Meeting Conflict algorithms - algorithm

Question

I had a interview today and was asked to check whether two meeting conflicts with each other or not. Each meeting has start time and end time.
I tried to answer the question but not that specific..can somebody throw some idea?
bool IsConflict(Datetime s1, Datetime e1, Datetime s2, Datetime e2)
should return true if Conflict is there and false if no conflict.
E.g
True if:
(s1, e1)= 8,10
(s2, e2) = 9, 11
(s1, e1)= 7,10
(s2, e2) = 8, 9
(s1, e1)= 8,11
(s2, e2) = 9, 11
and so on

Answer 1

This is basic interval algebra, see my answer here for more details, but the code would look like this:
bool IsConflict(Datetime s1, Datetime e1, Datetime s2, Datetime e2)
{
return (s1 < e2) && (e1 > s2);
}
I am assuming that two meetings where one start where the other ends are not in conflict.

Answer 2

In the simple case of two intervals I think this will work (untested pseudocode ahead):
bool IsConflict(Datatime s1, Datatime e1, Datatime s2, Datatime e2) {
if( s1 < s2 ) {
// meeting 1 starts first
if( e1 > s2 ) return true; // overlap
}
else {
// meeting 2 starts first
if( e2 > s1 ) return true; // overlap
}
return false;
}

Answer 3

The meetings overlap if and only if max(s1, s2) < min(e1, e2). This intersection based approach assumes that intervals (s, e) are open, and implies (rightly or wrongly) that an empty meeting s = e cannot have an overlap with another meeting.

Answer 4

Complexity of following algorithm is O (nlogn)
public boolean isConflicts(float startTime[], float endTime[])
{
TreeMap<Float, Integer> map = new TreeMap<Float, Integer>();
for (int i = 0; i < startTime.length; i++)
{
map.put(startTime[i], -1);
map.put(endTime[i], 1);
}
Iterator<Integer>iter = map.values().iterator();
while (iter.hasNext())
{
if ((iter.next() + iter.next()) != 0)
{
System.out.println ("Conflicts...");
return true;
}
}
return false;
}

Answer 5

The plan
There are three cases to check for with this problem.
Case 1: Does s1 lie within the interval [s2,e2]
(s1 >= s2) && (s1 <= e2)
Case 2: Does e1 lie within the interval [s2, e2]
(e1 >= s2) && (e2 <= e2))
Case 3: Does the point (s2, e2) lie within [s1, e1]
(s1 <= s2) && (e1 >= e2)
So here is the answer. I apologize; It's not the most readable lines of code.
The code (pseudo):
bool isConflict(Datetime s1, Datetime e1, Datetime s2, Datetime e2){
return ((s1 >= s2) && (s1 <= e2)) || ((e1 >= s2) && (e2 <= e2)) || (s1 <= s2) && (e1 >= e2));
}