How to find nearest future date from now with specific weekday? - algorithm

Question

I have an appointment management scenario in which I have a field List of availableWeekDays of a person on which weekdays they are available for booking. For example if they are available on monday I have a value in availableWeekDays list as 1.
I have currentDate stored in a DateTime variable using DateTime.now().
What I have to do is first look if today's week day matches the availableWeekDay of person. If yes then bookingDate is valid else I have to look for nearest future date with weekday of person. I have tried following solution but it works for one week day as I'm using first index.
void main() {
DateTime bookingDate = DateTime.now();
int availableWeekDays = [1,4,7];
getDate(){
if(bookingDate.weekday == availableWeekDays[0]){
print("Available on $bookingDate ");
} else {
bookingDate = bookingDate.add(Duration(days: 1));
getDate();
}
}
getDate();
}
I want to find the nearest future date among all weekdays present in list.

Answer 1

DateTime findNearestWeekday(List<int> availableWeekdays) {
assert(availableWeekdays.isNotEmpty);
var date = DateTime.now();
final weekday = date.weekday;
for (var num in availableWeekdays) {
if (num == date.weekday) {
return date;
} else if (date.weekday < num) {
return date.add(Duration(days: num - weekday));
}
}
/// I have assumed that you will always have a sorted list of available
/// weekdays, in case you do not have a sorted list, either you can sort
/// the list first or just use this line (instead of the last one),
///
/// ```dart
/// import 'dart:math'; /// Add on Top of file.
///
/// return date.add(Duration(days: 7 - weekday + availableWeekdays.reduce(min)))
/// ```
///
/// I will prefer to pre sort the array before storing it somewhere,
/// because it will add an additional dependency dart:math
return date.add(Duration(days: 7 - weekday + availableWeekdays[0]));
}

Answer 2

A more efficient solution would be:
DateTime getDate() {
var currentDay = bookingDate.weekday;
var offset = (availableWeekDay - currentDay).remainder(7);
return offset == 0 ? bookingDate : bookingDate.add(Duration(days: offset));
}
Do be aware that adding "days" (multiples of 24 hours) to a date may run into daylight savings issues if the time of the bookingDate is close to midnight.

Answer 3

I forgot to reassign the result to the variable. This works now.
void main() {
DateTime bookingDate = DateTime.now();
int availableWeekDay = 1;
getDate(){
if(bookingDate.weekday == availableWeekDay){
print("Available on $bookingDate ");
} else {
bookingDate = bookingDate.add(Duration(days: 1));
getDate();
}
}
getDate();
}