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2446. Determine if Two Events Have Conflict

Description

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.

Solutions

Solution 1: String Comparison

If the start time of $event1$ is later than the end time of $event2$, or the end time of $event1$ is earlier than the start time of $event2$, then the two events will not conflict. Otherwise, the two events will conflict.

The time complexity is $O(1)$, and the space complexity is $O(1)$.

  • 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 not (event1[0] > event2[1] or 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]);
    }
    
    
  • impl Solution {
        pub fn have_conflict(event1: Vec<String>, event2: Vec<String>) -> bool {
            !(event1[1] < event2[0] || event1[0] > event2[1])
        }
    }
    
    

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