
  
          if validate_solution(PLACEHOLDER_FN_NAME(n), expected)
              return "PASSED";
          end
          return "FAILED";
      catch exception_obj
          return string(Base.typename(typeof(exception_obj)).wrapper);
      end
  end
  
  function main()
      
      result = driver(27, 64);
      @printf("TEST-0...%s\n",result);
      
      result = driver(64, 450);
      @printf("TEST-1...%s\n",result);
      
      result = driver(93, 1250);
      @printf("TEST-2...%s\n",result);
      
      result = driver(90, 1152);
      @printf("TEST-3...%s\n",result);
      
      result = driver(85, 972);
      @printf("TEST-4...%s\n",result);
      
      result = driver(86, 1000);
      @printf("TEST-5...%s\n",result);
      
      result = driver(72, 625);
      @printf("TEST-6...%s\n",result);
      
      result = driver(86, 1000);
      @printf("TEST-7...%s\n",result);
      
      result = driver(32, 90);
      @printf("TEST-8...%s\n",result);
      
      result = driver(1, 1);
      @printf("TEST-9...%s\n",result);
      
  end
  
  main()
source_py: def ugly_numbers(n):
      ugly = ([0] * n)
      ugly[0] = 1
      i2 = i3 = i5 = 0
      next_multiple_of_2 = 2
      next_multiple_of_3 = 3
      next_multiple_of_5 = 5
      for l in range(1, n):
          ugly[l] = min(next_multiple_of_2, next_multiple_of_3, next_multiple_of_5)
          if (ugly[l] == next_multiple_of_2):
              i2 += 1
              next_multiple_of_2 = (ugly[i2] * 2)
          if (ugly[l] == next_multiple_of_3):
              i3 += 1
              next_multiple_of_3 = (ugly[i3] * 3)
          if (ugly[l] == next_multiple_of_5):
              i5 += 1
              next_multiple_of_5 = (ugly[i5] * 5)
      return ugly[(- 1)]
source_cpp: unsigned ugly_numbers(unsigned n) {
    unsigned ugly [ n ];
    unsigned i2 = 0, i3 = 0, i5 = 0;
    unsigned next_multiple_of_2 = 2;
    unsigned next_multiple_of_3 = 3;
    unsigned next_multiple_of_5 = 5;
    unsigned next_ugly_no = 1;
    ugly [ 0 ] = 1;
    for ( int i = 1;
    i < n;
    i ++ ) {
      next_ugly_no = min ( next_multiple_of_2, min ( next_multiple_of_3, next_multiple_of_5 ) );
      ugly [ i ] = next_ugly_no;
      if ( next_ugly_no == next_multiple_of_2 ) {
        i2 = i2 + 1;
        next_multiple_of_2 = ugly [ i2 ] * 2;
      }
      if ( next_ugly_no == next_multiple_of_3 ) {
        i3 = i3 + 1;
        next_multiple_of_3 = ugly [ i3 ] * 3;
      }
      if ( next_ugly_no == next_multiple_of_5 ) {
        i5 = i5 + 1;
        next_multiple_of_5 = ugly [ i5 ] * 5;
      }
    }
    return next_ugly_no;
  }