Easy
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]
andevent2 = [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
HH:MM
format.public class Solution {
public boolean haveConflict(String[] event1, String[] event2) {
int aStart = getTimeSerial(event1[0]);
int aEnd = getTimeSerial(event1[1]);
int bStart = getTimeSerial(event2[0]);
int bEnd = getTimeSerial(event2[1]);
return (bStart >= aStart && bStart <= aEnd) || (bStart <= aStart && bEnd >= aStart);
}
private int getTimeSerial(String timestamp) {
int hours = 0;
int minutes = 0;
boolean isHours = true;
for (int i = 0; i < timestamp.length(); i += 1) {
if (timestamp.charAt(i) == ':') {
isHours = false;
} else if (isHours) {
hours = hours * 10 + Character.getNumericValue(timestamp.charAt(i));
} else {
minutes = minutes * 10 + Character.getNumericValue(timestamp.charAt(i));
}
}
return 60 * hours + minutes;
}
}