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Meeting Rooms

EasyGreedy

Given meeting time intervals, return true if a person could attend all meetings without overlaps.

Open problem
Constraints
  • 0 <= intervals.length <= 10^4
  • intervals[i] = [start, end]

Sort and check adjacent

Time O(n log n)Space O(1)

Sort meetings by start time. If any meeting starts before the previous one ends, there is a conflict; otherwise all meetings can be attended.

#include <bits/stdc++.h>
using namespace std;

bool canAttendMeetings(vector<vector<int>>& intervals) {
    sort(intervals.begin(), intervals.end());
    for (int i = 1; i < (int)intervals.size(); i++)
        if (intervals[i][0] < intervals[i - 1][1]) return false;
    return true;
}