How to memoize/cache fun with varying arguments in JavaScript | Imp Question for top company Interw

Memoizing a function is one of trickiest question that is generally asked in premium company interview. There are some questions which you will never be to answer, if you have not seen it before. This is one such kind. I have tried explaining it very much in detail in this video. Please watch the video till the end.

[Sangeeta]

Thanks a lot Vasanth for sharing nice stuff. You're helping unseen ways. Looking forward to your next post and best of luck with it

[somyaranjandas]

You bro doing very nobel work, respect ++

[itumandal]

Great content. Please bring more interview question in depth like this. Thankyou

[AARP]

Helpful. Please add more interview contents

[yashgupta-rsro]

Thanks vasant. Again great explaination 🔥

[deepakchandani]

Thank you for making these videos on interview preparation.

[himabindubanoth]

Well explained. Thank you, this really helped.

[manishkagathi]

Hi Vasanth, This made me to think unique ways, Thanks a lot.
I have a question, what if we pass more number of arguments like add(10, 20, 30) this code wont work Right?
Can u please make a video such that it should be able to accept any number of arguments.

[ayushjain]

Plz continue creating amazing content like this, and don't forget to share it in LinkedIn 😉🎉🎉
More power to you

[sawanpatodia]

Why is there need for adding function name while generating unique id? Doesn't the cache belong to each memoized function and hence only concatenation of arguments will be fine?

[daveraj]

Hello sir,
I tried my own way to do this after watching this.

the entire code snippet is below.

// Write a function that will return cache o/p if present

// function add(a, b) - first time run 2nd time onwards from cache

function add(...args) {
sum = 0
for(let i of args){
sum = sum + i;
}
return sum;

}

function mul(...args) {
multiply = 0
for(let i of args){
multiply = multiply * i;
}
return multiply;

}

cache = {};
function memoize(fn, args){
key = fn.name + args.join("|");
if(cache[key]){
console.log("From cache");
return cache[key];
}
else{
console.log("Not from cache");
cache[key] = fn(...args);
return cache[key];
}
}

console.log(memoize(add, [121, 34, 4, 4]));
console.log(memoize(add, [1, 2, 35, 6]));
console.log(memoize(add, [1, 2, 35, 6]));
console.log(memoize(mul, [121, 34, 4, 4]));
console.log(memoize(mul, [1, 2, 35, 6]));
console.log(memoize(mul, [1, 2, 35, 6]));

// console.log(memoize(add, 10, 30));
// console.log(memoize(add, 10, 30));
// console.log(memoize(add, 10, 30));
// console.log(memoize(add, 10, 30));




Thank you for educating these types of best practices.

[ayushbisht]

Thanks for sharing, Although I was able to solve this question.
But instead of putting callback during initialization of memo function. I keep it at the calling time.
In this way, we no longer have to create another instance of the memo function for memoizing the function with its argument.

[adwaithks]

Is it some extension that you are using for sticky function names...? Can you please share the name of extension?

[gogulbharathisubbaraj]

What if order of arguments changes?

add(2, 3) === add(3, 2), but this will again calculate the results right?