Valid Sudoku - Leetcode 36 - Hashmaps & Sets (Python)

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[@GregHogg]

Master Data Structures & Algorithms For FREE at AlgoMap.io!

[@karthikistired]

if someone doesn't wanna use variable starts, you can modify the for loop like this:
for row in range(0,9,3):
for col in range(0,9,3):
s=set()
for i in range(3):
for j in range(3):
if board[row+i][col+j] in s:
return False
elif board[row+i][col+j] != '.':
s.add(board[row+i][col+j])

[@nguyenkhiem2318]

Thanks Greg, you really have a natural talent for teaching. Was struggling with this one

[@egg5687]

I usually solve the problem and then compare to your solution and for the case of the medium difficulty problems, my solution is often somewhat more convoluted than yours; there's always some detail that I miss that makes my solution significantly more complicated. But for once, I am still pretty happy with my solution even after seeing yours! So I felt compelled to share it:
class Solution(object):
def isValidSudoku(self, board):
"""
:type board: List[List[str]]
:rtype: bool
"""
sudoku_dict = {}
# The format of this dictionary will be number: [{rows},{cols},{box}]
for i in range(9):
for j in range(9):
val = board[i][j]
if val == ".":
continue
box_num = 3 * (i // 3) + (j // 3)
if val not in sudoku_dict:
sudoku_dict[val] = [{i}, {j}, {box_num}]
elif (i in sudoku_dict[val][0]) or (j in sudoku_dict[val][1]) or (box_num in sudoku_dict[val][2]):
return False
else:
sudoku_dict[val][0].add(i)
sudoku_dict[val][1].add(j)
sudoku_dict[val][2].add(box_num)
return True

[@punnettsquares]

Took me 3 years to understand this question

[@JSH1994]

your explanations are so clear and intuitive, thank you!

[@mercytum7045]

Thanks, Greg. This was really helpful!

[@GregHogg]

Glad it was helpful!

[@nathanaelcheramlak9332]

Helpful content, Keep going

[@GregHogg]

Thanks a ton 😊

[@HussainAli-hn2ef]

very helpful, thanks!

[@GregHogg]

Glad to hear it!!

[@alek002]

Thanks!

[@arsheyajain7055]

Super helpful!!!!!

[@GregHogg]

Awesome 😎😎

[@therockriders2759]

Super solution

[@anshpradhan1954]

"Easily the hardest part" lol

[@saedf0]

Hey Greg, thanks for the content. Great solution, although i do wonder, how is it BIG O(1) when we have multiple loops? I'm not the best with the complexity identification.

[@GregHogg]

Thank you! And that's a great question (at any level). Technically, these loops are going to run a constant amount of times (not dependent on the size of the grid) because WE KNOW the grid will be a 9*9. To scan the grid, it takes exactly 81 positions. Not dependent on some N or something. So O(81) is just O(1) in big o talk. I hope this helps!

[@saedf0]

@@GregHogg ohhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh /*click moment*/
Man, I cant rely on explanations all the time, do you have any resources/ or easy tricks to remember this stuff.
I've tried a few videos and identifying it kindve goes over my head
I get that N^2 (for everytime we do this, run these steps)
I also understand O(N), Do this
Log N (Half each time, divide and conquer
but when the code gets a little longer, kindve difficult to track
but yh thanks for getting back :)

[@user-jm6gp2qc8x]

greg anna vera level

[@krishkchavakrishnakarthik4162]

super anna

[@midhunskani]

Neat solution

[@lakku153]

very nice

[@kidyoshi007]

Why are you not "s.clear()" clearing the set after each check for row, col, and box??

[@michaelleeser-smith9342]

S is being reinitialized at each step. It's not global so it doesn't need to be cleared