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N Queens
- 🔗 Leetcode Link: https://leetcode.com/problems/n-queens/
- 💡 Difficulty: Hard
- ⏰ Time to complete: 20 mins
- 🛠️ Topics: Backtracking
- 🗒️ Similar Questions: N-Queens II
Understand what the interviewer is asking for by using test cases and questions about the problem.
- Established a set (2-3) of test cases to verify their own solution later.
- Established a set (1-2) of edge cases to verify their solution handles complexities.
- Have fully understood the problem and have no clarifying questions.
- Have you verified any Time/Space Constraints for this problem?
- Where can a queen move?
- A queen can attack horizontally, vertically, or diagonally. A queen can move any number of steps in any direction. The only constraint is that it can’t change its direction while it’s moving.
- Can a queen be in the same row or column?
- No two queens can be in the same row or column. That allows us to place only one queen in each row and each column.
- How can we check if we've reached the end?
- If row == N, we've filled in all our rows successfully which implies the current board state is a valid combination
Run through a set of example cases:
Example 1:
Input: n = 4
Output: [[".Q..","...Q","Q...","..Q."],["..Q.","Q...","...Q",".Q.."]]
Example 2:
Input: n = 1
Output: [["Q"]]Match what this problem looks like to known categories of problems, e.g. Linked List or Dynamic Programming, and strategies or patterns in those categories.
- How can N queens be placed on an NxN chessboard so that no two of them attack each other? Backtracking can be used since we start with one possible move out of many available moves. The three general steps include: the general steps of backtracking are: start with a sub-solution check if this sub-solution will lead to the solution or not. If not, then come back and change the sub-solution and continue again.
The way we try to solve this is by placing a queen at a position and trying to rule out the possibility of it being under attack. We place one queen in each row/column. If we see that the queen is under attack at its chosen position, we try the next position. If a queen is under attack at all the positions in a row, we backtrack and change the position of the queen placed prior to the current position. We repeat this process of placing a queen and backtracking until all the N queens are placed successfully.
Plan the solution with appropriate visualizations and pseudocode.
General Idea: Try a promising Queen position, see how it goes. If it fails, undo that Queen and try again somewhere else.
Check if we've reached the end:
If row == N, we've filled in all our rows successfully which implies the current board state is a valid combination. Let's add it to our output list.
Loop through each column in the current row.
3. If we can't add a Queen at this position, skip this col value.
If we can, add a Queen at this position and adjust our bitmasks respectively.
Continue to the next row (call backtrack for row+1).
Undo our changes so we can try other col values.-
What are some common pitfalls students might have when implementing this solution?
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For an 8×8 board, we have to choose positions for 8 identical queens from 64 different squares, which can be done in 64C8 = 4426165368 ways. You can tighten this up by enforcing one queen in each row through the use of 8 nested loop. Using such nested loops, we would only have to look at 44=256 configurations for a 4×4 board and 88 configurations for an 8×8 board --- that's still far too many.
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Pay attention to the last inserted value. It can conflict with previous rows if the column difference is 0 (the Queens are on the same column line) or if column difference = row difference (Queens are on one diagonal)
Implement the code to solve the algorithm.
class Solution:
def solveNQueens(self, n):
# Making use of a helper function to get the
# solutions in the correct output format
def create_board(state):
board = []
for row in state:
board.append(".join(row))
return board
def backtrack(row, diagonals, anti_diagonals, cols, state):
# Base case - N queens have been placed
if row == n:
ans.append(create_board(state))
return
for col in range(n):
curr_diagonal = row - col
curr_anti_diagonal = row + col
# If the queen is not placeable
if (col in cols
or curr_diagonal in diagonals
or curr_anti_diagonal in anti_diagonals):
continue
# "Add" the queen to the board
cols.add(col)
diagonals.add(curr_diagonal)
anti_diagonals.add(curr_anti_diagonal)
state[row][col] = "Q"
# Move on to the next row with the updated board state
backtrack(row + 1, diagonals, anti_diagonals, cols, state)
# "Remove" the queen from the board since we have already
# explored all valid paths using the above function call
cols.remove(col)
diagonals.remove(curr_diagonal)
anti_diagonals.remove(curr_anti_diagonal)
state[row][col] = "."
ans = []
empty_board = [["."] * n for _ in range(n)]
backtrack(0, set(), set(), set(), empty_board)
return ansclass Solution {
private int size;
private List<List<String>> solutions = new ArrayList<List<String>>();
public List<List<String>> solveNQueens(int n) {
size = n;
char emptyBoard[][] = new char[size][size];
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
emptyBoard[i][j] = '.';
}
}
backtrack(0, new HashSet<>(), new HashSet<>(), new HashSet<>(), emptyBoard);
return solutions;
}
// Making use of a helper function to get the
// solutions in the correct output format
private List<String> createBoard(char[][] state) {
List<String> board = new ArrayList<String>();
for (int row = 0; row < size; row++) {
String current_row = new String(state[row]);
board.add(current_row);
}
return board;
}
private void backtrack(int row, Set<Integer> diagonals, Set<Integer> antiDiagonals, Set<Integer> cols, char[][] state) {
// Base case - N queens have been placed
if (row == size) {
solutions.add(createBoard(state));
return;
}
for (int col = 0; col < size; col++) {
int currDiagonal = row - col;
int currAntiDiagonal = row + col;
// If the queen is not placeable
if (cols.contains(col) || diagonals.contains(currDiagonal) || antiDiagonals.contains(currAntiDiagonal)) {
continue;
}
// "Add" the queen to the board
cols.add(col);
diagonals.add(currDiagonal);
antiDiagonals.add(currAntiDiagonal);
state[row][col] = 'Q';
// Move on to the next row with the updated board state
backtrack(row + 1, diagonals, antiDiagonals, cols, state);
// "Remove" the queen from the board since we have already
// explored all valid paths using the above function call
cols.remove(col);
diagonals.remove(currDiagonal);
antiDiagonals.remove(currAntiDiagonal);
state[row][col] = '.';
}
}
}Review the code by running specific example(s) and recording values (watchlist) of your code's variables along the way.
- Trace through your code with an input to check for the expected output
- Catch possible edge cases and off-by-one errors
Evaluate the performance of your algorithm and state any strong/weak or future potential work.
Given n as the number of queens (which is the same as the width and height of the board),
- Time Complexity: O(n!), because we are selecting if we can put or not put a Queen at that place
- Space Complexity: O(n^2), extra memory used includes the 3 sets used to store board state, as well as the recursion call stack