Problem
Annotate a rectangular board. A star (*) is a mine and a dot (.) is a safe cell. For each safe cell, count mines in the eight surrounding cells, including diagonals. Replace it with the count from 1 to 8, or keep a dot if the count is zero. Preserve every mine. Cells outside the board do not exist.
Implement minesweeper_neighbor_counts(board).
Input
board: The rows from top to bottom; characters run left to right.
Output
Return the annotated rows in the same order and dimensions.
Constraints
1 ≤ rows, columns ≤ 200. Every row has the same nonzero length and contains only * and . .
Examples
Input:
board = ["*..", "...", "..*"]
Expected output:
expected = ["*1.", "121", ".1*"]Input:
std::vector<std::string> board = {"*..", "...", "..*"};
Expected output:
std::vector<std::string> expected = {"*1.", "121", ".1*"};Input:
string[] board = new string[] {"*..", "...", "..*"};
Expected output:
string[] expected = new string[] {"*1.", "121", ".1*"};Explanation
The center touches both mines, so its count is 2. The top-right and bottom-left touch none and stay dots. Other safe cells touch one mine.
Input:
board = ["..."]
Expected output:
expected = ["..."]Input:
std::vector<std::string> board = {"..."};
Expected output:
std::vector<std::string> expected = {"..."};Input:
string[] board = new string[] {"..."};
Expected output:
string[] expected = new string[] {"..."};Explanation
This one-row board has no mines. All counts are zero, so every dot remains a dot.
How to approach it
Try on your own first, then open one hint at a time.
Hint 1
Only safe cells change; mines stay exactly where they are.
Hint 2
For each dot, inspect the eight surrounding positions and count nearby mines.
Hint 3
Be careful near edges and corners so you do not read outside the board.
Expected complexity
O(r*c) time, O(r*c) extra space
References
This challenge is adapted for CodeUtility. Follow the statement and constraints above.