LeetCode #2627 — MEDIUM

Debounce

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

The Problem

Problem Statement

Given a function fn and a time in milliseconds t, return a debounced version of that function.

A debounced function is a function whose execution is delayed by t milliseconds and whose execution is cancelled if it is called again within that window of time. The debounced function should also receive the passed parameters.

For example, let's say t = 50ms, and the function was called at 30ms, 60ms, and 100ms.

The first 2 function calls would be cancelled, and the 3rd function call would be executed at 150ms.

If instead t = 35ms, The 1st call would be cancelled, the 2nd would be executed at 95ms, and the 3rd would be executed at 135ms.

The above diagram shows how debounce will transform events. Each rectangle represents 100ms and the debounce time is 400ms. Each color represents a different set of inputs.

Please solve it without using lodash's _.debounce() function.

Example 1:

Input: 
t = 50
calls = [
  {"t": 50, inputs: [1]},
  {"t": 75, inputs: [2]}
]
Output: [{"t": 125, inputs: [2]}]
Explanation:
let start = Date.now();
function log(...inputs) { 
  console.log([Date.now() - start, inputs ])
}
const dlog = debounce(log, 50);
setTimeout(() => dlog(1), 50);
setTimeout(() => dlog(2), 75);

The 1st call is cancelled by the 2nd call because the 2nd call occurred before 100ms
The 2nd call is delayed by 50ms and executed at 125ms. The inputs were (2).

Example 2:

Input: 
t = 20
calls = [
  {"t": 50, inputs: [1]},
  {"t": 100, inputs: [2]}
]
Output: [{"t": 70, inputs: [1]}, {"t": 120, inputs: [2]}]
Explanation:
The 1st call is delayed until 70ms. The inputs were (1).
The 2nd call is delayed until 120ms. The inputs were (2).

Example 3:

Input: 
t = 150
calls = [
  {"t": 50, inputs: [1, 2]},
  {"t": 300, inputs: [3, 4]},
  {"t": 300, inputs: [5, 6]}
]
Output: [{"t": 200, inputs: [1,2]}, {"t": 450, inputs: [5, 6]}]
Explanation:
The 1st call is delayed by 150ms and ran at 200ms. The inputs were (1, 2).
The 2nd call is cancelled by the 3rd call
The 3rd call is delayed by 150ms and ran at 450ms. The inputs were (5, 6).

Constraints:

0 <= t <= 1000
1 <= calls.length <= 10
0 <= calls[i].t <= 1000
0 <= calls[i].inputs.length <= 10

Step 01

Brute Force Baseline

Problem summary: Given a function fn and a time in milliseconds t, return a debounced version of that function. A debounced function is a function whose execution is delayed by t milliseconds and whose execution is cancelled if it is called again within that window of time. The debounced function should also receive the passed parameters. For example, let's say t = 50ms, and the function was called at 30ms, 60ms, and 100ms. The first 2 function calls would be cancelled, and the 3rd function call would be executed at 150ms. If instead t = 35ms, The 1st call would be cancelled, the 2nd would be executed at 95ms, and the 3rd would be executed at 135ms. The above diagram shows how debounce will transform events. Each rectangle represents 100ms and the debounce time is 400ms. Each color represents a different set of inputs. Please solve it without using lodash's _.debounce() function.

Baseline thinking

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

Pattern signal: General problem-solving

Example 1

50
[{"t":50,"inputs":[1]},{"t":75,"inputs":[2]}]

Example 2

20
[{"t":50,"inputs":[1]},{"t":100,"inputs":[2]}]

Example 3

150
[{"t":50,"inputs":[1,2]},{"t":300,"inputs":[3,4]},{"t":300,"inputs":[5,6]}]

Core Insight

What unlocks the optimal approach

You execute code with a delay with "ref = setTimeout(fn, delay)". You can abort the execution of that code with "clearTimeout(ref)"
Whenever you call the function, you should abort any existing scheduled code. Then, you should schedule code to be executed after some delay.

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 #2627: Debounce
// Auto-generated Java example from ts.
class Solution {
    public void exampleSolution() {
    }
}
// Reference (ts):
// // Accepted solution for LeetCode #2627: Debounce
// type F = (...p: any[]) => any;
// 
// function debounce(fn: F, t: number): F {
//     let timeout: ReturnType<typeof setTimeout> | undefined;
// 
//     return function (...args) {
//         if (timeout !== undefined) {
//             clearTimeout(timeout);
//         }
//         timeout = setTimeout(() => {
//             fn.apply(this, args);
//         }, t);
//     };
// }
// 
// /**
//  * const log = debounce(console.log, 100);
//  * log('Hello'); // cancelled
//  * log('Hello'); // cancelled
//  * log('Hello'); // Logged at t=100ms
//  */

// Accepted solution for LeetCode #2627: Debounce
// Auto-generated Go example from ts.
func exampleSolution() {
}
// Reference (ts):
// // Accepted solution for LeetCode #2627: Debounce
// type F = (...p: any[]) => any;
// 
// function debounce(fn: F, t: number): F {
//     let timeout: ReturnType<typeof setTimeout> | undefined;
// 
//     return function (...args) {
//         if (timeout !== undefined) {
//             clearTimeout(timeout);
//         }
//         timeout = setTimeout(() => {
//             fn.apply(this, args);
//         }, t);
//     };
// }
// 
// /**
//  * const log = debounce(console.log, 100);
//  * log('Hello'); // cancelled
//  * log('Hello'); // cancelled
//  * log('Hello'); // Logged at t=100ms
//  */

# Accepted solution for LeetCode #2627: Debounce
# Auto-generated Python example from ts.
def example_solution() -> None:
    return
# Reference (ts):
# // Accepted solution for LeetCode #2627: Debounce
# type F = (...p: any[]) => any;
# 
# function debounce(fn: F, t: number): F {
#     let timeout: ReturnType<typeof setTimeout> | undefined;
# 
#     return function (...args) {
#         if (timeout !== undefined) {
#             clearTimeout(timeout);
#         }
#         timeout = setTimeout(() => {
#             fn.apply(this, args);
#         }, t);
#     };
# }
# 
# /**
#  * const log = debounce(console.log, 100);
#  * log('Hello'); // cancelled
#  * log('Hello'); // cancelled
#  * log('Hello'); // Logged at t=100ms
#  */

// Accepted solution for LeetCode #2627: Debounce
// 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 #2627: Debounce
// type F = (...p: any[]) => any;
// 
// function debounce(fn: F, t: number): F {
//     let timeout: ReturnType<typeof setTimeout> | undefined;
// 
//     return function (...args) {
//         if (timeout !== undefined) {
//             clearTimeout(timeout);
//         }
//         timeout = setTimeout(() => {
//             fn.apply(this, args);
//         }, t);
//     };
// }
// 
// /**
//  * const log = debounce(console.log, 100);
//  * log('Hello'); // cancelled
//  * log('Hello'); // cancelled
//  * log('Hello'); // Logged at t=100ms
//  */

// Accepted solution for LeetCode #2627: Debounce
type F = (...p: any[]) => any;

function debounce(fn: F, t: number): F {
    let timeout: ReturnType<typeof setTimeout> | undefined;

    return function (...args) {
        if (timeout !== undefined) {
            clearTimeout(timeout);
        }
        timeout = setTimeout(() => {
            fn.apply(this, args);
        }, t);
    };
}

/**
 * const log = debounce(console.log, 100);
 * log('Hello'); // cancelled
 * log('Hello'); // cancelled
 * log('Hello'); // Logged 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.