How would I get the amount of weekdays in a month with LocalDate - Java

Question

How would I go about getting the number of weekdays (Monday though Friday) in a month with LocalDate? I've never used java.time before so I don't know all of its workings. I've been looking on this site to no avail along with searching for an answer. I also do not want to use any external libraries.

Example: As of this month, April of 2018, there are 21 weekdays. And next month there is 23 weekdays.

Answer 1

If desired, see optimized brute-force solution at the end

Here is a non-brute-force implementation to calculate week days (Mon-Fri) in a month.

It uses YearMonth instead of LocalDate, since the day-of-month value is meaningless to the calculation.

public static int weekDaysInMonth(YearMonth yearMonth) {
    int len = yearMonth.lengthOfMonth(); // 28-31, supporting leap year
    int dow = yearMonth.atDay(1).getDayOfWeek().getValue(); // 1=Mon, 7=Sun
    return (dow <= 5 ? Math.min(len - 8, 26 - dow) : Math.max(len + dow - 16, 20));
}

Here is an overload taking a LocalDate, so it's easy to call if that's what you have.

public static int weekDaysInMonth(LocalDate date) {
    return weekDaysInMonth(YearMonth.from(date));
}

Test

System.out.println(weekDaysInMonth(LocalDate.parse("2018-04-15"))); // April 15, 2018
System.out.println(weekDaysInMonth(YearMonth.of(2018, 5)));         // May 2018

Output

21
23

---

Explanation of Formula

The formula in the return statement was created by examining the expected return value for every combination of len (number of days in month, 28 - 31) and dow (day-of-week of first day of month, 1=Mon - 7=Sun):

   |  1   2   3   4   5    6   7
   | Mo  Tu  We  Th  Fr   Sa  Su
---+----------------------------
28 | 20  20  20  20  20   20  20
29 | 21  21  21  21  21   20  20
30 | 22  22  22  22  21   20  21
31 | 23  23  23  22  21   21  22

Explanation for dow <= 5 (Mon-Fri)

Initially there are len - 8 weekdays, i.e. we subtract the 4 weekends that always exist in a month.

As we get to Thursday and Friday, we need to cap that for the 1 or 2 weekend days we lose. If you look at the 31-day row, we cap it at 26 - dow, i.e. for Friday (dow=5) we cap at 21, and for Thursday (dow=4) we cap at 22. For Monday-Wednesday, we also cap, but cap is equal to or higher than initial calculation, so it doesn't matter.

Capping is done using min(xxx, cap) method, so we get:

min(len - 8, 26 - dow)

Explanation for dow >= 6 (Sat-Sun)

You can see a small triangle in the lower-right corner. If we extend that pattern, we get:

   |  4   5   6   7
---+---------------
28 | 16  17  18  19
29 | 17  18  19  20
30 | 18  19  20  21
31 | 19  20  21  22

As a formula, that is len + dow - 16.

Comparing that to original grid, numbers bottom out at 20, which is done using max(xxx, bottom) method, so we get:

max(len + dow - 16, 20)

Conclusion

Finally we combine the two using ternary conditional operator:

dow <= 5  ?  min(len - 8, 26 - dow)  :  max(len + dow - 16, 20)

Full Java statement is then:

return (dow <= 5 ? Math.min(len - 8, 26 - dow) : Math.max(len + dow - 16, 20));

---

Brute-Force Solution

If you prefer a brute-force solution, you can ease it by skipping the first 4 weeks that always exists in a month:

public static int weekDaysInMonth(LocalDate refDate) {
    LocalDate firstOfMonth = refDate.withDayOfMonth(1);
    LocalDate nextMonth = firstOfMonth.plusMonths(1);
    int days = 20;
    for (LocalDate date = firstOfMonth.plusDays(28); date.isBefore(nextMonth); date = date.plusDays(1))
        if (date.getDayOfWeek().getValue() <= 5) // 1=Mon - 5=Fri, i.e. not 6=Sat and 7=Sun
            days++;
    return days;
}