Leetcode #2056: Number of Valid Move Combinations On Chessboard
In this guide, we solve Leetcode #2056 Number of Valid Move Combinations On Chessboard in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
There is an 8 x 8 chessboard containing n pieces (rooks, queens, or bishops). You are given a string array pieces of length n, where pieces[i] describes the type (rook, queen, or bishop) of the ith piece.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Array, String, Backtracking, Simulation
Intuition
We must explore combinations of choices, but many branches can be pruned early.
Backtracking enumerates valid candidates while keeping the search space under control.
Approach
Use DFS to build candidates step by step, and backtrack when constraints are violated.
Pruning keeps the exploration practical for typical constraints.
Steps:
- Define the decision tree.
- DFS through choices and backtrack.
- Prune invalid paths early.
Example
Input: pieces = ["rook"], positions = [[1,1]]
Output: 15
Explanation: The image above shows the possible squares the piece can move to.
Python Solution
rook_dirs = [(1, 0), (-1, 0), (0, 1), (0, -1)]
bishop_dirs = [(1, 1), (1, -1), (-1, 1), (-1, -1)]
queue_dirs = rook_dirs + bishop_dirs
def get_dirs(piece: str) -> List[Tuple[int, int]]:
match piece[0]:
case "r":
return rook_dirs
case "b":
return bishop_dirs
case _:
return queue_dirs
class Solution:
def countCombinations(self, pieces: List[str], positions: List[List[int]]) -> int:
def check_stop(i: int, x: int, y: int, t: int) -> bool:
return all(dist[j][x][y] < t for j in range(i))
def check_pass(i: int, x: int, y: int, t: int) -> bool:
for j in range(i):
if dist[j][x][y] == t:
return False
if end[j][0] == x and end[j][1] == y and end[j][2] <= t:
return False
return True
def dfs(i: int) -> None:
if i >= n:
nonlocal ans
ans += 1
return
x, y = positions[i]
dist[i][:] = [[-1] * m for _ in range(m)]
dist[i][x][y] = 0
end[i] = (x, y, 0)
if check_stop(i, x, y, 0):
dfs(i + 1)
dirs = get_dirs(pieces[i])
for dx, dy in dirs:
dist[i][:] = [[-1] * m for _ in range(m)]
dist[i][x][y] = 0
nx, ny, nt = x + dx, y + dy, 1
while 1 <= nx < m and 1 <= ny < m and check_pass(i, nx, ny, nt):
dist[i][nx][ny] = nt
end[i] = (nx, ny, nt)
if check_stop(i, nx, ny, nt):
dfs(i + 1)
nx += dx
ny += dy
nt += 1
n = len(pieces)
m = 9
dist = [[[-1] * m for _ in range(m)] for _ in range(n)]
end = [(0, 0, 0) for _ in range(n)]
ans = 0
dfs(0)
return ans
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.