Leetcode 253: Meeting Rooms II

https://leetcode.com › problems › meeting-rooms-ii

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

NOTE: input types have been changed on April 15, 2019. Please reset to default code definition to get new method signature.

Notes:

This question is a very popular interview question, since that it can be solved by many methods.

However, this question can be solved by a very intuitive and surprisingly very powerful algorithm: sweeping-line algorithm.

The basic idea is that we just need to sort the start time and end time first. Then scan them from left to right.

1): if we meet with a start time, it means we need to start a new meeting; so meeting room needs +1;

2): if we meet with a end time, it means one meeting is over; so meeting room - 1.

Thus, this question is equivalent to ask: what is the maximum number of meeting that are hold at the same time?

See the code below:

class Solution {
public:
    int minMeetingRooms(vector<vector<char>>& intervals) {
        int res = 0;
        vector<pair<int, int>> startEndTimes; //<time, star/end>
        for(auto &a : intervals) {
            startEndTimes.push_back({a[0], 1});//start time
            startEndTimes.push_back({a[0], 0});//end time
        }
        sort(startEndTimes.begin(), startEndTimes.end());
        int count = 0;
        for(auto &a : startEndTimes) {
            if(a.second == 1) ++count;
            else --count;
            res = max(res, count);
        }
        return res;
    }
};

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