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n-queens.py
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n-queens.py
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# Time: O(n!)
# Space: O(n)
#
# The n-queens puzzle is the problem of placing n queens on
# an nxn chess board such that no two queens attack each other.
#
# Given an integer n, return all distinct solutions to the n-queens puzzle.
#
# Each solution contains a distinct board configuration of the n-queens' placement,
# where 'Q' and '.' both indicate a queen and an empty space respectively.
#
# For example,
# There exist two distinct solutions to the 4-queens puzzle:
#
# [
# [".Q..", // Solution 1
# "...Q",
# "Q...",
# "..Q."],
#
# ["..Q.", // Solution 2
# "Q...",
# "...Q",
# ".Q.."]
# ]
# quick solution for checking if it is diagonally legal
class Solution:
# @return an integer
def solveNQueens(self, n):
self.cols = [False] * n
self.main_diag = [False] * (2 * n)
self.anti_diag = [False] * (2 * n)
self.solutions = []
self.solveNQueensRecu([], 0, n)
return self.solutions
def solveNQueensRecu(self, solution, row, n):
if row == n:
self.solutions.append(map(lambda x: '.' * x + "Q" + '.' * (n - x - 1), solution))
else:
for i in xrange(n):
if not self.cols[i] and not self.main_diag[row + i] and not self.anti_diag[row - i + n]:
self.cols[i] = self.main_diag[row + i] = self.anti_diag[row - i + n] = True
self.solveNQueensRecu(solution + [i], row + 1, n)
self.cols[i] = self.main_diag[row + i] = self.anti_diag[row - i + n] = False
# slower solution
class Solution2:
# @return an integer
def solveNQueens(self, n):
self.solutions = []
self.solveNQueensRecu([], 0, n)
return self.solutions
def solveNQueensRecu(self, solution, row, n):
if row == n:
self.solutions.append(map(lambda x: '.' * x + "Q" + '.' * (n - x - 1), solution))
else:
for i in xrange(n):
if i not in solution and reduce(lambda acc, j: abs(row - j) != abs(i - solution[j]) and acc, xrange(len(solution)), True):
self.solveNQueensRecu(solution + [i], row + 1, n)
if __name__ == "__main__":
print Solution().solveNQueens(8)