How do I determine if a given date is Nth weekday of the month?

Sorry if the title is awkward, but I can't think of a better summary.

Here is what I am trying to do: Given a date, a day of a week and an integer n, is the date the nth day of the month?

For example:

input of 1/1/2009,Monday,2 would be false because 1/1/2009 is not the second Monday

input of 11/13/2008,Thursday,2 would return true because it is the second Thursday

can anybody think of a better implementation than this

private bool NthDayOfMonth(DateTime date, DayOfWeek dow, int n)
     {
      int d = date.Day;
      return date.DayOfWeek == dow && (d/ 7 == n || (d/ 7 == (n - 1) && d % 7 > 0));
     }

From stackoverflow

• You could change the check of the week:

  int d = date.Day;
  return date.DayOfWeek == dow && (d-1)/7 == (n-1);
  

  Other than that, it looks pretty good and efficient.

  Kevin : i think that should be (d-1)/7 == (n-1)

  Robert Wagner : Thanks for that. Thinking in 0 based counting again.

• Here is what the MSDN have to say. Its VB, but it translates easily.

  Kevin : that kinda does the opposite of what I want

  Andrew Bullock : yeah, but its easily reversed ;)

• The answer is from this website. Copy/pasted here in case that site is ever lost.

  public static DateTime FindTheNthSpecificWeekday(int year, int month,int nth, System.DayOfWeek day_of_the_week)
  {
  // validate month value
  if(month < 1 || month > 12)
  {
  throw new ArgumentOutOfRangeException(”Invalid month value.”);
  }
  
  // validate the nth value
  if(nth < 0 || nth > 5)
  {
  throw new ArgumentOutOfRangeException(”Invalid nth value.”);
  }
  
  // start from the first day of the month
  DateTime dt = new DateTime(year, month, 1);
  
  // loop until we find our first match day of the week
  while(dt.DayOfWeek != day_of_the_week)
  {
  dt = dt.AddDays(1);
  }
  
  if(dt.Month != month)
  {
  // we skip to the next month, we throw an exception
  throw new ArgumentOutOfRangeException(”The given month has less than ” nth.ToString() ” ”
  day_of_the_week.ToString() “s”);
  }
  
  // Complete the gap to the nth week
  dt = dt.AddDays((nth - 1) * 7);
  
  return dt;
  }
  

  Kevin : this also does the opposite of what I want

• It looks like the language supplies date/day methods for a given date. If anybody was interested you can read about Zeller's congruence.

  I don't think that's what they wanted you to do but you could find the day of week of the first day of a month from that. Now that I thought about it you could find the day of week for the given day as N and get that modulo 7.

  Oh wait, is that the Nth occurance of a day of the week (like Sunday) or like the Nth weekday of the month! Okay I see the examples.

  Maybe it would make a difference if you could construct a date such as the 1st of a month..

  Given that it is Nth occurance of a day of the week, and that you can't fiddle with whatever datetime datatype, and that you have access to both a get day of week and get day of month functions. Would Sunday be a zero?

  1) First, the day of the week would have to match the day of the week given.
  2) N would have to be at least 1 and at most 4.
  3) The day of the month would range between n*7*dayOfWeek + 1 and n*7*dayOfWeek + 6 for the same n.
  - Let me think about that. If Sunday was the first.. 0*7*0+1 = 1 and Saturday the 6th would be 0*7*0+6.

  Think 1 and 3 above are sufficient since a get day of month function shouldn't violate 2.

  (* first try, this code sucks *)
  
  function isNthGivenDayInMonth(date : dateTime;
                                dow : dayOfWeek;
                                N : integer) : boolean;
      var B, A : integer (* on or before and after day of month *)
      var Day : integer (* day of month *)
      begin
      B := (N-1)*7 + 1; A := (N-1)*7 + 6;
      D := getDayOfMonth(date);
      if (dow <> getDayOfWeek(date) 
          then return(false)
          else return( (B <= Day) and (A >= Day) );
      end; (* function *)
  

  Hope there isn't a bug in that lol!
  [edit: Saturday would have been the 7th, and the upper bound above (N-1)*7 + 7.]
  Your solution looks like it would match 2 different weeks? Looks like it would always return zero for Sundays? Should have done pseudocode in C#.. short circuit && is like my if.. hey shouldn't Sunday the first match for N = 1 in months that start on Sunday?

   d/ 7 == n
  

  That would result in (either 0 or 1)/7 == 1, that can't be right! Your || catches the (n-1) also, Robert has that. Go with Robert Wagner's answer! It's only 2 lines, short is good! Having (Day-1) mod 7
  [edit: (Day-1) div 7] eliminates my unnecessary variables and 2 lines of setup.

  For the record this should be checked for boundary cases and so forth like what if August 31st was a Sunday or Saturday.
  [edit: Should have checked the end of week case too. Sorry!]

• In the above example the following code needs to be flipped.

  // Complete the gap to the nth week dt = dt.AddDays((nth - 1) * 7);

  if(dt.Month != month) { // we skip to the next month, we throw an exception throw new ArgumentOutOfRangeException(”The given month has less than ” nth.ToString() ” ” day_of_the_week.ToString() “s”); }

• You can find a function which returns date for nth occurrence of particular week day in any month. http://chiragrdarji.wordpress.com/2010/08/23/find-second-saturday-and-fourth-saturday-of-month/