Leetcode #2627: Debounce
In this guide, we solve Leetcode #2627 Debounce in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
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.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: JavaScript
Intuition
The constraints allow a direct scan that keeps only the essential state.
By translating the requirements into a clean loop, the logic stays easy to reason about.
Approach
Iterate through the data once, updating the state needed to compute the answer.
Return the final state after the traversal is complete.
Steps:
- Parse the input.
- Iterate and update state.
- Return the computed answer.
Example
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).
Python Solution
# TODO: add Python solution
Complexity
The time complexity is O(n). The space complexity is O(1) to O(n).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.