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;




Comments

Popular posts from this blog

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

Alternatives to Using UUIDs

  Alternatives to Using UUIDs UUIDs are valuable for several reasons: Global Uniqueness : UUIDs are designed to be globally unique across systems, ensuring that no two identifiers collide unintentionally. This property is crucial for distributed systems, databases, and scenarios where data needs to be uniquely identified regardless of location or time. Standardization : UUIDs adhere to well-defined formats (such as UUIDv4) and are widely supported by various programming languages and platforms. This consistency simplifies interoperability and data exchange. High Collision Resistance : The probability of generating duplicate UUIDs is extremely low due to the combination of timestamp, random bits, and other factors. This collision resistance is essential for avoiding data corruption. However, there are situations where UUIDs may not be the optimal choice: Length and Readability : UUIDs are lengthy (typically 36 characters in their canonical form) and may not be human-readable. In URLs,

Bellman Ford Graph Algorithm

The Shortest path algorithms so you go to google maps and you want to find the shortest path from one city to another.  Two algorithms can help you, they both calculate the shortest distance from a source node into all other nodes, one node can handle negative weights with cycles and another cannot, Dijkstra cannot and bellman ford can. One is Dijkstra if you run the Dijkstra algorithm on this map its input would be a single source node and its output would be the path to all other vertices.  However, there is a caveat if Elon mask comes and with some magic creates a black hole loop which makes one of the edges negative weight then the Dijkstra algorithm would fail to give you the answer. This is where bellman Ford algorithm comes into place, it's like the Dijkstra algorithm only it knows to handle well negative weight in edges. Dijkstra has an issue handling negative weights and cycles Bellman's ford algorithm target is to find the shortest path from a single node in a graph t