How do you solve Memoize on LeetCode?

Memoize (LeetCode #2623) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Caching results can drastically improve performance.

What is the brute force solution for Memoize?

The brute force approach for Memoize is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves calling the original function every time with the provided arguments, regardless of whether those arguments have been used before. This is straightforward but inefficient as it recalculates results for the same inputs. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Memoize?

Explain your thought process clearly. Discuss edge cases, such as large inputs or invalid arguments. Practice coding the solution without looking at references.

What are common mistakes when solving Memoize?

Not handling different argument orders correctly. Failing to use a cache effectively.

#2623

Memoize

Medium

Hash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses a cache to store results of function calls based on their arguments, ensuring that each unique set of arguments only triggers the function once. This significantly reduces redundant calculations.

⚙️

Algorithm

4 steps

1Step 1: Initialize a cache object to store results.
2Step 2: Create a counter to track the number of function calls.
3Step 3: Define the memoized function that checks the cache before calling the original function.
4Step 4: Return the cached result if it exists; otherwise, compute, cache, and return the result.

solution.py12 lines

1def memoize(fn):
2    cache = {}
3    call_count = 0
4    def memoized_fn(*args):
5        nonlocal call_count
6        key = str(args)
7        if key not in cache:
8            cache[key] = fn(*args)
9            call_count += 1
10        return cache[key]
11    memoized_fn.getCallCount = lambda: call_count
12    return memoized_fn

ℹ

Complexity note: The time complexity is linear because each unique input is processed once, and results are cached, preventing redundant calculations.

1Caching results can drastically improve performance.
2Unique inputs require separate cache entries.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.