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2636. Promise Pool πŸ”’

Description

Given an arrayΒ of asynchronous functionsΒ functionsΒ and a pool limitΒ n, return an asynchronous functionΒ promisePool. It should returnΒ a promise that resolves when all the inputΒ functions resolve.

Pool limit is defined as the maximum number promises that can be pending at once.Β promisePoolΒ should begin execution of as many functions as possible and continue executing new functions when old promisesΒ resolve.Β promisePoolΒ should execute functions[i]Β then functions[i + 1]Β then functions[i + 2], etc. When the last promise resolves,Β promisePoolΒ should also resolve.

For example, ifΒ n = 1, promisePoolΒ will execute one function atΒ a time inΒ series. However, ifΒ n = 2, it first executes two functions. When either of the two functions resolve, a 3rd function should be executed (if available), and so on until there are no functions left to execute.

You can assume allΒ functionsΒ never reject. It is acceptable forΒ promisePoolΒ to return a promise that resolves any value.

Β 

Example 1:

Input: 
functions = [
Β  () => new Promise(res => setTimeout(res, 300)),
Β  () => new Promise(res => setTimeout(res, 400)),
Β  () => new Promise(res => setTimeout(res, 200))
]
n = 2
Output: [[300,400,500],500]
Explanation:
Three functions are passed in. They sleep for 300ms, 400ms, and 200ms respectively.
They resolve at 300ms, 400ms, and 500ms respectively. The returned promise resolves at 500ms.
At t=0, the first 2 functions are executed. The pool size limit of 2 is reached.
At t=300, the 1st function resolves, and the 3rd function is executed. Pool size is 2.
At t=400, the 2nd function resolves. There is nothing left to execute. Pool size is 1.
At t=500, the 3rd function resolves. Pool size is zero so the returned promise also resolves.

Example 2:

Input:
functions = [
Β  () => new Promise(res => setTimeout(res, 300)),
Β  () => new Promise(res => setTimeout(res, 400)),
Β  () => new Promise(res => setTimeout(res, 200))
]
n = 5
Output: [[300,400,200],400]
Explanation:
The three input promises resolve at 300ms, 400ms, and 200ms respectively.
The returned promise resolves at 400ms.
At t=0, all 3 functions are executed. The pool limit of 5 is never met.
At t=200, the 3rd function resolves. Pool size is 2.
At t=300, the 1st function resolved. Pool size is 1.
At t=400, the 2nd function resolves. Pool size is 0, so the returned promise also resolves.

Example 3:

Input:
functions = [
Β  () => new Promise(res => setTimeout(res, 300)),
Β  () => new Promise(res => setTimeout(res, 400)),
Β  () => new Promise(res => setTimeout(res, 200))
]
n = 1
Output: [[300,700,900],900]
Explanation:
The three input promises resolve at 300ms, 700ms, and 900ms respectively.
The returned promise resolves at 900ms.
At t=0, the 1st function is executed. Pool size is 1.
At t=300, the 1st function resolves and the 2nd function is executed. Pool size is 1.
At t=700, the 2nd function resolves and the 3rd function is executed. Pool size is 1.
At t=900, the 3rd function resolves. Pool size is 0 so the returned promise resolves.

Β 

Constraints:

  • 0 <= functions.length <= 10
  • 1 <= n <= 10

Solutions

Solution 1

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type F = () => Promise<any>;

function promisePool(functions: F[], n: number): Promise<any> {
    const wrappers = functions.map(fn => async () => {
        await fn();
        const nxt = waiting.shift();
        nxt && (await nxt());
    });

    const running = wrappers.slice(0, n);
    const waiting = wrappers.slice(n);
    return Promise.all(running.map(fn => fn()));
}

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