pwc259 solution in python

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diff --git a/challenge-259/pokgopun/python/ch-1.py b/challenge-259/pokgopun/python/ch-1.py
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+### https://theweeklychallenge.org/blog/perl-weekly-challenge-259/
+"""
+
+Task 1: Banking Day Offset
+
+Submitted by: [42]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'
+
+Task 2: Line Parser
+"""
+### solution by pokgopun@gmail.com
+
+from datetime import date,datetime,timedelta
+
+def str2date(datestr: str):                                      ### convert date string to date object
+    return datetime.strptime(datestr,"%Y-%m-%d").date()
+
+def isWeekday(d: date):                                          ### check if date object is a weekday, that is less than Saturday (i.e. 5 in python)
+    return d.weekday() < 5 
+
+def BDO(start: str, ofst: int, hols: tuple = ()):
+    dstart = str2date(start)
+    if isWeekday(dstart) == False:                               ### if startdate is a weekend day, move back to last friday
+        dstart -= timedelta(days = dstart.weekday() - 4)         ### move back number of day from friday
+
+    ### convert weekday offset (ofst) to everyday offset (dur), that is 7 * number of week (i.e. ofset // 5) and the remaing partial-weekday offset (ofst % 5)
+    dur = timedelta(days = 7*(ofst // 5) + ofst % 5)
+    ### now we need to account for partial-weekday offset in the everyday offset
+    if dstart.weekday() + ofst % 5 > 4:                          ### if partial-weekday offset move startdate over the current weekday, add weekend offset to everyday offset
+        dur += timedelta(days=2)
+    dend = dstart + dur                                          ### calculate enddate
+
+    ### now account for holidays one by one, need to start from the earliest holiday first so we sort holidays before processing
+    for hol in sorted(set(hols)):
+        dhol = str2date(hol)
+        if dhol > dstart and dhol <= dend and isWeekday(dhol):   ### only offset holiday if it is a weekday and is later than startdate but not later than current enddate
+            dend += timedelta(days=1)
+            if isWeekday(dend) == False:                         ### if a holiday offset move enddate to a weekend day, add weekend offset
+                dend += timedelta(days=2)                        ### this is ok as we account for holiday offset one by one
+    return dend.strftime("%Y-%m-%d")
+
+import unittest
+
+class TestBDO(unittest.TestCase):
+    def test(self):
+        for output, (start_date, offset, bank_holidays) in {
+                "2018-07-04": ("2018-06-28", 3, ("2018-07-03",)),
+                "2018-07-03": ("2018-06-28", 3, ()),
+                "2018-07-05": ("2018-06-28", 3, ("2018-07-02", "2018-07-03")),
+                "2018-07-05": ("2018-06-28", 3, ("2018-07-06", "2018-07-03", "2018-07-02")),
+                "2018-07-06": ("2018-06-28", 3, ("2018-07-02", "2018-07-03", "2018-07-05")),
+                "2018-07-06": ("2018-06-28", 3, ("2018-07-02", "2018-07-03", "2018-07-05","2024-03-05")),
+                "2018-07-06": ("2018-06-28", 3, ("2018-07-02", "2018-07-03", "2018-07-05","1979-07-09")),
+                "2018-07-13": ("2018-06-28", 8, ("2018-07-02", "2018-07-03", "2018-07-05","1979-07-09")),
+                "2018-07-16": ("2018-06-28", 9, ("2018-07-02", "2018-07-03", "2018-07-05","1979-07-09")),
+                "2018-07-23": ("2018-06-28", 14, ("2018-07-02", "2018-07-03", "2018-07-05","1979-07-09")),
+                "2018-07-25": ("2018-06-28", 14, ("2018-07-02", "2018-07-03", "2018-07-05","2018-07-11","2018-07-18")),
+                "2018-07-05": ("2018-06-29", 3, ("2018-07-03",)),
+                "2018-07-05": ("2018-06-30", 3, ("2018-07-03",)),
+                }.items():
+            self.assertEqual(output, BDO(start_date, offset, bank_holidays))
+
+unittest.main()
diff --git a/challenge-259/pokgopun/python/ch-2.py b/challenge-259/pokgopun/python/ch-2.py
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index 0000000000..ac56114e24
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+++ b/challenge-259/pokgopun/python/ch-2.py
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+### https://theweeklychallenge.org/blog/perl-weekly-challenge-259/
+"""
+
+Task 2: Line Parser
+
+Submitted by: [43]Gabor Szabo
+     __________________________________________________________________
+
+   You are given a line like below:
+{%  id   field1="value1"    field2="value2"  field3=42 %}
+
+   Where
+a) "id" can be \w+.
