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Formatted question description: https://leetcode.ca/all/2446.html
2446. Determine if Two Events Have Conflict
 Difficulty: Easy.
 Related Topics: Array, String.
 Similar Questions: Merge Intervals, Nonoverlapping Intervals, My Calendar I.
Problem
You are given two arrays of strings that represent two inclusive events that happened on the same day, event1
and event2
, where:

event1 = [startTime1, endTime1]
and 
event2 = [startTime2, endTime2]
.
Event times are valid 24 hours format in the form of HH:MM
.
A conflict happens when two events have some nonempty intersection (i.e., some moment is common to both events).
Return true
** if there is a conflict between two events. Otherwise, return **false
.
Example 1:
Input: event1 = ["01:15","02:00"], event2 = ["02:00","03:00"]
Output: true
Explanation: The two events intersect at time 2:00.
Example 2:
Input: event1 = ["01:00","02:00"], event2 = ["01:20","03:00"]
Output: true
Explanation: The two events intersect starting from 01:20 to 02:00.
Example 3:
Input: event1 = ["10:00","11:00"], event2 = ["14:00","15:00"]
Output: false
Explanation: The two events do not intersect.
Constraints:

evnet1.length == event2.length == 2.

event1[i].length == event2[i].length == 5

startTime1 <= endTime1

startTime2 <= endTime2

All the event times follow the
HH:MM
format.
Solution (Java, C++, Python)

class Solution { public boolean haveConflict(String[] event1, String[] event2) { return event1[0].compareTo(event2[1]) <= 0 && event1[1].compareTo(event2[0]) >= 0; } }

class Solution { public: bool haveConflict(vector<string>& event1, vector<string>& event2) { return event1[0] <= event2[1] && event1[1] >= event2[0]; } };

class Solution: def haveConflict(self, event1: List[str], event2: List[str]) > bool: return event1[0] <= event2[1] and event1[1] >= event2[0]

func haveConflict(event1 []string, event2 []string) bool { return event1[0] <= event2[1] && event1[1] >= event2[0] }

function haveConflict(event1: string[], event2: string[]): boolean { return event1[0] <= event2[1] && event1[1] >= event2[0]; }
Explain:
nope.
Complexity:
 Time complexity : O(n).
 Space complexity : O(n).