2630. Memoize II
Description
Given a function fn
, return a memoized version of that function.
A memoized function is a function that will never be called twice with the same inputs. Instead it will return a cached value.
fn
can be any function and there are no constraints on what type of values it accepts. Inputs are considered identical if they are ===
to each other.
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}] Explanation: 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}] Explanation: 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}] Explanation: 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
Solutions
Solution 1
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