Runtime: 104 ms (Top 89.12%) | Memory: 14 MB (Top 35.48%)

Author: AnasImloul

scripts/algorithms/V/Valid Sudoku/Valid Sudoku.py

@@ -1,73 +1,64 @@
+# Runtime: 104 ms (Top 89.12%) | Memory: 14 MB (Top 35.48%)
 class Solution:
     def isValidSudoku(self, board: List[List[str]]) -> bool:
-        
+
         hrow = {}
         hcol = {}
         hbox = defaultdict(list)
-        
-        
+
         #CHECK FOR DUPLICATES ROWWISE
-        for  i in range(9):
+        for i in range(9):
             for j in range(9):
-                
-                
+
                 #JUST THAT THE DUPLICATE SHOULDNT BE ","
                 if board[i][j] != '.':
-                    
+
                     if board[i][j] not in hrow:
                         hrow[board[i][j]] = 1
-                        
+
                     else:
                         return False
-                    
+
             #CLEAR HASHMAP FOR THIS ROW
             hrow.clear()
         print("TRUE1")
         #CHECK FOR DUPLICATES COLUMNWISE
-        
-        for  i in range(9):
+
+        for i in range(9):
             for j in range(9):
-                
-                
+
                 #JUST THAT THE DUPLICATE SHOULDNT BE ","
                 if board[j][i] != '.':
-                    
+
                     if board[j][i] not in hcol:
                         hcol[board[j][i]] = 1
-                        
+
                     else:
                         return False
-           
+
         #CLEAR HASHMAP FOR THIS COL
-            
+
             hcol.clear()
-        
+
         print('TRUE2')
-            
-            
+
         #CHECK DUPLICATE IN BOX, THIS IS WHERE KEY DESIGN SKILLS COME INTO PLAY, FOR SUDOKU YOU COMBINE ROW INDICES AND COL INDICES
-        
-        
+
         for i in range(9):
             for j in range(9):
-                
+
                 i_3 = i //3
                 j_3 = j//3
-                
+
                 # print(hbox)
                 if board[i][j] != '.':
-                    
+
                 #CHECK ELEMENT OF ORIGINAL INDICE present in key i_3 , j_3
                     if board[i][j] not in hbox[i_3 , j_3]:
-#                       #CHECKED IN NEW KEY
-                        hbox[i_3 ,j_3  ]=  hbox[i_3 ,j_3  ] + [board[i][j]]
-                    
-                    
+# #CHECKED IN NEW KEY
+                        hbox[i_3 ,j_3 ]= hbox[i_3 ,j_3 ] + [board[i][j]]
+
                     else:
                         return False
-            
-            
-            
-        
+
         return True
-