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# Task 1: Banking Day Offset
# Submitted by: Lee Johnson
#
# You are given a start date and offset counter. Optionally you also get bank holiday date list.
#
# Given a number (of days) and a start date, return the number (of days) adjusted to take into account non-banking days. In other words: convert a banking day offset to a calendar day offset.
#
# Non-banking days are:
# a) Weekends
# b) Bank holidays
#
# Example 1
# Input: $start_date = '2018-06-28', $offset = 3, $bank_holidays = ['2018-07-03']
# Output: '2018-07-04'
#
# Thursday bumped to Wednesday (3 day offset, with Monday a bank holiday)
#
# Example 2
# Input: $start_date = '2018-06-28', $offset = 3
# Output: '2018-07-03'
require 'date'
def banking_day_offset(start_date, offset, bank_holidays)
dt = Date.strptime(start_date, '%Y-%m-%d')
(1..offset).each do |i|
dt += 1
while no_banking_day(dt, bank_holidays) == true
dt += 1
end
end
printf "(start %s offset %s Bank holidays [%s]) -> %s\n",
start_date,
offset,
bank_holidays.join(" / "),
dt.strftime('%Y-%m-%d');
end
def no_banking_day(dt, bank_holidays)
if dt.strftime("%u").to_i >= 6 or bank_holidays.include?( dt.strftime('%Y-%m-%d') )
return true
end
false
end
start_date = '2018-06-28'
offset = 3
bank_holidays = ['2018-07-03']
banking_day_offset(start_date, offset, bank_holidays)
start_date = '2018-06-28';
offset = 3;
bank_holidays = [];
banking_day_offset(start_date, offset, bank_holidays);
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