Determine if Two Events Have Conflict

Difficulty: Easy | Acceptance: 53.20% | Paid: No Topics: Array, String

You are given two arrays of strings event1 and event2 where:

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

startTime is always strictly less than endTime.

The format of time is “HH:MM”.

Two events have a conflict if the start time of one event is less than or equal to the end time of the other event, and the end time of that event is greater than or equal to the start time of the other event. Basically, if they overlap or touch.

Return true if there is a conflict, otherwise false.

• Examples
• Constraints
• String Comparison
• Integer Conversion

Examples

Input: event1 = ["01:15","02:00"], event2 = ["02:00","03:00"]
Output: true
Explanation: The two events overlap at time 2:00.

Input: event1 = ["01:00","02:00"], event2 = ["01:20","03:00"]
Output: true
Explanation: The two events overlap from 01:20 to 02:00.

Input: event1 = ["10:00","11:00"], event2 = ["12:00","13:00"]
Output: false
Explanation: The two events do not overlap.

Constraints

event1.length == event2.length == 2
event1[i].length == event2[i].length == 5
startTime1 <= endTime1
startTime2 <= endTime2
All the event times follow the HH:MM format.

String Comparison

Intuition Since the time format is strictly “HH:MM”, lexicographical string comparison works perfectly to determine chronological order (e.g., “01:15” < “02:00”). We can check for non-overlapping conditions and negate the result.

Steps

• Check if event1 ends before event2 starts (event1[1] < event2[0]).
• Check if event2 ends before event1 starts (event2[1] < event1[0]).
• If neither is true, there is a conflict.

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

Complexity

• Time: O(1)
• Space: O(1)
• Notes: Very efficient due to fixed string length and direct comparison.

Integer Conversion

Intuition Convert the “HH:MM” time strings into total minutes since midnight. This transforms the problem into a standard integer interval overlap check.

Steps

• Parse hours and minutes for start and end times of both events.
• Calculate total minutes: hours * 60 + minutes.
• Check if the maximum start time is less than or equal to the minimum end time.

class Solution:
    def haveConflict(self, event1: List[str], event2: List[str]) -> bool:
        def get_minutes(t):
            h, m = map(int, t.split(':'))
            return h * 60 + m
        s1, e1 = get_minutes(event1[0]), get_minutes(event1[1])
        s2, e2 = get_minutes(event2[0]), get_minutes(event2[1])
        return max(s1, s2) <= min(e1, e2)

Complexity

• Time: O(1)
• Space: O(1)
• Notes: Slightly more verbose but conceptually clear for interval arithmetic.