What is the main trick in Determine if Two Events Have Conflict?

Treat the two events as inclusive time intervals and check whether they overlap. The key boundary case is that matching endpoints still count as conflict.

Do I need to convert HH:MM to minutes for this problem?

Not strictly. Since HH:MM is zero-padded, direct string comparison works, but converting to minutes is often easier to explain and less error-prone.

Why does event1 = ["01:15","02:00"] and event2 = ["02:00","03:00"] return true?

Because the intervals are inclusive, both events contain the minute 02:00. Their intersection is not empty, so there is a conflict.

What overlap formula should I remember for this Array plus String pattern?

A reliable rule is max(start1, start2) <= min(end1, end2). If that condition holds, the intervals share at least one moment.

What is the most common bug on LeetCode 2446?

Using strict inequality and treating endpoint contact as non-overlap. In this problem, that mistake fails the exact-boundary examples.

#2446

Easy

auto_awesome数组·string

LeetCode 题解工作台

判断两个事件是否存在冲突

给你两个字符串数组 event1 和 event2 ，表示发生在同一天的两个闭区间时间段事件，其中： event1 = [startTime 1 , endTime 1 ] 且 event2 = [startTime 2 , endTime 2 ] 事件的时间为有效的 24 小时制且按 HH:MM …

数组字符串

题目描述

给你两个字符串数组 event1 和 event2 ，表示发生在同一天的两个闭区间时间段事件，其中：

event1 = [startTime₁, endTime₁] 且
event2 = [startTime₂, endTime₂]

事件的时间为有效的 24 小时制且按 HH:MM 格式给出。

当两个事件存在某个非空的交集时（即，某些时刻是两个事件都包含的），则认为出现冲突。

如果两个事件之间存在冲突，返回 true ；否则，返回 false 。

示例 1：

输入：event1 = ["01:15","02:00"], event2 = ["02:00","03:00"]
输出：true
解释：两个事件在 2:00 出现交集。

示例 2：

输入：event1 = ["01:00","02:00"], event2 = ["01:20","03:00"]
输出：true
解释：两个事件的交集从 01:20 开始，到 02:00 结束。

示例 3：

输入：event1 = ["10:00","11:00"], event2 = ["14:00","15:00"]
输出：false
解释：两个事件不存在交集。

提示：

event1.length == event2.length == 2
event1[i].length == event2[i].length == 5
startTime₁ <= endTime₁
startTime₂ <= endTime₂
所有事件的时间都按照 HH:MM 格式给出

lightbulb

解题思路

方法一：字符串比较

如果 $event1$ 的开始时间大于 $event2$ 的结束时间，或者 $event1$ 的结束时间小于 $event2$ 的开始时间，那么两个事件不会有冲突。否则，两个事件存在冲突。

时间复杂度 $O(1)$ ，空间复杂度 $O(1)$ 。

1

2

3

4

class Solution:
    def haveConflict(self, event1: List[str], event2: List[str]) -> bool:
        return not (event1[0] > event2[1] or event1[1] < event2[0])

speed

复杂度分析

指标	值
时间	Depends on the final approach
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
They mention parsing time format into an integer interval first, which points to converting HH:MM into total minutes.
question_mark
They ask whether events that touch at one endpoint should count, which tests whether you noticed the intervals are inclusive.
question_mark
They push on whether parsing is required, which is a hint to discuss why fixed-width HH:MM strings can be compared directly.

warning

常见陷阱

外企场景

error
Using a strict overlap check with < instead of <=, which incorrectly returns false when one event ends exactly when the other starts.
error
Overcomplicating an Easy problem with sorting or minute-by-minute simulation instead of comparing just two intervals.
error
Assuming all string times compare safely without explaining that the HH:MM format is zero-padded and therefore lexicographically ordered.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Return the length of the overlap in minutes instead of a boolean conflict flag.
arrow_right_alt
Given many events on the same day, detect whether any pair conflicts after sorting by start time.
arrow_right_alt
Handle times that can cross midnight, which breaks the simple same-day interval assumption in this problem.

help

常见问题

外企场景

继续练习

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