Problem of the Day
A new programming or logic puzzle every Mon-Fri

Comments:

• JtMach - 9 years, 3 months ago

  Lucky Numbers: 911

  Javascript

  this.window.onload = function () {
    var luckyNumbers = 0;
    for (var index = 0; index <= 1000000; index++) {
      if(new IsNumberLucky(index.toString())) {
        this.document.writeln(index);
        luckyNumbers = luckyNumbers + 1;
      }
    }
  
    this.document.writeln("Lucky Numbers: " + luckyNumbers);
  };
  
  function IsNumberLucky(chkStr)
  {
    if (chkStr.length == 3 || chkStr.length == 7) {
      return true;
    }
  
    var validChar = "";
    if (chkStr.charAt(0) == "3" || chkStr.charAt(0) == "7") {
      validChar = chkStr.charAt(0);
    }
    else {
      return false;
    }
  
    for (var chkStrPosition = 1; chkStrPosition < chkStr.length; chkStrPosition++) {
      if (chkStr.charAt(chkStrPosition) != validChar) {
        return false;
      }
    }
  
    return true;
  }
  

  reply permalink

• Kevin Benton - 9 years, 3 months ago

  I only got 900...

  nums = [x for x in xrange(1000000) if set(str(x)) in [set([3]), set([7])] or len(str(x)) in (7,3)]
  print len(nums)
  900
  

  reply permalink

• Aaron Frase - 9 years, 3 months ago

  You're comparing a set of strings with a set of integers so you missed 10.

  This fixes it.

  nums = [x for x in xrange(1000000) if set(str(x)) in [set(['3']), set(['7'])] or len(str(x)) in (7,3)]

  Here was my original (much longer) version

  lucky_numbers = []
  
  for x in range(1000000):
      lucky = False
      nums = [int(i) for i in str(x)]
      s = set(nums)
      if len(s) == 1 and (s.pop() in (3, 7)):
          lucky = True
      elif len(nums) in (3, 7):
          lucky = True
      if lucky:
          lucky_numbers.append(x)
  
  print(len(lucky_numbers))
  

  reply permalink

• slp - 9 years, 3 months ago

  Lucky Numbers: 911

  PHP

  $count = 0;
  for ($i = 0; $i <= 1000000; $i++)
  {
      $length     = strlen($i);
      $str        = (string) $i;
      $firstChar  = $str[0];
  
      if ($length == 3 || $length == 7)
          $count++;
  
      elseif (($firstChar == '3' || $firstChar == '7') && substr_count($i, $firstChar) == $length)
          $count++;
  }
  
  echo $count;
  

  reply permalink

• Eddy - 9 years, 3 months ago

  My solution in C code using only basic arithmetic and as few lines of code as I could.

  reply permalink