LeetCode #2725 — EASY

Interval Cancellation

Build confidence with an intuition-first walkthrough focused on core interview patterns fundamentals.

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The Problem

Problem Statement

Given a function fn, an array of arguments args, and an interval time t, return a cancel function cancelFn.

After a delay of cancelTimeMs, the returned cancel function cancelFn will be invoked.

setTimeout(cancelFn, cancelTimeMs)

The function fn should be called with args immediately and then called again every t milliseconds until cancelFn is called at cancelTimeMs ms.

Example 1:

Input: fn = (x) => x * 2, args = [4], t = 35
Output: 
[
   {"time": 0, "returned": 8},
   {"time": 35, "returned": 8},
   {"time": 70, "returned": 8},
   {"time": 105, "returned": 8},
   {"time": 140, "returned": 8},
   {"time": 175, "returned": 8}
]
Explanation: 
const cancelTimeMs = 190;
const cancelFn = cancellable((x) => x * 2, [4], 35);
setTimeout(cancelFn, cancelTimeMs);

Every 35ms, fn(4) is called. Until t=190ms, then it is cancelled.
1st fn call is at 0ms. fn(4) returns 8.
2nd fn call is at 35ms. fn(4) returns 8.
3rd fn call is at 70ms. fn(4) returns 8.
4th fn call is at 105ms. fn(4) returns 8.
5th fn call is at 140ms. fn(4) returns 8.
6th fn call is at 175ms. fn(4) returns 8.
Cancelled at 190ms

Example 2:

Input: fn = (x1, x2) => (x1 * x2), args = [2, 5], t = 30
Output: 
[
   {"time": 0, "returned": 10},
   {"time": 30, "returned": 10},
   {"time": 60, "returned": 10},
   {"time": 90, "returned": 10},
   {"time": 120, "returned": 10},
   {"time": 150, "returned": 10}
]
Explanation: 
const cancelTimeMs = 165; 
const cancelFn = cancellable((x1, x2) => (x1 * x2), [2, 5], 30) 
setTimeout(cancelFn, cancelTimeMs)

Every 30ms, fn(2, 5) is called. Until t=165ms, then it is cancelled.
1st fn call is at 0ms 
2nd fn call is at 30ms 
3rd fn call is at 60ms 
4th fn call is at 90ms 
5th fn call is at 120ms 
6th fn call is at 150ms
Cancelled at 165ms

Example 3:

Input: fn = (x1, x2, x3) => (x1 + x2 + x3), args = [5, 1, 3], t = 50
Output: 
[
   {"time": 0, "returned": 9},
   {"time": 50, "returned": 9},
   {"time": 100, "returned": 9},
   {"time": 150, "returned": 9}
]
Explanation: 
const cancelTimeMs = 180;
const cancelFn = cancellable((x1, x2, x3) => (x1 + x2 + x3), [5, 1, 3], 50)
setTimeout(cancelFn, cancelTimeMs)

Every 50ms, fn(5, 1, 3) is called. Until t=180ms, then it is cancelled. 
1st fn call is at 0ms
2nd fn call is at 50ms
3rd fn call is at 100ms
4th fn call is at 150ms
Cancelled at 180ms

Constraints:

  • fn is a function
  • args is a valid JSON array
  • 1 <= args.length <= 10
  • 30 <= t <= 100
  • 10 <= cancelTimeMs <= 500

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: Given a function fn, an array of arguments args, and an interval time t, return a cancel function cancelFn. After a delay of cancelTimeMs, the returned cancel function cancelFn will be invoked. setTimeout(cancelFn, cancelTimeMs) The function fn should be called with args immediately and then called again every t milliseconds until cancelFn is called at cancelTimeMs ms.

Baseline thinking

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

Pattern signal: General problem-solving

Example 1

(x) => x * 2
[4]
35
190

Example 2

(x1, x2) => (x1 * x2)
[2,5]
30
165

Example 3

(x1, x2, x3) => (x1 + x2 + x3)
[5,1,3]
50
180
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

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. 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 #2725: Interval Cancellation
// Auto-generated Java example from ts.
class Solution {
    public void exampleSolution() {
    }
}
// Reference (ts):
// // Accepted solution for LeetCode #2725: Interval Cancellation
// function cancellable(fn: Function, args: any[], t: number): Function {
//     fn(...args);
//     const time = setInterval(() => fn(...args), t);
//     return () => clearInterval(time);
// }
// 
// /**
//  *  const result = []
//  *
//  *  const fn = (x) => x * 2
//  *  const args = [4], t = 20, cancelT = 110
//  *
//  *  const log = (...argsArr) => {
//  *      result.push(fn(...argsArr))
//  *  }
//  *
//  *  const cancel = cancellable(fn, args, t);
//  *
//  *  setTimeout(() => {
//  *     cancel()
//  *     console.log(result) // [
//  *                         //      {"time":0,"returned":8},
//  *                         //      {"time":20,"returned":8},
//  *                         //      {"time":40,"returned":8},
//  *                         //      {"time":60,"returned":8},
//  *                         //      {"time":80,"returned":8},
//  *                         //      {"time":100,"returned":8}
//  *                         //  ]
//  *  }, cancelT)
//  */
Step 06

Interactive Study Demo

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

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.