Skip to main content

LeetCode 51 N Queens

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;


Popular posts from this blog

Dev OnCall Patterns

Introduction Being On-Call is not easy. So does writing software. Being On-Call is not just a magic solution, anyone who has been On-Call can tell you that, it's a stressful, you could be woken up at the middle of the night, and be undress stress, there are way's to mitigate that. White having software developers as On-Calls has its benefits, in order to preserve the benefits you should take special measurements in order to mitigate the stress and lack of sleep missing work-life balance that comes along with it. Many software developers can tell you that even if they were not being contacted the thought of being available 24/7 had its toll on them. But on the contrary a software developer who is an On-Call's gains many insights into troubleshooting, responsibility and deeper understanding of the code that he and his peers wrote. Being an On-Call all has become a natural part of software development. Please note I do not call software development software engineering b

SQL Window functions (OVER, PARTITION_BY, ...)

Introduction When you run an SQL Query you select rows, but what if you want to have a summary per multiple rows, for example you want to get the top basketball for each country, in this case we don't only group by country, but we want also to get the top player for each of the country.  This means we want to group by country and then select the first player.  In standard SQL we do this with joining with same table, but we could also use partition by and windowing functions. For each row the window function is computed across the rows that fall into the same partition as the current row.  Window functions are permitted only in the  SELECT  list and the  ORDER BY  clause of the query They are forbidden elsewhere, such as in  GROUP BY ,  HAVING  and  WHERE  clauses. This is because they logically execute after the processing of those clauses Over, Partition By So in order to do a window we need this input: - How do we want to group the data which windows do we want to have? so  def c

Functional Programming in Scala for Working Class OOP Java Programmers - Part 1

Introduction Have you ever been to a scala conf and told yourself "I have no idea what this guy talks about?" did you look nervously around and see all people smiling saying "yeah that's obvious " only to get you even more nervous? . If so this post is for you, otherwise just skip it, you already know fp in scala ;) This post is optimistic, although I'm going to say functional programming in scala is not easy, our target is to understand it, so bare with me. Let's face the truth functional programmin in scala is difficult if is difficult if you are just another working class programmer coming mainly from java background. If you came from haskell background then hell it's easy. If you come from heavy math background then hell yes it's easy. But if you are a standard working class java backend engineer with previous OOP design background then hell yeah it's difficult. Scala and Design Patterns An interesting point of view on scala, is