252. Meeting Rooms

https://leetcode.com/problems/meeting-rooms/#/description

Given an array of meeting time intervals consisting of start and end times [[s1,e1],[s2,e2],...] (si < ei), determine if a person could attend all meetings.

For example,
Given [[0, 30],[5, 10],[15, 20]],
return false.

Sol:

The end time of the previous interval should be less or equal than the start time of the start time of the next interval....But wait, sort out first! by ascending order of the start time. 

# 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
        """
        if len(intervals) < 2:
            return True
        intervals = sorted(intervals, key = lambda x: x.start)
        for i in range(len(intervals)-1):
            if intervals[i].end > intervals[i+1].start:
                return False
        return True
        
        

Note:

1 The input list is composed of tuples. 

ex. s = [ [11,22], [33, 44] ]

len(s) = 2

2 The problem pre-defines the class of the tuple object.  Tuples are [s, e] and they are defined as:

self.start = s

self.end = e

So when calling the first element of the tuple, we should use intervals.start.  

3 Edge Check! 

4 One way to sort by certain element of a turple. 

intervals = sorted(intervals, key = lambda x: x.start)

1) sorted, key = lambda are the build-in python chars. x can be any name.

2) The sentence above means sort intervals by the start/first element.

3)Two more examples:

ex1.

s = [['b', 2], ['a', 1], ['c', 0]]

s = sorted(s, key = lambda any_name: any_name[0])

print s

==>  [['a', 1], ['b', 2], ['c', 0]]

This is sorted by the first element of the tuple. 

ex2. 

s = [['b', 2], ['a', 1], ['c', 0]]

s = sorted(s, key = lambda any_name: any_name[1])

print s

==>  [['c', 0], ['a', 1], ['b', 2]]

This is sorted by the second element of the tuple. 

 5 Pay attention to the end condition of the index in for loop, especially the loop contains index + 1 or index - 1.  Otherwise it will get  "Line 17: IndexError: list index out of range".