# Meeting Rooms II in C++

Suppose there is an array of meeting time intervals. There are two times start and end times [[s1,e1],[s2,e2],...] and each pair satisfies the rule (si < ei), We have to find the minimum number of conference rooms required.

So, if the input is like [[0, 30], [5, 10], [15, 20]], then the output will be 2.

To solve this, we will follow these steps −

• define one priority queue pq

• sort the intervals array

• ret := 0

• for initialize i := 0, when i < size of intervals, update (increase i by 1), do −

• while (not pq is empty and top element of pq <= intervals[i, 0]), do −

• delete element from pq

• insert intervals[i] into pq

• ret := maximum of ret and size of pq

• return ret

## Example

Let us see the following implementation to get a better understanding −

Live Demo

#include <bits/stdc++.h>
using namespace std;
struct Comparator{
bool operator()(vector <int<& a, vector <int<& b){
return !(a < b);
}
};
class Solution {
public:
static bool cmp(vector <int< a, vector <int< b){
return (a < b);
}
int minMeetingRooms(vector<vector<int<>& intervals) {
priority_queue<vector<int<, vector<vector<int< >, Comparator> pq;
sort(intervals.begin(), intervals.end());
int ret = 0;
for (int i = 0; i < intervals.size(); i++) {
while (!pq.empty() && pq.top() <= intervals[i])
pq.pop();
pq.push(intervals[i]);
ret = max(ret, (int)pq.size());
}
return ret;
}
};
main(){
vector<vector<int<> v = {{0, 30}, {5, 10}, {15, 20}};
Solution ob;
cout << (ob.minMeetingRooms(v));
}

## Input

{{0, 30}, {5, 10}, {15, 20}}

## Output

2

Updated on: 18-Nov-2020

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