LeetCode #2637 — MEDIUM

Promise Time Limit

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

The Problem

Problem Statement

Given an asynchronous function fn and a time t in milliseconds, return a new time limited version of the input function. fn takes arguments provided to the time limited function.

The time limited function should follow these rules:

If the fn completes within the time limit of t milliseconds, the time limited function should resolve with the result.
If the execution of the fn exceeds the time limit, the time limited function should reject with the string "Time Limit Exceeded".

Example 1:

Input: 
fn = async (n) => { 
  await new Promise(res => setTimeout(res, 100)); 
  return n * n; 
}
inputs = [5]
t = 50
Output: {"rejected":"Time Limit Exceeded","time":50}
Explanation:
const limited = timeLimit(fn, t)
const start = performance.now()
let result;
try {
   const res = await limited(...inputs)
   result = {"resolved": res, "time": Math.floor(performance.now() - start)};
} catch (err) {
   result = {"rejected": err, "time": Math.floor(performance.now() - start)};
}
console.log(result) // Output

The provided function is set to resolve after 100ms. However, the time limit is set to 50ms. It rejects at t=50ms because the time limit was reached.

Example 2:

Input: 
fn = async (n) => { 
  await new Promise(res => setTimeout(res, 100)); 
  return n * n; 
}
inputs = [5]
t = 150
Output: {"resolved":25,"time":100}
Explanation:
The function resolved 5 * 5 = 25 at t=100ms. The time limit is never reached.

Example 3:

Input: 
fn = async (a, b) => { 
  await new Promise(res => setTimeout(res, 120)); 
  return a + b; 
}
inputs = [5,10]
t = 150
Output: {"resolved":15,"time":120}
Explanation:
The function resolved 5 + 10 = 15 at t=120ms. The time limit is never reached.

Example 4:

Input: 
fn = async () => { 
  throw "Error";
}
inputs = []
t = 1000
Output: {"rejected":"Error","time":0}
Explanation:
The function immediately throws an error.

Constraints:

0 <= inputs.length <= 10
0 <= t <= 1000
fn returns a promise

Step 01

Brute Force Baseline

Problem summary: Given an asynchronous function fn and a time t in milliseconds, return a new time limited version of the input function. fn takes arguments provided to the time limited function. The time limited function should follow these rules: If the fn completes within the time limit of t milliseconds, the time limited function should resolve with the result. If the execution of the fn exceeds the time limit, the time limited function should reject with the string "Time Limit Exceeded".

Baseline thinking

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

Pattern signal: General problem-solving

Example 1

async (n) => { await new Promise(res => setTimeout(res, 100)); return n * n; }
[5]
50

Example 2

async (n) => { await new Promise(res => setTimeout(res, 100)); return n * n; }
[5]
150

Example 3

async (a, b) => { await new Promise(res => setTimeout(res, 120)); return a + b; }
[5,10]
150

Core Insight

What unlocks the optimal approach

You can return a copy of a function with: function outerFunction(fn) { return function innerFunction(...params) { return fn(...params); }; }
Inside the inner function, you will need to return a new Promise.
You can create a new promise like: new Promise((resolve, reject) => {}).
You can execute code with a delay with "setTimeout(fn, delay)"
To reject a promise after a delay, "setTimeout(() => reject('err'), delay)"
You can resolve and reject when the passed promise resolves or rejects with: "fn(...params).then(resolve).catch(reject)"

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 #2637: Promise Time Limit
// Auto-generated Java example from ts.
class Solution {
    public void exampleSolution() {
    }
}
// Reference (ts):
// // Accepted solution for LeetCode #2637: Promise Time Limit
// type Fn = (...params: any[]) => Promise<any>;
// 
// function timeLimit(fn: Fn, t: number): Fn {
//     return async function (...args) {
//         return Promise.race([
//             fn(...args),
//             new Promise((_, reject) => setTimeout(() => reject('Time Limit Exceeded'), t)),
//         ]);
//     };
// }
// 
// /**
//  * const limited = timeLimit((t) => new Promise(res => setTimeout(res, t)), 100);
//  * limited(150).catch(console.log) // "Time Limit Exceeded" at t=100ms
//  */

// Accepted solution for LeetCode #2637: Promise Time Limit
// Auto-generated Go example from ts.
func exampleSolution() {
}
// Reference (ts):
// // Accepted solution for LeetCode #2637: Promise Time Limit
// type Fn = (...params: any[]) => Promise<any>;
// 
// function timeLimit(fn: Fn, t: number): Fn {
//     return async function (...args) {
//         return Promise.race([
//             fn(...args),
//             new Promise((_, reject) => setTimeout(() => reject('Time Limit Exceeded'), t)),
//         ]);
//     };
// }
// 
// /**
//  * const limited = timeLimit((t) => new Promise(res => setTimeout(res, t)), 100);
//  * limited(150).catch(console.log) // "Time Limit Exceeded" at t=100ms
//  */

# Accepted solution for LeetCode #2637: Promise Time Limit
# Auto-generated Python example from ts.
def example_solution() -> None:
    return
# Reference (ts):
# // Accepted solution for LeetCode #2637: Promise Time Limit
# type Fn = (...params: any[]) => Promise<any>;
# 
# function timeLimit(fn: Fn, t: number): Fn {
#     return async function (...args) {
#         return Promise.race([
#             fn(...args),
#             new Promise((_, reject) => setTimeout(() => reject('Time Limit Exceeded'), t)),
#         ]);
#     };
# }
# 
# /**
#  * const limited = timeLimit((t) => new Promise(res => setTimeout(res, t)), 100);
#  * limited(150).catch(console.log) // "Time Limit Exceeded" at t=100ms
#  */

// Accepted solution for LeetCode #2637: Promise Time Limit
// 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 #2637: Promise Time Limit
// type Fn = (...params: any[]) => Promise<any>;
// 
// function timeLimit(fn: Fn, t: number): Fn {
//     return async function (...args) {
//         return Promise.race([
//             fn(...args),
//             new Promise((_, reject) => setTimeout(() => reject('Time Limit Exceeded'), t)),
//         ]);
//     };
// }
// 
// /**
//  * const limited = timeLimit((t) => new Promise(res => setTimeout(res, t)), 100);
//  * limited(150).catch(console.log) // "Time Limit Exceeded" at t=100ms
//  */

// Accepted solution for LeetCode #2637: Promise Time Limit
type Fn = (...params: any[]) => Promise<any>;

function timeLimit(fn: Fn, t: number): Fn {
    return async function (...args) {
        return Promise.race([
            fn(...args),
            new Promise((_, reject) => setTimeout(() => reject('Time Limit Exceeded'), t)),
        ]);
    };
}

/**
 * const limited = timeLimit((t) => new Promise(res => setTimeout(res, t)), 100);
 * limited(150).catch(console.log) // "Time Limit Exceeded" at t=100ms
 */

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.