L2 · Working engineerRecursion~8 min · 5 tests#49

Fibonacci With Memoization

Turn exponential O(2ⁿ) recursive Fibonacci into O(n) with memoization using a dict or functools.lru_cache. See the difference live in Python.

The problem

Return the n-th Fibonacci number (fib(0) = 0, fib(1) = 1) using recursion with memoization. It must handle n = 300 instantly.

Examples

Example 1
Input
```
fib(10)
```
Expected output
```
55
```
Example 2
Input
```
fib(0)
```
Expected output
```
0
```

+ 3 hidden tests on Submit — must be memoized.

Edge cases to ask about

n = 0 and 1
Large n (300)

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 · fib

def fib(n, memo=None):
    """n-th Fibonacci number, recursive + memoized."""
    # Write your solution here
    pass

⌘/Ctrl + Enter runs the examples

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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
Plain recursion	O(2ⁿ)	O(n)	fib(n−2) is recomputed inside fib(n−1) over and over.
Memoized recursion	O(n)	O(n)	Each fib(k) is computed once.
bestTwo variables, bottom-up	O(n)	O(1)	a, b = b, a + b.

Walkthrough of the optimal approach (try it yourself first)

Plain recursion recomputes the same subproblems exponentially many times (O(2ⁿ)). A memo dict stores fib(k) the first time; every later request is O(1), so total work becomes O(n).

Note the memo=None default: a mutable default memo={} would work here but is the trap from question 66. @functools.lru_cache(maxsize=None) is the idiomatic version.

Complexity: O(n) time, O(n) space. With the memo every fib(k) for k ≤ n is computed exactly once; the memo and the call stack each hold O(n).

Reveal the reference solution

def fib(n, memo=None):
    if memo is None:
        memo = {}
    if n < 2:
        return n
    if n not in memo:
        memo[n] = fib(n - 1, memo) + fib(n - 2, memo)
    return memo[n]

Follow-ups interviewers ask

Bottom-up with O(1) space.
O(log n) with matrix exponentiation.

Frequently asked interview questions

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

What is the time complexity of Fibonacci With Memoization in Python?

The optimal solution runs in O(n) time and O(n) auxiliary space. With the memo every fib(k) for k ≤ n is computed exactly once; the memo and the call stack each hold O(n).

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

Plain recursion: O(2ⁿ) time, O(n) space. fib(n−2) is recomputed inside fib(n−1) over and over. Memoized recursion: O(n) time, O(n) space. Each fib(k) is computed once. Two variables, bottom-up: O(n) time, O(1) space. a, b = b, a + b.

What follow-up questions do interviewers ask about Fibonacci With Memoization?

Bottom-up with O(1) space. O(log n) with matrix exponentiation.