LeetCode #2721 — MEDIUM

Execute Asynchronous Functions in Parallel

Move from brute-force thinking to an efficient approach using core interview patterns strategy.

The Problem

Problem Statement

Given an array of asynchronous functions functions, return a new promise promise. Each function in the array accepts no arguments and returns a promise. All the promises should be executed in parallel.

promise resolves:

When all the promises returned from functions were resolved successfully in parallel. The resolved value of promise should be an array of all the resolved values of promises in the same order as they were in the functions. The promise should resolve when all the asynchronous functions in the array have completed execution in parallel.

promise rejects:

When any of the promises returned from functions were rejected. promise should also reject with the reason of the first rejection.

Please solve it without using the built-in Promise.all function.

Example 1:

Input: functions = [
  () => new Promise(resolve => setTimeout(() => resolve(5), 200))
]
Output: {"t": 200, "resolved": [5]}
Explanation: 
promiseAll(functions).then(console.log); // [5]

The single function was resolved at 200ms with a value of 5.

Example 2:

Input: functions = [
    () => new Promise(resolve => setTimeout(() => resolve(1), 200)), 
    () => new Promise((resolve, reject) => setTimeout(() => reject("Error"), 100))
]
Output: {"t": 100, "rejected": "Error"}
Explanation: Since one of the promises rejected, the returned promise also rejected with the same error at the same time.

Example 3:

Input: functions = [
    () => new Promise(resolve => setTimeout(() => resolve(4), 50)), 
    () => new Promise(resolve => setTimeout(() => resolve(10), 150)), 
    () => new Promise(resolve => setTimeout(() => resolve(16), 100))
]
Output: {"t": 150, "resolved": [4, 10, 16]}
Explanation: All the promises resolved with a value. The returned promise resolved when the last promise resolved.

Constraints:

functions is an array of functions that returns promises
1 <= functions.length <= 10

Step 01

Brute Force Baseline

Problem summary: Given an array of asynchronous functions functions, return a new promise promise. Each function in the array accepts no arguments and returns a promise. All the promises should be executed in parallel. promise resolves: When all the promises returned from functions were resolved successfully in parallel. The resolved value of promise should be an array of all the resolved values of promises in the same order as they were in the functions. The promise should resolve when all the asynchronous functions in the array have completed execution in parallel. promise rejects: When any of the promises returned from functions were rejected. promise should also reject with the reason of the first rejection. Please solve it without using the built-in Promise.all function.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: General problem-solving

Example 1

[() => new Promise(resolve => setTimeout(() => resolve(5), 200))]

Example 2

[() => new Promise(resolve => setTimeout(() => resolve(1), 200)), () => new Promise((resolve, reject) => setTimeout(() => reject("Error"), 100))]

Example 3

[() => new Promise(resolve => setTimeout(() => resolve(4), 50)), () => new Promise(resolve => setTimeout(() => resolve(10), 150)), () => new Promise(resolve => setTimeout(() => resolve(16), 100))]

Step 02

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2721: Execute Asynchronous Functions in Parallel
// Auto-generated Java example from ts.
class Solution {
    public void exampleSolution() {
    }
}
// Reference (ts):
// // Accepted solution for LeetCode #2721: Execute Asynchronous Functions in Parallel
// async function promiseAll<T>(functions: (() => Promise<T>)[]): Promise<T[]> {
//     return new Promise<T[]>((resolve, reject) => {
//         let cnt = 0;
//         const ans = new Array(functions.length);
//         for (let i = 0; i < functions.length; ++i) {
//             const f = functions[i];
//             f()
//                 .then(res => {
//                     ans[i] = res;
//                     cnt++;
//                     if (cnt === functions.length) {
//                         resolve(ans);
//                     }
//                 })
//                 .catch(err => {
//                     reject(err);
//                 });
//         }
//     });
// }
// 
// /**
//  * const promise = promiseAll([() => new Promise(res => res(42))])
//  * promise.then(console.log); // [42]
//  */

// Accepted solution for LeetCode #2721: Execute Asynchronous Functions in Parallel
// Auto-generated Go example from ts.
func exampleSolution() {
}
// Reference (ts):
// // Accepted solution for LeetCode #2721: Execute Asynchronous Functions in Parallel
// async function promiseAll<T>(functions: (() => Promise<T>)[]): Promise<T[]> {
//     return new Promise<T[]>((resolve, reject) => {
//         let cnt = 0;
//         const ans = new Array(functions.length);
//         for (let i = 0; i < functions.length; ++i) {
//             const f = functions[i];
//             f()
//                 .then(res => {
//                     ans[i] = res;
//                     cnt++;
//                     if (cnt === functions.length) {
//                         resolve(ans);
//                     }
//                 })
//                 .catch(err => {
//                     reject(err);
//                 });
//         }
//     });
// }
// 
// /**
//  * const promise = promiseAll([() => new Promise(res => res(42))])
//  * promise.then(console.log); // [42]
//  */

# Accepted solution for LeetCode #2721: Execute Asynchronous Functions in Parallel
# Auto-generated Python example from ts.
def example_solution() -> None:
    return
# Reference (ts):
# // Accepted solution for LeetCode #2721: Execute Asynchronous Functions in Parallel
# async function promiseAll<T>(functions: (() => Promise<T>)[]): Promise<T[]> {
#     return new Promise<T[]>((resolve, reject) => {
#         let cnt = 0;
#         const ans = new Array(functions.length);
#         for (let i = 0; i < functions.length; ++i) {
#             const f = functions[i];
#             f()
#                 .then(res => {
#                     ans[i] = res;
#                     cnt++;
#                     if (cnt === functions.length) {
#                         resolve(ans);
#                     }
#                 })
#                 .catch(err => {
#                     reject(err);
#                 });
#         }
#     });
# }
# 
# /**
#  * const promise = promiseAll([() => new Promise(res => res(42))])
#  * promise.then(console.log); // [42]
#  */

// Accepted solution for LeetCode #2721: Execute Asynchronous Functions in Parallel
// Rust example auto-generated from ts reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (ts):
// // Accepted solution for LeetCode #2721: Execute Asynchronous Functions in Parallel
// async function promiseAll<T>(functions: (() => Promise<T>)[]): Promise<T[]> {
//     return new Promise<T[]>((resolve, reject) => {
//         let cnt = 0;
//         const ans = new Array(functions.length);
//         for (let i = 0; i < functions.length; ++i) {
//             const f = functions[i];
//             f()
//                 .then(res => {
//                     ans[i] = res;
//                     cnt++;
//                     if (cnt === functions.length) {
//                         resolve(ans);
//                     }
//                 })
//                 .catch(err => {
//                     reject(err);
//                 });
//         }
//     });
// }
// 
// /**
//  * const promise = promiseAll([() => new Promise(res => res(42))])
//  * promise.then(console.log); // [42]
//  */

// Accepted solution for LeetCode #2721: Execute Asynchronous Functions in Parallel
async function promiseAll<T>(functions: (() => Promise<T>)[]): Promise<T[]> {
    return new Promise<T[]>((resolve, reject) => {
        let cnt = 0;
        const ans = new Array(functions.length);
        for (let i = 0; i < functions.length; ++i) {
            const f = functions[i];
            f()
                .then(res => {
                    ans[i] = res;
                    cnt++;
                    if (cnt === functions.length) {
                        resolve(ans);
                    }
                })
                .catch(err => {
                    reject(err);
                });
        }
    });
}

/**
 * const promise = promiseAll([() => new Promise(res => res(42))])
 * promise.then(console.log); // [42]
 */

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.