36. Valid Sudoku - DevCodeEx

Explanation

To solve this problem, we need to validate if a given Sudoku board is valid according to the rules mentioned in the problem description. We can iterate through each row, each column, and each 3x3 sub-box to check for duplicates. We can use three sets to keep track of seen digits in rows, columns, and sub-boxes. If we encounter a digit that is already in the set, the board is invalid. If we finish iterating without finding any duplicates, the board is valid.

Algorithm:
1. Initialize three sets for rows, columns, and sub-boxes.
2. Iterate through each cell in the board.
3. Check if the current digit is a duplicate in the corresponding row, column, or sub-box.
4. If it is not a duplicate, add it to the corresponding set.
5. If a duplicate is found, return false.
6. If no duplicates are found, return true.
Time Complexity: O(1) since the board is always 9x9.
Space Complexity: O(1) since the sets will always have at most 9 unique elements.

View Leetcode Problem

class Solution {
    public boolean isValidSudoku(char[][] board) {
        Set<String> seen = new HashSet<>();
        
        for (int i = 0; i < 9; i++) {
            for (int j = 0; j < 9; j++) {
                char currentVal = board[i][j];
                if (currentVal != '.') {
                    if (!seen.add(currentVal + " in row " + i) ||
                        !seen.add(currentVal + " in column " + j) ||
                        !seen.add(currentVal + " in sub-box " + i/3 + "-" + j/3)) {
                        return false;
                    }
                }
            }
        }
        
        return true;
    }
}

Code Editor (Testing phase)

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