# Leetcode: N-Queens

N-Queens

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 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.."]
]
```

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

```## Blog link: https://code.dennyzhang.com/n-queens
## Basic Ideas: backtracking.
##              Place queens row by row
##              Check if place in current position, examine the column and triangle
##
## Complexity: Time ?, Space ?
class Solution(object):
def solveNQueens(self, n):
"""
:type n: int
:rtype: List[List[str]]
"""
if n <= 0:
return None

self.board = []
for i in xrange(n):
self.board.append(['.']*n)

self.res = []
self.mySolveNQueens(n, 0)
return self.res

def mySolveNQueens(self, n, irow):
if irow == n:
item = []
for row in self.board:
item.append(''.join(row))
self.res.append(item)
return

for icol in xrange(n):
# place Q
if self.isNQuees(n, irow, icol):
self.board[irow][icol] = 'Q'
self.mySolveNQueens(n, irow+1)
self.board[irow][icol] = '.'

def isNQuees(self, n, irow, icol):
for index in xrange(n):
# check column
if index == irow: continue
if self.board[index][icol] == 'Q': return False

for i in xrange(n):
for j in xrange(n):
if irow == i and icol == j: continue
if abs(irow-i) == abs(icol-j) and self.board[i][j] == 'Q':
return False
return True

# s = Solution()
# print s.solveNQueens(8)
```

Share It, If You Like It.