Sunday, May 20, 2018

LeetCode 732 - My Calendar III


https://leetcode.com/problems/my-calendar-iii/description/
Implement a MyCalendarThree class to store your events. A new event can always be added.
Your class will have one method, book(int start, int end). Formally, this represents a booking on the half open interval [start, end), the range of real numbers x such that start <= x < end.
K-booking happens when K events have some non-empty intersection (ie., there is some time that is common to all K events.)
For each call to the method MyCalendar.book, return an integer K representing the largest integer such that there exists a K-booking in the calendar.
Your class will be called like this: MyCalendarThree cal = new MyCalendarThree(); MyCalendarThree.book(start, end)
Example 1:
MyCalendarThree();
MyCalendarThree.book(10, 20); // returns 1
MyCalendarThree.book(50, 60); // returns 1
MyCalendarThree.book(10, 40); // returns 2
MyCalendarThree.book(5, 15); // returns 3
MyCalendarThree.book(5, 10); // returns 3
MyCalendarThree.book(25, 55); // returns 3
Explanation: 
The first two events can be booked and are disjoint, so the maximum K-booking is a 1-booking.
The third event [10, 40) intersects the first event, and the maximum K-booking is a 2-booking.
The remaining events cause the maximum K-booking to be only a 3-booking.
Note that the last event locally causes a 2-booking, but the answer is still 3 because
eg. [10, 20), [10, 40), and [5, 15) are still triple booked.
Note:
  • The number of calls to MyCalendarThree.book per test case will be at most 400.
  • In calls to MyCalendarThree.book(start, end)start and end are integers in the range [0, 10^9].
  • X. Boundary Count
    When booking a new event [start, end), count delta[start]++ and delta[end]--. When processing the values of delta in sorted order of their keys, the largest such value is the answer.
    In Python, we sort the set each time instead, as there is no analog to TreeMap available.
    • Time Complexity: O(N^2), where N is the number of events booked. For each new event, we traverse delta in O(N) time. In Python, this is O(N^2 \log N) owing to the extra sort step.
    • Space Complexity: O(N), the size of delta.
        TreeMap<Integer, Integer> delta;

        public MyCalendarThree() {
            delta = new TreeMap();
        }

        public int book(int start, int end) {
            delta.put(start, delta.getOrDefault(start, 0) + 1);
            delta.put(end, delta.getOrDefault(end, 0) - 1);

            int active = 0, ans = 0;
            for (int d: delta.values()) {
                active += d;
                if (active > ans) ans = active;
            }
            return ans;
        }

    https://leetcode.com/problems/my-calendar-iii/discuss/109556/JavaC++-Clean-Code
    This is to find the maximum number of concurrent ongoing event at any time.
    We can log the start & end of each event on the timeline, each start add a new ongoing event at that time, each end terminate an ongoing event. Then we can scan the timeline to figure out the maximum number of ongoing event at any time.
    The most intuitive data structure for timeline would be array, but the time spot we have could be very sparse, so we can use sorted map to simulate the time line to save space.
    Java
        private TreeMap<Integer, Integer> timeline = new TreeMap<>();
        public int book(int s, int e) {
            timeline.put(s, timeline.getOrDefault(s, 0) + 1); // 1 new event will be starting at [s]
            timeline.put(e, timeline.getOrDefault(e, 0) - 1); // 1 new event will be ending at [e];
            int ongoing = 0, k = 0;
            for (int v : timeline.values())
                k = Math.max(k, ongoing += v);
            return k;
        }
    
    X. Segment Tree
    https://leetcode.com/problems/my-calendar-iii/discuss/109568/Java-Solution-O(n-log(len))-beats-100-Segment-Tree
    This solution is basically a segment tree solution.
    The qurey MyCalendarThree.book(start, end) can be treated as for(i = start; i < end; i++) nums[i] += 1. And the request becomes "find the maximum nums[i]".
    class MyCalendarThree {
     SegmentTree segmentTree;
        public MyCalendarThree() {
         segmentTree = new SegmentTree(0, 1000000000);
        }
        public int book(int start, int end) {
            segmentTree.add(start, end, 1);
            return segmentTree.getMax();
        }
    }
    
    class SegmentTree {
        TreeNode root;
        public SegmentTree(int left, int right) {
            root = new TreeNode(left, right);
        }
        public void add(int start, int end, int val) {
            TreeNode event = new TreeNode(start, end);
         add(root, event, val);
        }
        private void add(TreeNode root, TreeNode event, int val) {
            if(root == null) {
                return ;
            }
            /**
             * If current node's range lies completely in update query range.
             */
            if(root.inside(event)) {
                root.booked += val;
                root.savedres += val;
            }
            /**
             * If current node's range overlaps with update range, follow the same approach as above simple update.
             */
            if(root.intersect(event)) {
             // Recur for left and right children.
                int mid = (root.start + root.end) / 2;
                if(root.left == null) {
                    root.left = new TreeNode(root.start, mid);
                }
                add(root.left, event, val);
                if(root.right == null) {
                    root.right = new TreeNode(mid, root.end);
                }
                add(root.right, event, val);
                // Update current node using results of left and right calls.
                root.savedres = Math.max(root.left.savedres, root.right.savedres) + root.booked;
            }
        }
        public int getMax() {
            return root.savedres;
        }
        /**
         * find maximum for nums[i] (start <= i <= end) is not required.
         * so i did not implement it. 
         */
        public int get(int start, int right) {return 0;}
     class TreeNode {
         int start, end;
         TreeNode left = null, right = null;
         /**
          * How much number is added to this interval(node)
          */
         int booked = 0;
         /**
          * The maximum number in this interval(node). 
          */
         int savedres = 0;
         public TreeNode(int s, int t) {
             this.start = s;
             this.end = t;
         }
         public boolean inside(TreeNode b) {
             if(b.start <= start && end <= b.end) {
                 return true;
             }
             return false;
         }
         public boolean intersect(TreeNode b) {
          if(inside(b) || end <= b.start || b.end <= start) {
                 return false;
             }
             return true;
         }
     }
    }
    
    Time Complexity:
    To query a range maximum, we process at most two nodes at every level and number of levels is O(logn)1. Thus the overall complexity is O(n log(len)), where len is the length of this segment (ie. 10^9 in this problem
    https://leetcode.com/problems/my-calendar-iii/discuss/109569/Question-explanation-please


    The problem asks you to return an integer K representing the largest integer such that there exists a K-booking in the calendar, i.e., the global max number of overlaps. A K-booking happens when there is some time that is common to K events.
    1. In the example when you book (5, 10), the booking itself only causes 2-booking locally, but the global max is still three (made by (10,20), (10,40), (5,15)).
    2. (25,55) causes 2-booking with (10,40) or (50,60), while the global max is still three (made by (10,20), (10,40), (5,15)).

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