NSDate get next occurring weekday that is in the future

Question

The logic for my problem is a mouthful and I have already come up with a solution. My solution is posted below. I am looking to see if someone can come up with a more efficient / simpler solution.

The method is supposed to return the next, future from now, occurring weekday given a date. Time should be preserved between the input date and the output date.

For all examples, today is Friday May 8, 2015, 4:00 PM: All inputs and outputs are in 2015:

+---------------------------+---------------------------+
|           Input           |          Output           |
+---------------------------+---------------------------+
| Tuesday April 7, 3:00 PM  | Monday May 11, 3:00 PM    |
| Thursday May 7, 3:00 PM   | Monday May 11, 3:00 PM    |
| Thursday May 7, 5:00 PM   | Friday May 8, 5:00 PM     |
| Tuesday May 12, 3:00 PM   | Wednesday May 13, 3:00 PM |
| Saturday June 20, 3:00 PM | Monday June 22, 3:00 PM   |
+---------------------------+---------------------------+

Here is some pseudo code for the logic:

do {
    date += 1 day;
} while(date.isWeekend || date.isInThePast)

Here is the solution I came up with, avoiding the use of loops to keep efficiency in mind:

- (NSDate *)nextWeekDayInFuture {
    NSDate *now = [NSDate date];
    NSDate *nextWeekDaydate;
    NSCalendar *calendar = [NSCalendar currentCalendar];
    nextWeekDaydate = [self dateByAddingDays:1]; // custom method, adds 1 day

    if ([nextWeekDaydate isLessThan:now]) {
        NSDateComponents *nowComponents = [calendar components:(NSCalendarUnitYear | NSCalendarUnitMonth | NSCalendarUnitDay) fromDate:now];
        NSDateComponents *dateComponents = [calendar components:(NSCalendarUnitHour | NSCalendarUnitMinute | NSCalendarUnitSecond) fromDate:nextWeekDaydate];

        [nowComponents setHour:dateComponents.hour];
        [nowComponents setMinute:dateComponents.minute];
        [nowComponents setSecond:dateComponents.second];

        nextWeekDaydate = [calendar dateFromComponents:nowComponents];
        if ([nextWeekDaydate isLessThan:now]) {
            nextWeekDaydate = [nextWeekDaydate dateByAddingDays:1];
        }
    }

    NSDateComponents *components = [calendar components:NSCalendarUnitWeekday fromDate:nextWeekDaydate];
    if (components.weekday == Saturday) {
        nextWeekDaydate = [nextWeekDaydate dateByAddingDays:2];
    } else if (components.weekday == Sunday) {
        nextWeekDaydate = [nextWeekDaydate dateByAddingDays:1];
    }

    return nextWeekDaydate;
}

Use the input / output table from above to test your logic before posting a solution.

Answer 1

There's no reason not to place the logic in a loop, since at most it will loop twice (If the current day is Saturday).

Here is a solution for working out if a current day is a weekend: How to find weekday from today's date using NSDate?

- (NSInteger)isWeekend:(NSDate*)inDate
{
    NSCalendar* cal = [NSCalendar currentCalendar];
    NSDateComponents* comp = [cal components:kCFCalendarUnitWeekday fromDate:inDate];
    return [comp weekday];
}

We can then call this method, checking the returned NSInteger to see if it is sat or sun and run it again:

NSDate *now = [NSDate date];
now = [now dateByAddingTimeInterval:60*60*24]; // Add 1 day's worth.

NSInteger curDay = [self isWeekend:now];

while ((curDay == 6) || (curDay == 1)) // Sat or Sund.
{
    now = [now dateByAddingTimeInterval:60*60*24]; // Add 1 day's worth.
    curDay = [self isWeekend:now];
}

If you really want to remove the loop, you can check if it's Sat and add 2 days worth, or if it's Sun and add 1 days worth.

NSDate *now = [NSDate date];
now = [now dateByAddingTimeInterval:60*60*24]; // Add 1 day's worth.
NSInteger curDay = [self isWeekend:now];

if (curDay == 6) // Sat
{
    now = [now dateByAddingTimeInterval:60*60*24*2]; // Add 2 day's worth.
}
else if (curDay == 1) // Sun
{
    now = [now dateByAddingTimeInterval:60*60*24]; // Add 1 day's worth.
}