## The problem

1. Input integer n
2. Place n queens on n x n board
3. Q means queen at cell
4. . Means empty space
5. Queen can move in any direction
1. Horizontally
2. Up
3. Down
4. Diagonal neg + pos

## The Trick, let's say for 4x4

1. Understanding
1. Notice that each and every queen has to be in a different row!
2. Notice that each and every queen has to be in a different column!
3. Notice that each and every queen has to be in a different positive diagonal!
4. Notice that each and every queen has to be in a different negative diagonal!
2. State - Set - is queen in column, posDiag (row-col), negDiag (row+col)
1. Which rows have queen - do not have to store in state we just loop the rows
2. Which columns have queen - store in state!
3. Which posDiag have queen - store in state!
4. Which negDiag have queen - store in state!
3. Trick
1. posDiag -> row - column = constant
1. Every time we increase row we increase column
2. negDiag -> row + column = constant
1. Every time we increase row we decrease column
4. Loop
1. 1st queen 1st row
1. Can put first queen in first column
2. Can put first queen in second column
3. Can put first queen in third column
4. Can put first queen in 4th column

class Solution(object): def solveNQueens(self, n): """ :type n: int :rtype: List[List[str]] """ cols = set() # Cannot put Q in this col. diag1 = set() # Cannot put Q in this diagonal. diag2 = set() # Cannot put Q in this diagonal. # We do not need to store cannot put Q in row because we just loop to next row. res = [] board = [['.'] * n for i in range(n)] # For each row try putting a queen start with row 0 def backtrack(r): # Do we have a solution? if r == n: copy = ["".join(row) for row in board] res.append(copy) # Try put Q in each col for c in range(n): if c in cols or (r + c) in diag1 or (r-c) in diag2: # Cannot put Q in this col continue to next col. continue # Put Q in this col mark following rows cannot use this col, diag1, diag2 cols.add(c) diag1.add(r+c) diag2.add(r-c) board[r][c] = 'Q' # We have placed our Q continue to next row. backtrack(r + 1) # We get here in two flows # 1. Finished backtrack(r+1) successfully we have a solution, now try the next column as a new solution. # 2. Finished backtrack(r+1) unsuccessfully we don't have a solution try next column. cols.remove(c) diag1.remove(r+c) diag2.remove(r-c) board[r][c] = '.' backtrack(0) return res;

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