valid-sudoku py3

0036_valid-sudoku/python3/solution.py

@@ -0,0 +1,54 @@
+class Solution:
+    def isValidSudoku(self, board: List[List[str]]) -> bool:
+        #
+        # NOTE: **READ THE PROBLEM DESCRIPTION ALWAYS!!**
+        #
+        # Even though it mentions Sudoku, it doesn't mean the
+        # rules are the same as the real world. Which is the case
+        # here — we need to only look at the filled cell and the rules
+        # provided.
+        
+        size = len(board)
+        
+        # 1. Check if rows and cols are valid
+        #
+        # We can iterate both rows and cols at the same time since
+        # the sudoku board has 9 rows and 9 cols
+        for i in range(size):
+            row = set()
+            col = set()
+            
+            for j in range(size):
+                # ith row check
+                if board[i][j] != '.':
+                    if board[i][j] in row:
+                        return False
+                    
+                    row.add(board[i][j])
+                
+                # ith col check
+                if board[j][i] != '.':
+                    if board[j][i] in col:
+                        return False
+                    
+                    col.add(board[j][i])
+        
+        # 2. Check if squares are valid
+        # 
+        # We can do normal row * col iteration and use `// 3` to figure
+        # out which square a cell belongs to
+        
+        squares = [[set(), set(), set()] for _ in range(3)]
+        for i in range(size):
+            for j in range(size):
+                cell = board[i][j]
+                if cell == '.':
+                    continue
+                
+                square = squares[i // 3][j // 3]
+                if cell in square:
+                    return False
+                
+                square.add(cell)
+        
+        return True