732. My Calendar III LeetCode Solution

In this guide, you will get 732. My Calendar III LeetCode Solution with the best time and space complexity. The solution to My Calendar III problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Table of Contents

1. Problem Statement
2. Complexity Analysis
3. My Calendar III solution in C++
4. My Calendar III solution in Java
5. My Calendar III solution in Python
6. Additional Resources

Problem Statement of My Calendar III

A k-booking happens when k events have some non-empty intersection (i.e., there is some time that is common to all k events.)
You are given some events [startTime, endTime), after each given event, return an integer k representing the maximum k-booking between all the previous events.
Implement the MyCalendarThree class:

MyCalendarThree() Initializes the object.
int book(int startTime, int endTime) Returns an integer k representing the largest integer such that there exists a k-booking in the calendar.

Example 1:

Input
[“MyCalendarThree”, “book”, “book”, “book”, “book”, “book”, “book”]
[[], [10, 20], [50, 60], [10, 40], [5, 15], [5, 10], [25, 55]]
Output
[null, 1, 1, 2, 3, 3, 3]

Explanation
MyCalendarThree myCalendarThree = new MyCalendarThree();
myCalendarThree.book(10, 20); // return 1
myCalendarThree.book(50, 60); // return 1
myCalendarThree.book(10, 40); // return 2
myCalendarThree.book(5, 15); // return 3
myCalendarThree.book(5, 10); // return 3
myCalendarThree.book(25, 55); // return 3

Constraints:

0 <= startTime < endTime <= 109
At most 400 calls will be made to book.

Complexity Analysis

• Time Complexity: O(n\log n)
• Space Complexity: O(n)

732. My Calendar III LeetCode Solution in C++

class MyCalendarThree {
 public:
  int book(int start, int end) {
    ++timeline[start];
    --timeline[end];

    int ans = 0;
    int activeEvents = 0;

    for (const auto& [_, count] : timeline) {
      activeEvents += count;
      ans = max(ans, activeEvents);
    }

    return ans;
  }

 private:
  map<int, int> timeline;
};
/* code provided by PROGIEZ */

732. My Calendar III LeetCode Solution in Java

class MyCalendarThree {
  public int book(int start, int end) {
    timeline.merge(start, 1, Integer::sum);
    timeline.merge(end, -1, Integer::sum);

    int ans = 0;
    int activeEvents = 0;

    for (final int count : timeline.values()) {
      activeEvents += count;
      ans = Math.max(ans, activeEvents);
    }

    return ans;
  }

  private Map<Integer, Integer> timeline = new TreeMap<>();
}
// code provided by PROGIEZ

732. My Calendar III LeetCode Solution in Python

from sortedcontainers import SortedDict


class MyCalendarThree:
  def __init__(self):
    self.timeline = SortedDict()

  def book(self, start: int, end: int) -> int:
    self.timeline[start] = self.timeline.get(start, 0) + 1
    self.timeline[end] = self.timeline.get(end, 0) - 1

    ans = 0
    activeEvents = 0

    for count in self.timeline.values():
      activeEvents += count
      ans = max(ans, activeEvents)

    return ans
 # code by PROGIEZ

Additional Resources

• Explore all LeetCode problem solutions at Progiez here
• Explore all problems on LeetCode website here

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