【LeetCode】36. Valid Sudoku

Difficulty: Medium

 More:【目录】LeetCode Java实现

Description

https://leetcode.com/problems/valid-sudoku/

Determine if a 9x9 Sudoku board is valid. Only the filled cells need to be validated according to the following rules:

1. Each row must contain the digits 1-9 without repetition.
2. Each column must contain the digits 1-9 without repetition.
3. Each of the 9 3x3 sub-boxes of the grid must contain the digits 1-9 without repetition.

A partially filled sudoku which is valid.

The Sudoku board could be partially filled, where empty cells are filled with the character '.'.

Example 1:

Input:
[
  ["5","3",".",".","7",".",".",".","."],
  ["6",".",".","1","9","5",".",".","."],
  [".","9","8",".",".",".",".","6","."],
  ["8",".",".",".","6",".",".",".","3"],
  ["4",".",".","8",".","3",".",".","1"],
  ["7",".",".",".","2",".",".",".","6"],
  [".","6",".",".",".",".","2","8","."],
  [".",".",".","4","1","9",".",".","5"],
  [".",".",".",".","8",".",".","7","9"]
]
Output: true

Example 2:

Input:
[
  ["8","3",".",".","7",".",".",".","."],
  ["6",".",".","1","9","5",".",".","."],
  [".","9","8",".",".",".",".","6","."],
  ["8",".",".",".","6",".",".",".","3"],
  ["4",".",".","8",".","3",".",".","1"],
  ["7",".",".",".","2",".",".",".","6"],
  [".","6",".",".",".",".","2","8","."],
  [".",".",".","4","1","9",".",".","5"],
  [".",".",".",".","8",".",".","7","9"]
]
Output: false
Explanation: Same as Example 1, except with the 5 in the top left corner being 
    modified to 8. Since there are two 8's in the top left 3x3 sub-box, it is invalid.

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.
• The given board contain only digits 1-9 and the character '.'.
• The given board size is always 9x9.

Intuition

Mehtod 1: Use three hashSet (rowSet, colSet, boxSet) to store  digits in each row/ column/ sub-box.

Details: 

　　i: 1 ->9   j: i ->9

　　rowSet: a[i][j]

　　colSet: a[j][i]

　　boxSet: 

　　　　For a given sub-box, we can traverse a 3*3 sub-box using just j: a[ j/3 ] [ j%3 ]

　　　　We has 9 sub-boxes, so we can use i to switch in different sub-boxes: a[i/3*3+j/3][i%3*3+j%3]

Mehtod 2: Use only one hashSet to store a[i][j] in three ways.

　　Encode a[i][j] as a[i][j]+"r"+i  to represent a[i][j] in row i;

　　Encode a[i][j] as a[i][j]+"c"+j  to represent a[i][j] in colum j; 

　　Encode a[i][j] as a[i][j]+"box"+i/3+j/3  to represent a[i][j] in box-i/3-j/3; 

 

Solution

    //Method 1
    public boolean isValidSudoku(char[][] board) {
        for(int i=0; i<9; i++){
            HashSet<Character> rowSet = new HashSet<>();
            HashSet<Character> colSet = new HashSet<>();
            HashSet<Character> boxSet = new HashSet<>();
            for(int j=0; j<9; j++){
                if(board[i][j]!='.' && !rowSet.add(board[i][j]))
                    return false;
                if(board[j][i]!='.' && !colSet.add(board[j][i]))
                    return false;
                if(board[i/3*3+j/3][i%3*3+j%3]!='.' && !boxSet.add(board[i/3*3+j/3][i%3*3+j%3]))
                    return false;
            }
        }
        return true;
    }
    
    //Method 2: easier to understand
    public boolean isValidSudoku(char[][] board) {
        HashSet<String> set = new HashSet<>();
        for(int i=0; i<9; i++){
            for(int j=0; j<9; j++){
                if(board[i][j]!='.'){
                    String num = String.valueOf(board[i][j]);
                    if(!set.add(num+"r"+i) || !set.add(num+"c"+j) || !set.add(num+"box"+i/3+j/3))
                        return false;                    
                }
            }
        }
        return true;
    }

　　

Complexity

Time complexity : O(n)

Space complexity : O(n)

 

What I've learned

1. Learn how to use i, j to represent elements in different rows and columns and sub-boxes.

2. For a given sub-box, we can traverse a 3*3 sub-box using just j: a[ j/3 ] [ j%3 ]

3. It's clever to store elements in a hashSet by convert elements to String.

 

 More:【目录】LeetCode Java实现