Python 0529.py
class Solution:
def updateBoard(self, board: List[List[str]], click: List[int]) -> List[List[str]]:
width = len(board[0])
height = len(board)
deltas = [ [ -1, -1 ], [ -1, 0 ], [ -1, +1 ],
[ 0, -1 ], [ 0, +1 ],
[ +1, -1 ], [ +1, 0 ], [ +1, +1 ] ]
def count(x, y):
result = 0
for dx, dy in deltas:
xx = x + dx
yy = y + dy
if 0 <= xx < width and 0 <= yy < height:
if board[yy][xx] in [ "M", "X" ]:
result += 1
return result
def reveal(x, y):
# detect mine
if board[y][x] == "M":
board[y][x] = "X"
return
# already visible
if board[y][x] != "E":
return
# count mines
board[y][x] = str(count(x, y))
# reveal surrounding squares if blank
if board[y][x] == "0":
board[y][x] = "B"
for dx, dy in deltas:
xx = x + dx
yy = y + dy
if 0 <= xx < width and 0 <= yy < height:
# recursion
reveal(xx, yy)
reveal(click[1], click[0])
return board
Difficulty
and has an acceptance rate of about 69% (300k+ submissions).
Typical patterns to solve this problem:
- Array
- Breadth First Search
- Depth First Search
- Matrix
AI Disclaimer
However, two years later I set up this website and used various AI tools to generate most of the walkthrough / explanations below.
See this dedicated page for more details.
Code Walkthrough
Overview
The provided implementation solves the classic Minesweeper reveal problem. Given a two-dimensional board of characters and a click coordinate, the algorithm updates the board according to the game rules and returns the new state. The board uses 'E' for unrevealed empty cells, 'M' for unrevealed mines, 'B' for revealed blank cells (no adjacent mines), digits '1'-'8' for revealed cells with adjacent mines, and 'X' for a revealed mine that caused the game to end. The solution follows a depth-first search (DFS) pattern, recursively expanding from the clicked cell while respecting board boundaries and already revealed cells.
Key Steps
Dimension Extraction
The first two statements compute the board dimensions, storing the horizontal size in width and the vertical size in height. These values are reused throughout the algorithm to keep boundary checks concise.
width = len(board[0])
height = len(board)
Neighbor Offsets
A static list named deltas encodes the eight possible directions (horizontal, vertical, and diagonal). By iterating over this list, the code can examine all adjacent squares without duplicating logic.
deltas = [ [ -1, -1 ], [ -1, 0 ], [ -1, +1 ],
[ 0, -1 ], [ 0, +1 ],
[ +1, -1 ], [ +1, 0 ], [ +1, +1 ] ]
Counting Adjacent Mines
The helper function count receives a coordinate (x, y) and iterates over deltas. For each neighbor it verifies that the neighbor lies inside the board limits and then checks whether the neighbor contains a mine ('M') or an already exploded mine ('X'). The function returns the total number of such neighbors.
def count(x, y):
result = 0
for dx, dy in deltas:
xx = x + dx
yy = y + dy
if 0 <= xx < width and 0 <= yy < height:
if board[yy][xx] in [ "M", "X" ]:
result += 1
return result
Recursive Reveal Logic
The core of the algorithm resides in the reveal function. It first checks whether the current cell is a mine. If so, the cell is turned into 'X' and the recursion stops, reflecting the game-over condition.
if board[y][x] == "M":
board[y][x] = "X"
return
If the cell is not unrevealed ('E'), the function returns immediately, preventing unnecessary work on already processed squares.
if board[y][x] != "E":
return
When the cell is a fresh empty square, the function computes the number of adjacent mines using count. The result is stored as a string in the board cell.
board[y][x] = str(count(x, y))
If the count is zero, the cell is transformed into a blank reveal 'B'. The function then iterates over all eight directions, recursively invoking reveal on each neighbor that lies within the board. This expands the blank region outward until cells with non-zero adjacent mine counts are encountered.
if board[y][x] == "0":
board[y][x] = "B"
for dx, dy in deltas:
xx = x + dx
yy = y + dy
if 0 <= xx < width and 0 <= yy < height:
reveal(xx, yy)
Entry Point
The algorithm starts the recursion from the user-provided click coordinate. Note that the click list is ordered as [row, column], but the reveal function expects (x, y). The call swaps the indices accordingly.
reveal(click[1], click[0])
return board
Edge Cases Handled
The implementation explicitly guards against out-of-bounds access by checking 0 <= xx < width and 0 <= yy < height before any board lookup. This ensures that clicks on border cells and recursive expansions that reach the edge do not raise index errors.
When the clicked cell contains a mine, the code immediately converts it to 'X' and halts further recursion, correctly modeling the loss condition without attempting to reveal surrounding cells.
If the clicked cell is already revealed (any character other than 'E'), the function returns without modifying the board. This covers scenarios where a user clicks on a previously opened square, a situation that can arise in interactive play.
The conversion of a zero-adjacent-mine count from '0' to 'B' before recursing guarantees that the board never displays a literal '0'. This matches the conventional Minesweeper visual where blank squares are shown without a numeric label.
The count function treats both 'M' (unexploded mines) and 'X' (exploded mines) as mines when tallying adjacent threats. This subtle handling ensures that after a mine has been revealed, any neighboring blank expansion still respects the presence of that exploded mine.
