L3 · FAANGBacktracking~20 min · 6 tests

N-Queens: Count All Solutions

Count the ways to place n queens on an n×n board with none attacking, using backtracking with column and diagonal sets for O(1) conflict checks.

The problem

Return how many ways n queens can be placed on an n × n chessboard so that no two attack each other (same row, column or diagonal).

Examples

Example 1
Input
```
n_queens(4)
```
Expected output
```
2
```
Example 2
Input
```
n_queens(1)
```
Expected output
```
1
```

+ 4 hidden tests on Submit.

Edge cases to ask about

n = 1
n = 2 and 3 (no solution)

Hints

0/3

How an interviewer scores this

0/9

1. Understand the problemRestate it, ask about edge cases and input size.
2. Write a brute-force solutionCorrect first, clever later.
3. Get the correct outputautoAll visible examples pass.
4. Analyse time complexityautoCommit to a Big-O in the Complexity Lab.
5. Analyse space complexityautoExtra memory, not the input.
6. Identify the bottleneckautoMeasure your code — where does it bend?
7. OptimiseUsually: a set or dict instead of a scan.
8. Explain why it worksSay it out loud to Solvi.
9. Test the edge casesautoSubmit passes every hidden test.

Python 3.13 · n_queens

def n_queens(n):
    """Number of non-attacking placements of n queens."""
    # Write your solution here
    pass

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Complexity Lab

What does this cost as n grows?

Interviewers score the analysis as much as the code. Commit to an answer first — then check it, and read why.

Time complexity of the optimal solution

Space complexity (extra memory)

Pick both to reveal the answer.

From brute force to optimal

The progression an interviewer wants to hear, one step at a time.

Approach	Time	Space	Idea
Try every placement of n queens	O(n²ⁿ)	O(n)
One queen per row + conflict sets	O(n!)	O(n)	row − col and row + col identify the two diagonals.
bestBitmasks	O(n!)	O(n)	Same search, much smaller constant.

Walkthrough of the optimal approach (try it yourself first)

Place one queen per row. Track the used columns and the two diagonal families: cells on the same "" diagonal share row − col, on the same "/" diagonal share row + col. Each conflict check is then an O(1) set lookup.

The skeleton — choose, recurse, un-choose — is the template for every backtracking problem. Pruning early (skipping attacked columns) is what keeps it tractable.

Complexity: O(n!) time, O(n) space. Row r has at most n − r safe columns left, so the search tree is bounded by n!; the sets and recursion hold O(n).

Reveal the reference solution

def n_queens(n):
    cols, diag, anti = set(), set(), set()

    def place(row):
        if row == n:
            return 1
        count = 0
        for c in range(n):
            if c in cols or row - c in diag or row + c in anti:
                continue
            cols.add(c); diag.add(row - c); anti.add(row + c)
            count += place(row + 1)
            cols.remove(c); diag.remove(row - c); anti.remove(row + c)
        return count

    return place(0)

Follow-ups interviewers ask

Return the boards themselves.
Rewrite with bitmasks.

Frequently asked interview questions

Core interview concepts, complexities, and follow-ups scored by hiring teams.

What is the time complexity of N-Queens: Count All Solutions in Python?

The optimal solution runs in O(n!) time and O(n) auxiliary space. Row r has at most n − r safe columns left, so the search tree is bounded by n!; the sets and recursion hold O(n).

What is the brute-force approach, and how do you optimise it?

Try every placement of n queens: O(n²ⁿ) time, O(n) space. One queen per row + conflict sets: O(n!) time, O(n) space. row − col and row + col identify the two diagonals. Bitmasks: O(n!) time, O(n) space. Same search, much smaller constant.

What follow-up questions do interviewers ask about N-Queens: Count All Solutions?

Return the boards themselves. Rewrite with bitmasks.

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