Interview Question: Count number of every week day in a year

Interview Question: With a single SELECT statement get the number of each week day in the current year.

Level: Intermediate

Expected Result:

Day                                  Days in Year
------------------------------------ ------------
SUNDAY                                         52
MONDAY                                         52
TUESDAY                                        52
WEDNESDAY                                      52
THURSDAY                                       52
FRIDAY                                         53
SATURDAY                                       53

Solution #1:

WITH x AS (
SELECT LEVEL-1+TRUNC(SYSDATE, 'YYYY') AS dd
FROM dual
CONNECT BY TRUNC(LEVEL-1+TRUNC(SYSDATE, 'YYYY'),'YYYY')=TRUNC(SYSDATE, 'YYYY')
)
SELECT TO_CHAR(dd, 'DAY') "Day", COUNT(*) "Days in Year"
FROM x
GROUP BY TO_CHAR(dd, 'DAY'), TO_CHAR(dd, 'D')
ORDER BY TO_CHAR(dd, 'D');

Explanation:

The WITH clause returns all days in the current year, this is a common trick used in majority of sql puzzle related to a calendar. The connect by query used in the WITH generated a date range which starts on TRUNC(SYSDATE, ‘YYYY’) – i.e. the 1st day of the year – and continues as long as the next day falls into the same year (see condition in the CONNECT BY clause). The main query groups by day name – TO_CHAR(dd, ‘DAY’) – and sorts by day number (in a week) – TO_CHAR(dd, ‘D’).

Solution #2:

WITH x AS (
SELECT TO_CHAR(ADD_MONTHS(TRUNC(SYSDATE,'YYYY'), 12)-1, 'DDD') days_in_year
FROM dual
)
SELECT TO_CHAR(LEVEL-1+TRUNC(SYSDATE, 'YYYY'),'DAY') "Day",
       CASE WHEN MOD(days_in_year,52)>=LEVEL THEN 53
            ELSE 52
       END "Days in Year"
FROM x
CONNECT BY LEVEL<=7
ORDER BY TO_CHAR(LEVEL-1+TRUNC(SYSDATE, 'YYYY'),'D');

Explanation:

The idea behind this solution is totally different than in the 1st one. A year has 52 weeks and 1 or 2 days depending on whether it is a  leap year or not. So each day of the week happens 52 times a year + first one or two days of the year make corresponding week days have 53 days in that same year. If we know the number of days in a year (365 or 366) we can find out which days of the week will happen 53 times. For that matter we can take MOD(days_in_year, 52) expression that will return either 1 or 2. If the day order number within a year is 1 (or 2 for the leap year) we know that the corresponding week day will occur 53 times, otherwise – 52.

The WITH clause returns number of days in the current year. We get that by taking the 1st day of the current year: TRUNC(SYSDATE,’YYYY’), adding 12 months to it and subtract 1 day to get the last day of the current year. Taking TO_CHAR(…, ‘DDD’) – gives us the order number of that day in the year which is exactly the number of days in the current year.

The main query generates the date range from Jan-1 to Jan-7 in the current year, and assigns 52 or 53 to the 2nd column based on the logic described above.

My Oracle Group on Facebook:

If you like this post, you may want to join my new Oracle group on Facebook: https://www.facebook.com/groups/sqlpatterns/

Would you like to read about many more tricks and puzzles?

For more tricks and cool techniques check my book “Oracle SQL Tricks and Workarounds” for instructions.

 

 

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s