Editorial

Solution walkthrough

Review the explanation, then compare the reference implementation.

Video explanation

Solution

Intuition

A filled cell belongs to three independent groups: its row, its column, and its 3-by-3 box. The board is invalid exactly when a digit repeats in any one of those groups. We only need to remember digits already encountered, not solve the unfinished puzzle.

Approach

Maintain a set of seen digits for every row, column, and box.
Visit every cell, skipping . because it represents an empty position.
Identify the box from the row and column divided by three, rounding down. Python uses the pair of box coordinates; JavaScript flattens that pair into an index.
If the digit is present in any of its three sets, return false. Otherwise, add it to all three sets.
Return true after the full scan. This certifies the current placements only; it does not guarantee a complete Sudoku solution exists.

class Solution:
    def isValidSudoku(self, board: List[List[str]]) -> bool:
        rows = defaultdict(set)
        cols = defaultdict(set)
        boxes = defaultdict(set)

        for i in range(9):
            for j in range(9):
                x = board[i][j]
                if x == ".":
                    continue
                if x in rows[i] or x in cols[j] or x in boxes[(i//3,j//3)]:
                    return False
                rows[i].add(x)
                cols[j].add(x)
                boxes[(i//3,j//3)].add(x)

        return True

Complexity analysis

The displayed code scans at most 81 cells of the fixed 9-by-9 board. Each filled cell is recorded in three sets, for at most 243 total set entries across 27 sets. Time and auxiliary space are both O(1) under the problem's fixed dimensions.