Intuition

Validity is purely about no duplicate digit within any row, column, or 3x3 box. Track seen digits for each of those 27 groups.

Approach

Maintain a set for each row, each column, and each box (box index = (r/3)*3 + c/3).
For each filled cell, check all three sets; if the digit is already present, it is invalid.
Otherwise record it in the three sets.

Complexity

Time — O(1): the board is a fixed 81 cells.
Space — O(1): 27 small sets.

Full Solution — all languages

class Solution:
    def isValidSudoku(self, board):
        rows=[set() for _ in range(9)]
        cols=[set() for _ in range(9)]
        boxes=[set() for _ in range(9)]
        for r in range(9):
            for c in range(9):
                v=board[r][c]
                if v==".": continue
                b=(r//3)*3+c//3
                if v in rows[r] or v in cols[c] or v in boxes[b]: return False
                rows[r].add(v); cols[c].add(v); boxes[b].add(v)
        return True

import java.util.*;

class Solution {
    public boolean isValidSudoku(char[][] board) {
        Set<Character>[] rows=new HashSet[9],cols=new HashSet[9],boxes=new HashSet[9]; for(int i=0;i<9;i++){rows[i]=new HashSet<>();cols[i]=new HashSet<>();boxes[i]=new HashSet<>();} for(int r=0;r<9;r++)for(int c=0;c<9;c++){char v=board[r][c]; if(v=='.')continue; int b=(r/3)*3+c/3; if(rows[r].contains(v)||cols[c].contains(v)||boxes[b].contains(v))return false; rows[r].add(v);cols[c].add(v);boxes[b].add(v);} return true;
    }
}

#include <bits/stdc++.h>
using namespace std;

class Solution {
public:
    bool isValidSudoku(vector<vector<char>>& board) {
        vector<set<char>> rows(9),cols(9),boxes(9); for(int r=0;r<9;r++)for(int c=0;c<9;c++){char v=board[r][c]; if(v=='.')continue; int b=(r/3)*3+c/3; if(rows[r].count(v)||cols[c].count(v)||boxes[b].count(v))return false; rows[r].insert(v);cols[c].insert(v);boxes[b].insert(v);} return true;
    }
};

/**
 * @param {character[][]} board
 * @return {boolean}
 */
var isValidSudoku = function(board) {
    const rows=[],cols=[],boxes=[]; for(let i=0;i<9;i++){rows.push(new Set());cols.push(new Set());boxes.push(new Set());} for(let r=0;r<9;r++)for(let c=0;c<9;c++){const v=board[r][c]; if(v==='.')continue; const b=((r/3)|0)*3+((c/3)|0); if(rows[r].has(v)||cols[c].has(v)||boxes[b].has(v))return false; rows[r].add(v);cols[c].add(v);boxes[b].add(v);} return true;
};

Valid Sudoku

Example 1

Example 2

Constraints