Performance Optimizations
The code reuses the pre-computed width, height, and deltas values throughout the recursion, avoiding repeated length calculations or reconstruction of direction vectors. This reduces overhead in tight loops that may be invoked many times during a large blank region expansion.
By checking board[y][x] != "E" early in the reveal function, the algorithm prevents redundant work on cells that have already been processed, effectively pruning the recursion tree. This early exit eliminates unnecessary calls to count and further neighbor exploration.
The recursion only proceeds when a cell’s adjacent mine count is zero. Cells with a non-zero count are revealed once and then become leaf nodes in the recursion, preventing deeper traversal into areas that would not change the board state. This selective expansion limits the number of recursive calls to the minimal necessary set.
The count helper iterates over the fixed eight directions and performs a single boundary check per neighbor. By consolidating the mine detection into a membership test against a short list ([ "M", "X" ]), the function avoids multiple conditional branches, keeping the inner loop lightweight.
Overall, the algorithm exhibits a depth-first search pattern that visits each cell at most once, and the constant-time neighbor checks keep the per-cell work bounded, leading to efficient handling even on larger boards.
ELI5 - Simplified Explanation
Imagine a giant grid that is like a game board filled with hidden surprises. Some spots have scary mines hidden under them, which are marked with the letter m. Other spots are empty and waiting to be cleared.
The code works like a magic wand that you tap on the board. When the wand touches a square, it first checks if there is a mine hiding there. If a mine is found, it changes to an x to show the game is finished.
If the spot is safe, the wand looks at all the squares touching it to see if any mines are nearby. It counts how many mines are hiding in those neighboring spots and writes that number down on the square. If the count is zero, it marks the square with a b and automatically taps on all the squares right next to it. This chain reaction continues until it finds spots that have numbers in them, clearing a nice path across the board as if it were uncovering a mystery.
Code Quality
Complexity Analysis
As a rule of thumb, the maximum input size for one second execution time is
- factorial: up to 10 elements
- exponential: up to 20 elements
- polynomial: a few thousand elements
- loglinear: up to 100,000 elements
- linear: a few million elements
- logarithmic: a billion elements
- constant: practically unlimited
Other Solutions
Obviously, I did that after I solved the problem myself.
In general, AlgoMonster and WalkCCC solutions are high-quality and always worth a look.
I haven't checked most of the other solutions, but they might be helpful as well.
And of course, the LeetCode forum is always a good place (or even the best) to look for alternative approaches.
(click on a column header to sort the table)
| author | ||||||||||
|---|---|---|---|---|---|---|---|---|---|---|
| author | solutions | |||||||||
| www | AlgoMonster | C++ | Java | Python | TS | |||||
| WalkCCC | C++ | Java | Python | |||||||
| leetcode.ca | C++ | Java | Python | Go | TS | |||||
| LeetcodeHelp | C++ | Java | Python | |||||||
| Progiez | C++ | Java | Python | |||||||
| github | BigEggStudy | C# | ||||||||
| black-shadows | C++ | |||||||||
| charles-wangkai | Java | |||||||||
| everthis | JS | |||||||||
| gouthampradhan | Java | |||||||||
| halfrost | Go | |||||||||
| JoshCrozier | JS | |||||||||
| keineahnung2345 | C++ | |||||||||
| Krazune | C++ | |||||||||
| linouk23 | C++ | |||||||||
| lkwq007 | Python | |||||||||
| martianoff | Scala | |||||||||
| ruoru | Rust | |||||||||
| suxrobGM | C# | |||||||||
| varunu28 | Java | |||||||||
| wisdompeak | C++ | |||||||||
| zfang399 | C++ | |||||||||
| AlgoMonster | C++, Python, Java, TypeScript | |||||||||
| WalkCCC | C++, Python, Java | |||||||||
| leetcode.ca | C++, Go, Java, Python, TypeScript | |||||||||
| LeetcodeHelp | C++, Java, Python | |||||||||
| Progiez | C++, Java, Python | |||||||||
| BigEggStudy | C# | |||||||||
| black-shadows | C++, C++ | |||||||||
| charles-wangkai | Java | |||||||||
| everthis | JavaScript | |||||||||
| gouthampradhan | Java | |||||||||
| halfrost | Go | |||||||||
| JoshCrozier | JavaScript | |||||||||
| keineahnung2345 | C++ | |||||||||
| Krazune | C++ | |||||||||
| linouk23 | C++ | |||||||||
| lkwq007 | Python | |||||||||
| martianoff | Scala | |||||||||
| ruoru | Rust | |||||||||
| suxrobGM | C# | |||||||||
| varunu28 | Java | |||||||||
| wisdompeak | C++ | |||||||||
| zfang399 | C++ |
Videos
Solved On The Same Day
- problem 524: Longest Word in Dictionary through Deleting
- problem 842: Split Array into Fibonacci Sequence
- problem 984: String Without AAA or BBB
- problem 1014: Best Sightseeing Pair
- problem 1017: Convert to Base -2
Extra
- Tech Stack - my development environment and tools
- AI - how AI helped me create this website (NOT solving the problems)
- Progress - progression over time
- Quality - code quality