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 <= 10001 <= calls.length <= 100 <= calls[i].t <= 10000 <= calls[i].inputs.length <= 10This approach involves using a timeout mechanism to delay the execution of the function by t milliseconds. If the function is called again within this delay period, the existing delay is cancelled and a new delay is started. This ensures that the function execution occurs only after the last call within the time window has passed.
The JavaScript implementation utilizes the setTimeout function to delay execution of fn by t milliseconds. If fn is called again before this delay expires, clearTimeout is used to cancel the previous delay and initiate a new one. This ensures that only the last call within the debounce window is eventually executed.
Time Complexity: O(1) per function call as we only set and clear timeouts.
Space Complexity: O(1) as we are only storing the timer id.
This approach takes advantage of closures to maintain the state of the timeout. This allows us to track whether a timeout is active and reset it if needed. This encapsulation within the closure ensures that each debounced function call properly manages the delay on its own.
In this JavaScript solution, a closure is created around a timerId variable that keeps track of the current timeout. Each invocation of the debounced function checks this variable; if a timeout is active, it is cleared before setting a new one. This method leverages closures to effectively manage state across function invocations.
Time Complexity: O(1) per function call because managing a single timeout involves constant-time operations.
Space Complexity: O(1) due to the single timerId variable maintained per debounced instance.
| Approach | Complexity |
|---|---|
| Using Timers for Debouncing | Time Complexity: O(1) per function call as we only set and clear timeouts. |
| Using a Closure to Manage State | Time Complexity: O(1) per function call because managing a single timeout involves constant-time operations. |
I HATE This Coding Question, but FAANG Loves it! | Majority Element - Leetcode 169 • Greg Hogg • 2,093,939 views views
Watch 9 more video solutions →Practice Debounce with our built-in code editor and test cases.
Practice on FleetCode