meeting room interval

Meeting room Interval

// Iterate through all the interval of input, let's call it itv,
// For each itv, find a Interval in the Set which ends is less than or equal to itv.start
// 	-> This one, we can use binary search, because the Set is sorted based on it's end
// If no such Interval exists in the Set, create a new Interval, and append it to the Set
// If there is such Interval, update it's start and end to itv.start and itv.end

Example:
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   Solution 2: provided by Julia            
  sweep line algorithm
  
  |______________________  0, 30
  |
  |  _____  ______
  |  
    5  10  15   20
    
    sort by start time, 0, 5, 15,  O(nlogn)
    sort by end time,  10, 20, 30, O(nlogn)
    two pointert -> lef right 
    
    0 - +1 
    5 - +1  -> maximum = 2
    10 - close -1 -> go back to 1
    15 - +1 , maximum = 2
    20 - - 1, maximum = 1
    30 -