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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'
+
+from dateutil.parser import *
+import datetime
+
+def banking_day_offset(start_date, offset, bank_holidays):
+    dt = parse(start_date, dayfirst = False, yearfirst = True)
+
+    for i in range(1,5):
+        dt += datetime.timedelta(days=1)
+        while no_banking_day(dt):
+            dt += datetime.timedelta(days=1)
+
+    print("(start %s offset %s Bank holidays [%s]) -> %s" %
+            (start_date, offset,
+             ' / '.join(bank_holidays),
+             dt.strftime('%Y-%m-%d')
+            )
+    )
+
+def no_banking_day(dt):
+    if (dt.weekday() >= 6) or (dt.strftime('%Y-%m-%d') in bank_holidays):
+        return 1
+    return 0
+
+if __name__ == "__main__":
+    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)