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LeetCode 999. Available Captures for Rook
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‎README.md

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@@ -294,6 +294,7 @@ Solutions to LeetCode problems. The first column links to the problem in LeetCod
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| [987. Vertical Order Traversal of a Binary Tree][lc987] | 🔴 Hard | [![python](res/py.png)][lc987py] |
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| [990. Satisfiability of Equality Equations][lc990] | 🟠 Medium | [![python](res/py.png)][lc990py] |
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| [994. Rotting Oranges][lc994] | 🟠 Medium | [![python](res/py.png)][lc994py] |
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| [999. Available Captures for Rook][lc999] | 🟢 Easy | [![python](res/py.png)][lc999py] |
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| [1008. Construct Binary Search Tree from Preorder Traversal][lc1008] | 🟠 Medium | [![python](res/py.png)][lc1008py] |
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| [1041. Robot Bounded In Circle][lc1041] | 🟠 Medium | [![python](res/py.png)][lc1041py] |
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| [1046. Last Stone Weight][lc1046] | 🟢 Easy | [![python](res/py.png)][lc1046py] |
@@ -908,6 +909,8 @@ Solutions to LeetCode problems. The first column links to the problem in LeetCod
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[lc990py]: leetcode/satisfiability-of-equality-equations.py
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[lc994]: https://leetcode.com/problems/rotting-oranges/
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[lc994py]: leetcode/rotting-oranges.py
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[lc999]: https://leetcode.com/problems/available-captures-for-rook/
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[lc999py]: leetcode/available-captures-for-rook.py
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[lc1008]: https://leetcode.com/problems/construct-binary-search-tree-from-preorder-traversal/
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[lc1008py]: leetcode/construct-binary-search-tree-from-preorder-traversal.py
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[lc1041]: https://leetcode.com/problems/robot-bounded-in-circle/

‎leetcode/available-captures-for-rook.py

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# 999. Available Captures for Rook
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# 🟢 Easy
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#
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# https://leetcode.com/problems/available-captures-for-rook/
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#
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# Tags: Array - Matrix - Simulation
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import timeit
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from typing import List
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# Iterate over the matrix rows and columns to find the rook, then travel
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# on the four directions of movement from the rook's position until we
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# either run out of cells to visit, find a bishop, or find a pawn, if
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# the latter, add one to the result.
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#
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# Time complexity: O(1) - We visit all cells, limited to 64, constant
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# time O(64) => O(1).
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# Space complexity: O(1) - Constant space.
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#
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# Runtime: 30 ms, faster than 96.32%
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# Memory Usage: 13.9 MB, less than 81.84%
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class Solution:
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def numRookCaptures(self, board: List[List[str]]) -> int:
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# Travel through the board to find the rook.
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for i in range(len(board)):
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for j in range(len(board[0])):
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if board[i][j] == "R":
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break
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else:
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continue
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break
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res, row, col = 0, i, j
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# Up.
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i = row - 1
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while i >= 0 and board[i][col] != "B":
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if board[i][col] == "p":
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res += 1
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break
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i -= 1
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i = row + 1
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while i < len(board) and board[i][col] != "B":
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if board[i][col] == "p":
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res += 1
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break
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i += 1
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j = col - 1
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while j >= 0 and board[row][j] != "B":
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if board[row][j] == "p":
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res += 1
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break
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j -= 1
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j = col + 1
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while j < len(board[0]) and board[row][j] != "B":
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if board[row][j] == "p":
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res += 1
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break
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j += 1
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return res
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def test():
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executors = [
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Solution,
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]
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tests = [
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[
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[
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[".", ".", ".", ".", ".", ".", ".", "."],
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[".", ".", ".", "p", ".", ".", ".", "."],
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[".", ".", ".", "R", ".", ".", ".", "p"],
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[".", ".", ".", ".", ".", ".", ".", "."],
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[".", ".", ".", ".", ".", ".", ".", "."],
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[".", ".", ".", "p", ".", ".", ".", "."],
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[".", ".", ".", ".", ".", ".", ".", "."],
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[".", ".", ".", ".", ".", ".", ".", "."],
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],
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3,
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],
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[
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[
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[".", ".", ".", ".", ".", ".", ".", "."],
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[".", "p", "p", "p", "p", "p", ".", "."],
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[".", "p", "p", "B", "p", "p", ".", "."],
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[".", "p", "B", "R", "B", "p", ".", "."],
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[".", "p", "p", "B", "p", "p", ".", "."],
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[".", "p", "p", "p", "p", "p", ".", "."],
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[".", ".", ".", ".", ".", ".", ".", "."],
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[".", ".", ".", ".", ".", ".", ".", "."],
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],
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0,
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],
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[
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[
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[".", ".", ".", ".", ".", ".", ".", "."],
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[".", ".", ".", "p", ".", ".", ".", "."],
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[".", ".", ".", "p", ".", ".", ".", "."],
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["p", "p", ".", "R", ".", "p", "B", "."],
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[".", ".", ".", ".", ".", ".", ".", "."],
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[".", ".", ".", "B", ".", ".", ".", "."],
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[".", ".", ".", "p", ".", ".", ".", "."],
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[".", ".", ".", ".", ".", ".", ".", "."],
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],
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3,
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],
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]
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for executor in executors:
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start = timeit.default_timer()
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for _ in range(1):
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for col, t in enumerate(tests):
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sol = executor()
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result = sol.numRookCaptures(t[0])
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exp = t[1]
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assert result == exp, (
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f"\033[93m» {result} <> {exp}\033[91m for"
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+ f" test {col} using \033[1m{executor.__name__}"
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)
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stop = timeit.default_timer()
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used = str(round(stop - start, 5))
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cols = "{0:20}{1:10}{2:10}"
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res = cols.format(executor.__name__, used, "seconds")
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print(f"\033[92m» {res}\033[0m")
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test()

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