Add solution for valid tic tac toe position

Author: Sigmanificient

src/python/katas/py6kyu/is_this_a_valid_tic_tac_toe_position.py

@@ -0,0 +1,41 @@
+"""Kata url: https://www.codewars.com/kata/68582183cf16719a5ba943df."""
+Line = tuple[str, str, str]
+Coord = tuple[int, int]
+
+
+def is_valid_position(board: tuple[Line, Line, Line]) -> bool:
+    flatten = [cell for line in board for cell in line]
+
+    p1 = flatten.count('O')
+    p2 = flatten.count('X')
+    if p2 != p1 and (p1 + 1 != p2):
+        return False
+
+    wins = set()
+    winner = None
+
+    for (x, y), (dx, dy) in (
+        ((0, 0), (1, 0)),
+        ((0, 1), (1, 0)),
+        ((0, 2), (1, 0)),
+        ((0, 0), (0, 1)),
+        ((1, 0), (0, 1)),
+        ((2, 0), (0, 1)),
+        ((0, 0), (1, 1)),
+        ((2, 0), (-1, 1)),
+    ):
+        chks = [(x + i * dx, y + i * dy) for i in range(3)]
+
+        for c in "OX":
+            if all(board[y][x] == c for (x, y) in chks):
+                wins.update(chks)
+                winner = c
+
+    if len(wins) not in {0,3,5}:
+        return False
+
+    return {
+        'O': p1 == p2,
+        'X': p1 != p2,
+        None: True
+    }[winner]