How do you solve Memoize II on LeetCode?

Memoize II (LeetCode #2630) 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: Memoization can significantly reduce the number of function calls by caching results.

What is the brute force solution for Memoize II?

The brute force approach for Memoize II is Brute Force, with O(n²) time complexity and O(n) space complexity. In the brute force approach, we will call the original function every time with the given inputs. We will keep track of previously seen inputs and their results in a list. If we encounter the same inputs again, we will return the cached result instead of calling the function again. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Memoize II?

Always clarify how you will handle different types of inputs. Discuss your approach to caching and how you will ensure inputs are unique. Be prepared to explain the trade-offs between time and space complexity.

What are common mistakes when solving Memoize II?

Not using a proper key for caching, leading to incorrect cache hits. Confusing deep equality with shallow equality when checking for previously seen inputs.

#2630

Memoize II

Hard

Hash MapMemoization

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

In the optimal solution, we will use a HashMap to store results based on the input arguments. This allows for constant time retrieval of cached results, significantly improving performance compared to the brute force approach.

⚙️

Algorithm

5 steps

1Step 1: Initialize a HashMap to store the results of function calls.
2Step 2: For each input, create a unique key based on the input arguments.
3Step 3: Check if the key exists in the HashMap.
4Step 4: If it exists, return the cached result and increment the call counter.
5Step 5: If it does not exist, call the original function, store the result in the HashMap, and increment the call counter.

solution.py13 lines

1def memoize(fn):
2    cache = {}
3    calls = 0
4    def memoized(*args):
5        nonlocal calls
6        key = str(args)
7        if key in cache:
8            return {'val': cache[key], 'calls': calls}
9        result = fn(*args)
10        cache[key] = result
11        calls += 1
12        return {'val': result, 'calls': calls}
13    return memoized

ℹ

Complexity note: The time complexity is O(n) because each input is processed in constant time due to the HashMap, resulting in linear performance. The space complexity remains O(n) for storing results.

1Memoization can significantly reduce the number of function calls by caching results.
2Identical inputs are determined by strict equality (===) in JavaScript.

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