Solving Leetcode Interviews in Seconds with AI: Count Days Without Meetings
Introduction
In this blog post, we will explore how to solve the LeetCode problem "3169" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.
Problem Statement
You are given a positive integer days representing the total number of days an employee is available for work (starting from day 1). You are also given a 2D array meetings of size n where, meetings[i] = [start_i, end_i] represents the starting and ending days of meeting i (inclusive). Return the count of days when the employee is available for work but no meetings are scheduled. Note: The meetings may overlap. Example 1: Input: days = 10, meetings = [[5,7],[1,3],[9,10]] Output: 2 Explanation: There is no meeting scheduled on the 4th and 8th days. Example 2: Input: days = 5, meetings = [[2,4],[1,3]] Output: 1 Explanation: There is no meeting scheduled on the 5th day. Example 3: Input: days = 6, meetings = [[1,6]] Output: 0 Explanation: Meetings are scheduled for all working days. Constraints: 1 <= days <= 109 1 <= meetings.length <= 105 meetings[i].length == 2 1 <= meetings[i][0] <= meetings[i][1] <= days
Explanation
Here's a breakdown of the optimal approach, complexity analysis, and Python code:
Approach:
- Sort the meetings by their start times to efficiently merge overlapping intervals.
- Merge overlapping meeting intervals into a consolidated list of busy periods.
- Iterate through the merged intervals, summing up the free days between the start and end of the work period and the consolidated meeting schedule.
Complexity:
- Runtime: O(n log n) due to sorting the meetings.
- Storage: O(n) in the worst case to store the merged intervals.
Code
def available_days(days: int, meetings: list[list[int]]) -> int:
"""
Calculates the number of days an employee is available for work given their total working days and meeting schedule.
Args:
days: The total number of days the employee is available for work.
meetings: A list of meetings, where each meeting is a list of two integers representing the start and end days.
Returns:
The number of days when the employee is available for work but no meetings are scheduled.
"""
meetings.sort() # Sort meetings by start time
merged_meetings = []
if meetings:
merged_meetings.append(meetings[0])
for i in range(1, len(meetings)):
current_meeting = meetings[i]
last_merged_meeting = merged_meetings[-1]
if current_meeting[0] <= last_merged_meeting[1]:
# Overlapping meetings, merge them
last_merged_meeting[1] = max(last_merged_meeting[1], current_meeting[1])
else:
# Non-overlapping meetings, add the current meeting to the merged list
merged_meetings.append(current_meeting)
available_days_count = days
for meeting in merged_meetings:
available_days_count -= (meeting[1] - meeting[0] + 1)
return available_days_count