Meta / Amazon / Google / Microsoft Interview Question: 36. Valid Sudoku | LeetCode-36: Medium | JavaScript Implementation

We will be seeing the solution to the LeetCode problem 36 in JavaScript. I hope you will find it useful.

Problem Statement:

Determine if a 9 x 9 Sudoku board is valid. Only the filled cells need to be validated according to the following rules:

Each row must contain the digits 1–9 without repetition.
Each column must contain the digits 1–9 without repetition.
Each of the nine 3 x 3 sub-boxes of the grid must contain the digits 1–9 without repetition.
Note:

A Sudoku board (partially filled) could be valid but is not necessarily solvable.
Only the filled cells need to be validated according to the mentioned rules.

Link to Problem: https://leetcode.com/problems/wildcard-matching/

Implementation:

var isValidSudoku = function (board) {
  let gridSize = board.length;
  const util = () => {
    for (let row = 0; row < gridSize; row++) {
      for (let col = 0; col < gridSize; col++) {
        let countMap = {
          1: false,
          2: false,
          3: false,
          4: false,
          5: false,
          6: false,
          7: false,
          8: false,
          9: false,
        };
for (let i = 0; i < 9; i++) {
          if (countMap[board[row][i]] && board[row][i] != ".") {
            return false;
          }
          countMap[board[row][i]] = true;
        }
        countMap = {
          1: false,
          2: false,
          3: false,
          4: false,
          5: false,
          6: false,
          7: false,
          8: false,
          9: false,
        };
        for (let i = 0; i < 9; i++) {
          if (countMap[board[i][col]] && board[i][col] != ".") {
            return false;
          }
          countMap[board[i][col]] = true;
        }
let subMatrixRow = Math.floor(row / 3) * 3;
        let subMatrixCol = Math.floor(col / 3) * 3;
        countMap = {
          1: false,
          2: false,
          3: false,
          4: false,
          5: false,
          6: false,
          7: false,
          8: false,
          9: false,
        };
        for (let i = subMatrixRow; i < subMatrixRow + 3; i++) {
          for (let j = subMatrixCol; j < subMatrixCol + 3; j++) {
            if (countMap[board[i][j]] && board[i][j] != ".") {
              return false;
            }
            countMap[board[i][j]] = true;
          }
        }
      }
    }
    return true;
  };
  return util();
};

Thank you for reading hope this code is helpful to you.