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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, Non-overlapping 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 non-empty 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.compareTo(event2) <= 0 && event1.compareTo(event2) >= 0;
}
}

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

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


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

• function haveConflict(event1: string[], event2: string[]): boolean {
return event1 <= event2 && event1 >= event2;
}


• impl Solution {
pub fn have_conflict(event1: Vec<String>, event2: Vec<String>) -> bool {
!(event1 < event2 || event1 > event2)
}
}


Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).