#2630

Hard

auto_awesomeMemoize II core interview pattern

LeetCode Problem Workspace

Memoize II

Implementing a memoized function ensures that identical inputs avoid redundant calls, boosting efficiency and performance.

Problem Statement

In this problem, you are tasked with creating a memoized version of a given function fn. A memoized function caches the result of its calculations for identical inputs, avoiding redundant calls. This caching mechanism ensures that if the function is called again with the same input, the cached value is returned, improving performance.

You need to ensure that inputs are considered identical if they are strictly equal (===). The memoization strategy should handle diverse types of inputs, including non-primitive data types like objects, arrays, and functions. The challenge involves ensuring cache hits are properly detected and executed efficiently.

Examples

Example 1

Input: getInputs = () => [[2,2],[2,2],[1,2]] fn = function (a, b) { return a + b; }

Output: [{"val":4,"calls":1},{"val":4,"calls":1},{"val":3,"calls":2}]

const inputs = getInputs(); const memoized = memoize(fn); for (const arr of inputs) { memoized(...arr); }

For the inputs of (2, 2): 2 + 2 = 4, and it required a call to fn(). For the inputs of (2, 2): 2 + 2 = 4, but those inputs were seen before so no call to fn() was required. For the inputs of (1, 2): 1 + 2 = 3, and it required another call to fn() for a total of 2.

Example 2

Input: getInputs = () => [[{},{}],[{},{}],[{},{}]] fn = function (a, b) { return ({...a, ...b}); }

Output: [{"val":{},"calls":1},{"val":{},"calls":2},{"val":{},"calls":3}]

Merging two empty objects will always result in an empty object. It may seem like there should only be 1 call to fn() because of cache-hits, however none of those objects are === to each other.

Example 3

Input: getInputs = () => { const o = {}; return [[o,o],[o,o],[o,o]]; } fn = function (a, b) { return ({...a, ...b}); }

Output: [{"val":{},"calls":1},{"val":{},"calls":1},{"val":{},"calls":1}]

Merging two empty objects will always result in an empty object. The 2nd and 3rd third function calls result in a cache-hit. This is because every object passed in is identical.

Constraints

1 <= inputs.length <= 105
0 <= inputs.flat().length <= 105
inputs[i][j] != NaN

Solution Approach

Memoization Using a Cache

Create a cache to store previously computed results. Each time the memoized function is called, check if the inputs are already in the cache. If found, return the cached result; otherwise, compute the result and store it in the cache.

Handling Input Uniqueness

For objects and arrays, use reference equality (===) to detect identical inputs. This ensures that inputs which are structurally the same but not the same reference are handled correctly. For non-primitive values, make sure to differentiate between identical and different inputs.

Efficient Memory Usage

To manage memory usage efficiently, avoid storing unnecessary data in the cache. Use weak references for large inputs when possible, ensuring that only relevant data is kept in memory.

Complexity Analysis

Metric	Value
Time	O(N)
Space	O(NL)

Time complexity is O(N) as we are processing each input once, with the caching mechanism providing O(1) lookup time. Space complexity is O(NL), where N is the number of function calls and L is the size of the input, since the cache needs to store results for each unique input combination.

What Interviewers Usually Probe

Look for an efficient cache lookup mechanism.
Evaluate how the candidate handles object equality and memoization with non-primitive inputs.
Check for understanding of performance optimizations and memory management in memoization.

Common Pitfalls or Variants

Common pitfalls

Failing to account for object equality with non-primitive values like objects or arrays.
Incorrect cache management leading to memory overflow or incorrect cache hits.
Assuming that JSON.stringify can reliably determine input equality for non-primitive types.

Follow-up variants

Handle inputs with large datasets efficiently.
Implement the cache with weak references to avoid memory bloat.
Extend the solution to handle asynchronous functions.

FAQ

What is the Memoize II pattern?

The Memoize II pattern involves creating a function that caches results based on identical inputs to avoid redundant computations, improving efficiency.

How do you handle objects in memoization?

Objects and arrays are considered identical only if they are strictly equal (===). To handle non-primitive inputs correctly, you need to check reference equality.

How can I optimize memory usage in Memoize II?

You can optimize memory usage by using weak references for large inputs or by limiting the cache size to store only necessary values.

Can I memoize asynchronous functions in Memoize II?

Yes, you can extend the memoization to asynchronous functions by ensuring that the cache properly stores and returns the result once the asynchronous operation completes.

What are common mistakes in solving Memoize II?

Common mistakes include improper handling of object equality, inefficient cache management, and assuming that JSON.stringify can reliably determine input equality.

terminal

Solution

Solution 1

#### TypeScript

type Fn = (...params: any) => any;

function memoize(fn: Fn): Fn {
    const idxMap: Map<string, number> = new Map();
    const cache: Map<string, any> = new Map();

    const getIdx = (obj: any): number => {
        if (!idxMap.has(obj)) {
            idxMap.set(obj, idxMap.size);
        }
        return idxMap.get(obj)!;
    };

    return function (...params: any) {
        const key = params.map(getIdx).join(',');
        if (!cache.has(key)) {
            cache.set(key, fn(...params));
        }
        return cache.get(key)!;
    };
}

/**
 * let callCount = 0;
 * const memoizedFn = memoize(function (a, b) {
 *	 callCount += 1;
 *   return a + b;
 * })
 * memoizedFn(2, 3) // 5
 * memoizedFn(2, 3) // 5
 * console.log(callCount) // 1
 */

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