The n-queens puzzle is the problem of placing n queens on an n x n chessboard such that no two queens attack each other.
Given an integer n, return the number of distinct solutions to the n-queens puzzle.
https://leetcode.com/problems/n-queens-ii/
My solution:
class Solution:
def totalNQueens(self, n: int) -> int:
def genRestricted(restricted, r, c):
restricted = set(restricted)
for row in range(n): restricted.add((row, c))
for col in range(n): restricted.add((r, col))
movements = [[-1, -1], [-1, 1], [1, -1], [1, 1]]
for movement in movements:
row, col = r, c
while 0 <= row < n and 0 <= col < n:
restricted.add((row, col))
row += movement[0]
col += movement[1]
return restricted
def gen(row, col, curCount, restricted):
count, total_count = curCount, 0
for r in range(row, n):
for c in range(col, n):
if (r, c) not in restricted:
count += 1
if count == n: total_count += 1
total_count += gen(row + 1, 0, count, genRestricted(restricted, r, c))
count -= 1
return total_count
return gen(0, 0, 0, set())
It fails at n=8. I can't figure out why, and how to have less iterations. It seems I am already doing the minimum iterations possible.
The restricted
set seems wasteful, both time- and space-wise. At the end of the successful recursion, n
levels deep it grows to n^2
size, which drives the total complexity to O(n^3)
. And it is not really needed. It is much easier to check the availability of the square by looking at the queens already placed (please forgive the chess lingo; file
stand for vertical, and rank
for horizontal):
def square_is_safe(file, rank, queens_placed):
for queen_rank, queen_file in enumerate(queens_placed):
if queen_file == file: # vertical attack
return false
if queen_file - file == queen_rank - rank: # diagonal attack
return false
if queen_file - file == rank - queen_rank: # anti-diagonal attack
return false
return true
to be used in
def place_queen_at_rank(queens_placed, rank):
if rank == n:
total_count += 1
return
for file in range(0, n):
if square_is_safe(file, rank, queens_placed):
queens_placed.append(file)
place_queen_at_rank(queens_placed, rank + 1)
queens_placed.pop()
And there is a plenty of room for the optimization. For example, you may want to special-case the first rank: due to a symmetry, you only need to inspect a half of it (cutting execution time by the factor of 2).
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