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252. Meeting Rooms

Description

Given an array of meeting time intervals where intervals[i] = [starti, endi], determine if a person could attend all meetings.

 

Example 1:

Input: intervals = [[0,30],[5,10],[15,20]]
Output: false

Example 2:

Input: intervals = [[7,10],[2,4]]
Output: true

 

Constraints:

  • 0 <= intervals.length <= 104
  • intervals[i].length == 2
  • 0 <= starti < endi <= 106

Solutions

  1. Sort the Intervals: First, the list of intervals is sorted based on the start times of the meetings. Sorting is crucial because it allows for a linear comparison of consecutive meetings to check for any overlaps. The sorting uses a lambda function as the key, which sorts the intervals by their start time (x[0]).

     intervals.sort(key=lambda x: x[0])

  2. Iterate Through Sorted Intervals: The method then iterates through the sorted intervals, except for the last one (hence len(intervals) - 1). This is because the comparison is always made between the current interval and the next one, and there’s no next interval for the last one.

     for i in range(len(intervals) - 1):

  3. Check for Overlapping Intervals: For each pair of consecutive intervals, the code checks if the end time of the current interval (intervals[i][1]) is greater than the start time of the next interval (intervals[i + 1][0]). If this condition is true for any pair of intervals, it means there’s a conflict (one meeting does not end before the next one starts), and the person cannot attend all meetings. In such a case, the method immediately returns False.

     if intervals[i][1] > intervals[i + 1][0]:
     return False

  4. Return True if No Conflicts Found: If the method iterates through all interval pairs without finding any overlaps, it means all meetings can be attended without conflicts. Thus, the method returns True.

     return True

Example:

Given the input intervals = [[0, 30], [5, 10], [15, 20]]:

  • After sorting, intervals = [[0, 30], [5, 10], [15, 20]].
  • The method finds that the first interval [0, 30] overlaps with the next interval [5, 10] since 30 > 5, and it returns False.
  • class Solution {
    public boolean canAttendMeetings(int[][] intervals) {
        Arrays.sort(intervals, (a, b) -> a[0] - b[0]);
        for (int i = 1; i < intervals.length; ++i) {
            var a = intervals[i - 1];
            var b = intervals[i];
            if (a[1] > b[0]) {
                return false;
            }
        }
        return true;
    }
}

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

  • class Solution:
    def canAttendMeetings(self, intervals: List[List[int]]) -> bool:
        intervals.sort()
        return all(a[1] <= b[0] for a, b in pairwise(intervals))

############

from itertools import pairwise
class Solution:
    def canAttendMeetings(self, intervals: List[List[int]]) -> bool:
        intervals.sort(key=lambda x: x[0])
        return not any(a[1] > b[0] for a,b in pairwise(intervals))
        # any is True, then conflict found

############

class Solution:
    def canAttendMeetings(self, intervals: List[List[int]]) -> bool:
        intervals.sort(key=lambda x: x[0])
        for i in range(len(intervals) - 1):
            if intervals[i][1] > intervals[i + 1][0]:
                return False
        return True

############

# Definition for an interval.
# class Interval(object):
#     def __init__(self, s=0, e=0):
#         self.start = s
#         self.end = e

class Solution(object):
  def canAttendMeetings(self, intervals):
    """
    :type intervals: List[Interval]
    :rtype: bool
    """
    intervals = sorted(intervals, key=lambda x: x.start)
    for i in range(1, len(intervals)):
      if intervals[i].start < intervals[i - 1].end:
        return False
    return True



  • func canAttendMeetings(intervals [][]int) bool {
	sort.Slice(intervals, func(i, j int) bool {
		return intervals[i][0] < intervals[j][0]
	})
	for i := 1; i < len(intervals); i++ {
		if intervals[i][0] < intervals[i-1][1] {
			return false
		}
	}
	return true
}

  • function canAttendMeetings(intervals: number[][]): boolean {
    intervals.sort((a, b) => a[0] - b[0]);
    for (let i = 1; i < intervals.length; ++i) {
        if (intervals[i][0] < intervals[i - 1][1]) {
            return false;
        }
    }
    return true;
}


  • impl Solution {
    #[allow(dead_code)]
    pub fn can_attend_meetings(intervals: Vec<Vec<i32>>) -> bool {
        if intervals.len() == 1 {
            return true;
        }

        let mut intervals = intervals;

        // Sort the intervals vector
        intervals.sort_by(|lhs, rhs| { lhs[0].cmp(&rhs[0]) });

        let mut end = -1;

        // Begin traverse
        for p in &intervals {
            if end == -1 {
                // This is the first pair
                end = p[1];
                continue;
            }
            if p[0] < end {
                return false;
            }
            end = p[1];
        }

        true
    }
}

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