Most Meetings You Can Attend When Each Takes One Day Within Its Range
Company: Snowflake
Role: Software Engineer
Category: Coding & Algorithms
Difficulty: medium
Interview Round: Technical Screen
You are given a list of meetings, each of which can be attended on any one day within a range of days. Attending a meeting takes one whole day, and you can attend at most one meeting per day. Return the largest number of meetings you can attend.
### Function Signature
```python
def max_meetings_attended(meetings: list[list[int]]) -> int:
```
### Rules
- `meetings[i] = [start, end]` means meeting `i` can be attended on any single day `d` with `start <= d <= end`; both ends are inclusive.
- Each meeting counts at most once, and each day can be used for at most one meeting.
### Constraints
- `1 <= len(meetings) <= 10^5`
- `1 <= start <= end <= 10^5` for every meeting
### Examples
**Example 1**
```text
Input: meetings = [[2, 3], [1, 2], [1, 2], [3, 4]]
Output: 4
```
The two meetings over days 1 to 2 take days 1 and 2, the meeting over days 2 to 3 takes day 3, and the meeting over days 3 to 4 takes day 4.
**Example 2**
```text
Input: meetings = [[1, 1], [1, 1], [1, 2]]
Output: 2
```
Both meetings that are available only on day 1 need that day, so only one of them can be attended; the third takes day 2.
**Example 3**
```text
Input: meetings = [[1, 2], [1, 2], [1, 2], [3, 3]]
Output: 3
```
Only two of the three meetings over days 1 to 2 fit, and the last meeting takes day 3.
Overview: Each meeting can be attended on any single day within its range, and only one meeting fits on each day; return the largest number of meetings you can attend. It tests interval scheduling reasoning and an exchange argument for why a local choice stays optimal.
You are given a list of meetings. Meeting `i` is described by `meetings[i] = [start, end]` and can be attended on any single day `d` with `start <= d <= end`; both ends are inclusive. Attending a meeting takes one whole day, and you can attend at most one meeting per day. Each meeting counts at most once, and each day can be used for at most one meeting.
Return the largest number of meetings you can attend.
The meetings may be listed in any order, and the same `[start, end]` pair may appear more than once; every copy is a separate meeting.
### Function Signature
```python
def max_meetings_attended(meetings: list[list[int]]) -> int:
```
### Constraints
- `1 <= len(meetings) <= 10^5`
- `meetings[i] = [start, end]` with `1 <= start <= end <= 10^5` for every meeting
- No value can exceed 2^31 - 1: every day and the answer (at most `len(meetings)`) fit in a 32-bit signed integer.
### Example 1
```text
Input: meetings = [[2, 3], [1, 2], [1, 2], [3, 4]]
Output: 4
```
The two meetings over days 1 to 2 take days 1 and 2, the meeting over days 2 to 3 takes day 3, and the meeting over days 3 to 4 takes day 4.
### Example 2
```text
Input: meetings = [[1, 1], [1, 1], [1, 2]]
Output: 2
```
Both meetings that are available only on day 1 need that day, so only one of them can be attended; the third takes day 2.
### Example 3
```text
Input: meetings = [[1, 2], [1, 2], [1, 2], [3, 3]]
Output: 3
```
Only two of the three meetings over days 1 to 2 fit, and the last meeting takes day 3.
Constraints
- 1 <= len(meetings) <= 10^5
- meetings[i] = [start, end] with 1 <= start <= end <= 10^5 for every meeting; both ends are inclusive
- Each meeting counts at most once and each day can be used for at most one meeting
- No value can exceed 2^31 - 1: every day and the answer fit in a 32-bit signed integer
Examples
Input: ([[1, 1]],)
Expected Output: 1
Explanation: Minimum valid input: a single one-day meeting is attended; an exclusive-end reading would return 0.
Input: ([[2, 3], [1, 2], [1, 2], [3, 4]],)
Expected Output: 4
Explanation: Source Example 1: unsorted input with a duplicate pair; the meetings take days 1, 2, 3 and 4.
Hints
- Example 3 shows a capacity limit: when more meetings than days fall inside the same window, some of them must be skipped no matter how the rest are scheduled.
- When several meetings could use the same day, ask which choice keeps the most options open for the meetings that are still unscheduled.
- Both ends of a window are usable days, and identical [start, end] pairs are separate meetings that compete for the same days.