Valid Suduko Problem Solution

Valid Suduko Problem: Determine if a 9 x 9 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 nine 3 x 3 sub-boxes of the grid must contain the digits 1-9 without repetition.

Input: board = 
[["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

Valid Suduko Problem Solution

Problem Solution In Python

class Solution:
    def isValidSudoku(self, board: List[List[str]]) -> bool:
        
        def is_valid_row():
            for row in board:
                if not is_valid(row):
                    return False
            return True
        
        def is_valid_col():
            for col in zip(*board):
                if not is_valid(col):
                    return False
            return True
        
        def is_valid_square():
            for i in (0,3,6):
                for j in (0,3,6):
                    square = [board[x][y] for x in range(i,i+3) 
                                        for y in range(j,j+3)]
                    if not is_valid(square):
                        return False
            return True
    
        def is_valid(value):
            res=[i for i in value if i!="."]
            return len(res)==len(set(res))
        
        return is_valid_row() and is_valid_col() and is_valid_square()

Problem Solution In Java

class Solution {
    public boolean isValidSudoku(char[][] board) {
        int n = board.length;
        Set<Integer>[] rows = new Set[n];
        Set<Integer>[] cols = new Set[n];
        Set<Integer>[] blocks = new Set[n];
        
        for(int i = 0; i < n; i++) {
            rows[i] = new HashSet<Integer>();
            cols[i] = new HashSet<Integer>();
            blocks[i] = new HashSet<Integer>();
        }
            
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {

                if (board[i][j] == '.') continue;                
                int k = 3 * (i /3) + (j / 3);
                
                int num = board[i][j] - '0';
                if (rows[i].contains(num)) return false;
                if (cols[j].contains(num)) return false;
                if (blocks[k].contains(num)) return false;
                
                rows[i].add(num);
                cols[j].add(num);
                blocks[k].add(num);
            }
        }                
        return true;
    }
}

Problem Solution In C++

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

Problem Solution In C

bool isValidSudoku(char** board, int boardSize, int* boardColSize){

int i=0, j=0;
int rowSet[9][10] = {0};
int colSet[9][10] = {0};
int box[3][3][10] = {0};    
for(i = 0; i<9; i++){
    for(j=0; j<9; j++){
        if(board[i][j] != '.'){
            int val = (int)board[i][j] - 48;//Converting to int
            if(rowSet[i][val] != 0)
                return false;
            if(colSet[j][val] != 0)
                return false;
            if(box[i/3][j/3][val] !=0)
                return false;   
            box[i/3][j/3][val] +=1;
            rowSet[i][val] +=1;
            colSet[j][val] +=1;
        }
    }
}
return true;
}