[Day 6 of 100] Same as yesterday but more complicated

If you’ve read the day 5 of this series, then today’s exercise is easy for you. Let’s see it.

1. About the problem

Problem

Description: In this problem, we have a calendar and the events in this calendar. Our mission is to allow the inserting of new event such that we prevent the triple booking. A triple booking happens when three events have some non-empty intersection (in other words, when some moment is common to all the three events).
Input: the array of events
Output: the array of bool mentioning either the event can be added or not

2. Solution

public class MyCalendarTwo {
    private List<int[]> _books = [];
    
    public bool Book(int start, int end)
    {
        var intersect = new List<int[]>();
        foreach (var book in _books)
        {
            if (start >= book[1] || book[0] >= end) continue;
            var overlap = new[] { Math.Max(start, book[0]), Math.Min(end, book[1]) };
            if (intersect.Any(i => overlap[0] < i[1] && i[0] < overlap[1])) return false;
            intersect.Add(overlap);
        }
        
        _books.Add([start, end]);
        return true;
    }
}

Result

This exercise is easy and quite similar to yesterday. If you are interested, this is the link. Then that’s it for today. I see ya tomorrow!