+b) There can be 0  or more field-value pairs.
+c) The name of the fields are \w+.
+b) The values are either number in which case we don't need parentheses or strin
+g in
+   which case we need parentheses around them.
+
+   The line parser should return structure like below:
+{
+       name => id,
+       fields => {
+           field1 => value1,
+           field2 => value2,
+           field3 => value3,
+       }
+}
+
+   It should be able to parse the following edge cases too:
+{%  youtube title="Title \"quoted\" done" %}
+
+   and
+{%  youtube title="Title with escaped backslash \\" %}
+
+   BONUS: Extend it to be able to handle multiline tags:
+{% id  filed1="value1" ... %}
+LINES
+{% endid %}
+
+   You should expect the following structure from your line parser:
+{
+       name => id,
+       fields => {
+           field1 => value1,
+           field2 => value2,
+           field3 => value3,
+       }
+       text => LINES
+}
+     __________________________________________________________________
+
+   Last date to submit the solution 23:59 (UK Time) Sunday 10th March
+   2024.
+     __________________________________________________________________
+
+SO WHAT DO YOU THINK ?
+"""
+### solution by pokgopun@gmail.com
+
+import re
+
+rgxLine = re.compile(r'{%\s(?P<name>\w+)(?P<kv>(?:\s\w+=(?:\d+|"(?:\\(?:"|\\)|[^"\\])+?"))+)?\s%}(?:\n(?P<tag>[\d\D]+?)\n\{%\sendid\s%})?')
+rgxKV = re.compile(r'\s(?P<key>\w+)=(?P<value>\d+|"(?:\\(?:"|\\)|[^"\\])+?")')
+ 
+def lineParser(msg: str):
+    res = rgxLine.match(msg)
+    parsed = { "name": res.group("name") }
+    resKV = res.group("kv")
+    if resKV is not None:
+        parsed["fields"] = {}
+        for resKV in rgxKV.finditer(res.group("kv")):
+            k = resKV.group("key")
+            v = resKV.group("value")
+            if v.isnumeric():
+                v = int(v)
+            else:
+                v = v[1:-1].encode("utf-8").decode('unicode_escape')
+            parsed["fields"][k] = v
+    tag = res.group("tag")
+    if tag is not None:
+        parsed["text"] = tag
+    return parsed
+
+import unittest
+
+class TestLineParser(unittest.TestCase):
+    def test(self):
+        for inpt, otpt in {
+                r'{% youtube id=1234 title="Title \"quoted\" done" %}': {
+                    'name': 'youtube',
+                    'fields': {
+                        'id': 1234,
+                        'title': 'Title "quoted" done',
+                        },
+                    },
+                r'{% youtube title="Title with escaped backslash \\" id=4321 %}': {
+                    'name': 'youtube',
+                    'fields': {
+                        'id': 4321,
+                        'title': 'Title with escaped backslash \\',
+                        },
+                    },
+                r'''{% id field1="value1" field2=1324 %}
+LINE1
+LINE2
+{% endid %}''': {
+    'name': 'id',
+    'fields': {
+        'field2': 1324,
+        'field1': 'value1',
+        },
+    'text': '''LINE1
+LINE2''',
+},
+}.items():
+            self.assertEqual(lineParser(inpt),otpt)
+
+unittest.main()