Given an array of meeting time intervals consisting of start and end times[[s1,e1],[s2,e2],...](si< ei), find the minimum number of conference rooms required.
Example 1:
Input:
[[0, 30],[5, 10],[15, 20]]
Output:
2
Example 2:
Input:
[[7,10],[2,4]]
Output:
1
Hint 1
Think about how we would approach this problem in a very simplistic way. We will allocate rooms to meetings that occur earlier in the day v/s the ones that occur later on, right?
Hint 2
If you've figured out that we have to sort the meetings by their start time, the next thing to think about is how do we do the allocation?
There are two scenarios possible here for any meeting. Either there is no meeting room available and a new one has to be allocated, or a meeting room has freed up and this meeting can take place there.
Hint 3
An important thing to note is that we don't really care which room gets freed up while allocating a room for the current meeting. As long as a room is free, our job is done.
We already know the rooms we have allocated till now and we also know when are they due to get free because of the end times of the meetings going on in those rooms. We can simply check the room which is due to get vacated the earliest amongst all the allocated rooms.
Hint 4
Following up on the previous hint, we can make use of a min-heap to store the end times of the meetings in various rooms.
So, every time we want to check if any room is free or not, simply check the topmost element of the min heap as that would be the room that would get free the earliest out of all the other rooms currently occupied.
If the room we extracted from the top of the min heap isn't free, then no other room is. So, we can save time here and simply allocate a new room.
Analysis & Solution
Sweep Line 扫描线算法
/**
* Definition for an interval.
* public class Interval {
* int start;
* int end;
* Interval() { start = 0; end = 0; }
* Interval(int s, int e) { start = s; end = e; }
* }
*/
class Solution {
class Point {
int type; // start: 1, end: 0
int time;
Point (int time, int type) {
this.time = time;
this.type = type;
}
}
public int minMeetingRooms(Interval[] intervals) {
List<Point> points = new ArrayList<>(intervals.length * 2);
for (Interval i: intervals) {
points.add(new Point(i.start, 1));
points.add(new Point(i.end, 0));
}
Comparator<Point> cmp = new Comparator<Point>() {
@Override
public int compare (Point a, Point b) {
if (a.time == b.time) {
return a.type - b.type;
} else {
return a.time - b.time;
}
}
};
Collections.sort(points, cmp);
int maxOverlap = 0;
int ongoing = 0;
for (Point p: points) {
if (p.type == 1) {
ongoing++;
} else if (p.type == 0) {
ongoing--;
}
maxOverlap = Math.max(maxOverlap, ongoing);
}
return maxOverlap;
}
}
继续扫描，看到新的 end 的时候，存下这个 end 的时间戳，因为它是重合区间里 end 时间最早的一个；
二者之间，就是具体的 max overlap 区间。
其他方法：
Approach 1: Priority Queues
/**
* Definition for an interval. public class Interval { int start; int end; Interval() { start = 0;
* end = 0; } Interval(int s, int e) { start = s; end = e; } }
*/
class Solution {
public int minMeetingRooms(Interval[] intervals) {
// Check for the base case. If there are no intervals, return 0
if (intervals.length == 0) {
return 0;
}
// Min heap
PriorityQueue<Integer> allocator =
new PriorityQueue<Integer>(
intervals.length,
new Comparator<Integer>() {
public int compare(Integer a, Integer b) {
return a - b;
}
});
// Sort the intervals by start time
Arrays.sort(
intervals,
new Comparator<Interval>() {
public int compare(Interval a, Interval b) {
return a.start - b.start;
}
});
// Add the first meeting
allocator.add(intervals[0].end);
// Iterate over remaining intervals
for (int i = 1; i < intervals.length; i++) {
// If the room due to free up the earliest is free, assign that room to this meeting.
if (intervals[i].start >= allocator.peek()) {
allocator.poll();
}
// If a new room is to be assigned, then also we add to the heap,
// If an old room is allocated, then also we have to add to the heap with updated end time.
allocator.add(intervals[i].end);
}
// The size of the heap tells us the minimum rooms required for all the meetings.
return allocator.size();
}
}
Approach 2: Chronological Ordering
/**
* Definition for an interval. public class Interval { int start; int end; Interval() { start = 0;
* end = 0; } Interval(int s, int e) { start = s; end = e; } }
*/
class Solution {
public int minMeetingRooms(Interval[] intervals) {
// Check for the base case. If there are no intervals, return 0
if (intervals.length == 0) {
return 0;
}
Integer[] start = new Integer[intervals.length];
Integer[] end = new Integer[intervals.length];
for (int i = 0; i < intervals.length; i++) {
start[i] = intervals[i].start;
end[i] = intervals[i].end;
}
// Sort the intervals by end time
Arrays.sort(
end,
new Comparator<Integer>() {
public int compare(Integer a, Integer b) {
return a - b;
}
});
// Sort the intervals by start time
Arrays.sort(
start,
new Comparator<Integer>() {
public int compare(Integer a, Integer b) {
return a - b;
}
});
// The two pointers in the algorithm: e_ptr and s_ptr.
int startPointer = 0, endPointer = 0;
// Variables to keep track of maximum number of rooms used.
int usedRooms = 0;
// Iterate over intervals.
while (startPointer < intervals.length) {
// If there is a meeting that has ended by the time the meeting at `start_pointer` starts
if (start[startPointer] >= end[endPointer]) {
usedRooms -= 1;
endPointer += 1;
}
// We do this irrespective of whether a room frees up or not.
// If a room got free, then this used_rooms += 1 wouldn't have any effect. used_rooms would
// remain the same in that case. If no room was free, then this would increase used_rooms
usedRooms += 1;
startPointer += 1;
}
return usedRooms;
}
}
Simple version
class Solution {
public int minMeetingRooms(Interval[] intervals) {
int[] starts = new int[intervals.length];
int[] ends = new int[intervals.length];
for (int i = 0; i < intervals.length; i++) {
starts[i] = intervals[i].start;
ends[i] = intervals[i].end;
}
Arrays.sort(starts);
Arrays.sort(ends);
int rooms = 0, endsItr = 0;
for (int i = 0; i < starts.length; i++) {
if (starts[i] < ends[endsItr]) {
rooms++;
} else {
endsItr++;
}
}
return rooms;
}
}
Approach 3 TreeMap
和扫描线原理类似，但是用TreeMap来实现自动排序
public int minMeetingRooms(Interval[] intervals) {
TreeMap<Integer, Integer> tmap = new TreeMap<>();
int ans = 0, cnt = 0;
for (Interval i : intervals) {
int start = i.start, end = i.end;
tmap.put(start, tmap.getOrDefault(start, 0) + 1);
tmap.put(end, tmap.getOrDefault(end, 0) - 1);
}
for (int k : tmap.keySet()) {
cnt += tmap.get(k);
ans = Math.max(ans, cnt);
}
return ans;
}