2636. Promise Pool - Leetcode JavaScript Solution with Explanation

In this tutorial, you will find a JavaScript solution that tackles the specific problem "2636. Promise Pool" presented by Leetcode. The step-by-step explanation guides you through the thought process, illustrating how the solution is derived. By following along, you'll gain a deeper understanding of the problem-solving techniques and algorithms employed.

But first, what exactly is a Promise Pool? Essentially, a Promise Pool is a collection of Promises that are executed simultaneously, allowing for faster processing of asynchronous tasks. By leveraging Promise Pools, you can drastically improve the speed and scalability of your applications.

In this video, we'll be walking through the implementation of a Promise Pool using JavaScript, and exploring its use in a Leetcode problem "2636. Promise Pool". By the end of this video, you'll have a solid understanding of how to use Promise Pools in your own projects, and be well on your way to mastering asynchronous JavaScript programming.

This resource is designed to be accessible to programmers of varying skill levels. Whether you're a beginner seeking to enhance your problem-solving abilities or an experienced developer looking to brush up on your JavaScript skills, this tutorial offers valuable insights and practical knowledge.

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📌 Timestamps:
[00:00] Introduction to the Leetcode JavaScript 30 Day Challenge Day 13
[00:05] Problem explanation
[01:40] Understanding example
[03:10] Implementing the solution step by step
[06:47] Testing the solution with different test cases
[06:57] Dry Run and explaning code
[10:45] Closing Notes

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[lokeshgoyal]

You have the rough idea of approach you are explaning! So Let me point out the general angle of this approach, if anyone is interested!
The approach is to make a concurrent call for n functions, and then whenever any one of the concurrent call is resolved, make it call the next function in queue.

for example :
// Lets take only milliseconds to demonstrate this approach general idea
input = [300, 400, 200, 500]

make 2 concurrent call, which is fn(300), fn(400) !
The moment fn(300) completes, it will make a call to fn(200) which represent the next function in queue.
The moment fn(400) completes, it will make a call to fn(500) which represents the next function in queue.
(Note: Keep track of next function in the queue using an extra variable named index, which represents the index we are currently executing)

At this point, we don't have any more function left to execute, so if any above function resolves now, it will encounter that the index reaches to an end.
So simply wait for all the functions that are in execution to settele down and then return.

This is a sample approach when n=2!

My Imaginative solution for n=2:
Just imagine a pool where only 2 person can swim at a time, so when anyone leaves the pool, will have a responsibility of sending the next person waiting in the queue to the pool!
The pool will be closed the moment everyone leaves and there are no more person waiting in the queue!