Validate a Partial Sudoku

1. Step 1: Initialize Solution.isValidSudoku

   Replace the empty starter with the first real state owned by Solution.isValidSudoku. A small, named state is easier to verify than a complete algorithm. Establish it before adding the branch or loop that changes it.

2. Step 2: Assemble the primary transition

   Extend the initialized state with the next contiguous part of the popular solution. The transition explains how one input element or operation changes the state; boundaries are easier to reason about after this invariant is visible.

3. Step 3: Pass the Valid Partial case

   Complete the readable core algorithm for one representative Interview case. Encode each cell's three constraint groups and reject repeated membership.

4. Step 4: Harden the Box Duplicate boundary

   Repair the reviewed boundary and pass the complete submission contract. Each filled cell is inserted into exactly its row, column, and box sets. A repeated membership is exactly a Sudoku constraint violation; no repeat means all three constraint families hold.

• Box identity is (row // 3, column // 3), not row % 3 and column % 3.
• hashing and sets

• LeetCode 36 — Valid Sudoku

• Track Previously Seen Values(opens in a new tab)

  Track Previously Seen Values isolates before processing index i, seen contains exactly the distinct values from indices smaller than i. That focused state discipline is required when implementing valid sudoku as a complete Interview Problem.

Maintain one set per row, column, and box while scanning non-empty cells.

Why it works: Each filled cell is inserted into exactly its row, column, and box sets. A repeated membership is exactly a Sudoku constraint violation; no repeat means all three constraint families hold.

class Solution:
    def isValidSudoku(self, board):
        """
        Checkpoint 1: initialize the state owned by this Interview contract.
        Checkpoint 2: assemble the primary transition without hiding the boundary.
        """
        rows = [set() for _ in range(9)]
        columns = [set() for _ in range(9)]
        boxes = [set() for _ in range(9)]
        for row in range(9):
            for column in range(9):
                value = board[row][column]
                if value == '.':
                    continue
                box = (row // 3) * 3 + column // 3
                if value in rows[row] or value in columns[column] or value in boxes[box]:
                    return False
                rows[row].add(value)
                columns[column].add(value)
                boxes[box].add(value)
        return True