Max async tasks - interview practice solution (NOT for prod use)

index.ts

type AsyncFn = () => Promise<unknown>;

class AsyncWorkers {
  private queue: Array<AsyncFn> = [];
  private workersAvailable;

  constructor(workersCount: number) {
    this.workersAvailable = workersCount;
  }

  private doTask() {
    if (!this.workersAvailable || !this.queue.length) return;

    const task = this.queue.shift();
    if (!task) return;

    this.workersAvailable--;
    task().then(() => {
      this.workersAvailable++;
      this.doTask();
    });
  }

  pushTask(fn: AsyncFn) {
    this.queue.push(fn);
    if (this.workersAvailable) {
      this.doTask();
    }
  }
}

function asyncFn(v: number) {
  return function () {
    return new Promise((resolve) => {
      setTimeout(function () {
        console.log(v);
        resolve(v);
      }, v * 1000);
    });
  };
}

const WORKER_POOL_LENGTH = 2;
const w = new AsyncWorkers(WORKER_POOL_LENGTH);

for (let i = 0; i < 30; i++) {
  w.pushTask(asyncFn(1));
}

A sophisticated way to solve this can be found here https://github.com/ajitid/fzf-for-js/tree/experiments/with-web-workers/src/utils