Valid Sudoku LeetCode Solution | Easy Approach

Valid Sudoku | Given an array of integers nums sorted in non-decreasing order, find the starting and ending position of a given target value.

If target is not found in the array, return [-1, -1].

You must write an algorithm with O(log n) runtime complexity.

Example 1:

Input: nums = [5,7,7,8,8,10], target = 8
Output: [3,4]

Example 2:

Input: nums = [5,7,7,8,8,10], target = 6
Output: [-1,-1]

Example 3:

Input: nums = [], target = 0
Output: [-1,-1]

Constraints:

• 0 <= nums.length <= 105
• -109 <= nums[i] <= 109
• nums is a non-decreasing array.
• -109 <= target <= 109

Valid Sudoku LeetCode Solution

✅Time: O(1)
✅Space:O(1)

C++

 class Solution {
 public:
  bool isValidSudoku(vector<vector<char>>& board) {
    unordered_set<string> seen;

    for (int i = 0; i < 9; ++i)
      for (int j = 0; j < 9; ++j) {
        if (board[i][j] == '.')
          continue;
        const string c(1, board[i][j]);
        if (!seen.insert(c + "@row" + to_string(i)).second ||
            !seen.insert(c + "@col" + to_string(j)).second ||
            !seen.insert(c + "@box" + to_string(i / 3) + to_string(j / 3))
                 .second)
          return false;
      }

    return true;
  }
};

Java

 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) {
        if (board[i][j] == '.')
          continue;
        final char c = board[i][j];
        if (!seen.add(c + "@row" + i) ||
            !seen.add(c + "@col" + j) ||
            !seen.add(c + "@box" + i / 3 + j / 3))
          return false;
      }

    return true;
  }
}

Python

 class Solution:
  def isValidSudoku(self, board: List[List[str]]) -> bool:
    seen = set()

    for i in range(9):
      for j in range(9):
        c = board[i][j]
        if c == '.':
          continue
        if c + '@row ' + str(i) in seen or \
           c + '@col ' + str(j) in seen or \
           c + '@box ' + str(i // 3) + str(j // 3) in seen:
          return False
        seen.add(c + '@row ' + str(i))
        seen.add(c + '@col ' + str(j))
        seen.add(c + '@box ' + str(i // 3) + str(j // 3))

    return True

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