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

Comments:

• Anonymous - 9 years, 10 months ago

  Final: 168,570

  class Program
  {
      static void Main(string[] args)
      {
          int uniqueNumbers = 0;
  
          for (int i = 1; i <= 1000000; i++ )
          {
              HashSet<char> chars = new HashSet<char>(i.ToString());
              if (chars.Count == i.ToString().Length)
              {
                  uniqueNumbers++;
              }
          }
  
          Console.WriteLine("Unique Integers: " + uniqueNumbers);
      }
  }
  

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• David - 9 years, 9 months ago

  This has an off by one error for not including "0".

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• Kevin Benton - 9 years, 10 months ago

  168571 len([x for x in xrange(0, 1000000) if len(str(x)) == len(set(str(x)))]) Another one that doesn't need a big list

  reduce(lambda count, item: count + 1 if len(str(item)) == len(set(str(item))) else count, xrange(0, 1000000), 0)
  

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• Jtmach - 9 years, 10 months ago

  Final 168,571 don't forget about 0.

  Javascript

  this.window.onload = function () {
    var uniqueIntegers = 0;
    for (var index = 0; index < 1000000; index++) {
      if(checkString(index.toString())) {
        uniqueIntegers = uniqueIntegers + 1;
      }
    }
  
    this.document.writeln("Unique integers: " + uniqueIntegers);
  };
  
  function checkString(chkStr)
  {
    if (chkStr.length == 1) {
      return true;
    }
  
    for (var chkInt = 0; chkInt < 10; chkInt++)
    if (chkStr.indexOf(chkInt.toString()) != chkStr.lastIndexOf(chkInt.toString())) {
      return false;
    }
  
    return true;
  }
  

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• Eddy - 9 years, 10 months ago

  My solution in C using only arithmetic and bitwise operations.

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• David - 9 years, 9 months ago

  I decided to do this mathematically using permutations.
  One day I will get fancy and upload my scratch work up to git so I can just link it, like Eddy.
  Here is my counting in C.

  #include <stdio.h>   
  
  int factorial(int num) {   
      int prod = 1;   
      for(int i = 1; i <= num; i ++) {
              prod *= i;
          }
      return prod;
         }
  
  int permutation(int n, int r) {
      int perm = factorial(n);
      perm /= factorial(n-r);
      return perm;
  }   
  
  int main(void)
  {
      int uniques = 1;    //My default value accounts for 0 because of my method.
      for(int i = 1; i < 7; i++)
      {
          if(6 == i)
              uniques += permutation(9,i);    //Count the 6 digit permutations of [1,9]
          else
              uniques += (i + 1) * permutation(9,i);  //Count the i digit permutations of [1,9] and the i + 1 digit permutations that include 0
      }
      printf("%d\n", uniques);
  
      return 0;
  }   
  

  I needed to build my permutation function n!/(n-r)! so it took a little extra overhead in that regard. I ran the permutations of [1,9] for various numbers of digits and added in the cases for then 0 being inserted after any existing digit.

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