What is the brute force approach and complexity for Count Days Without Meetings?

Mark every day of every meeting in a boolean array, then count unmarked days with time complexity O(D + N · L) and space complexity O(D).

What is the optimal approach and complexity for Count Days Without Meetings?

Merge the meetings into disjoint blocks and add each block's length once with optimal time complexity O(N log N) and space complexity O(1).

How do you identify the pattern for Count Days Without Meetings?

Look for key signals in the prompt: "meetings may overlap" indicates Shared days must be counted once → merge overlapping ranges first.; "days up to 10^9" indicates A per-day array is impossible. Work with the ranges, not the days.; "start and end are both included" indicates A block holds `end − start + 1` days, not `end − start`..

Count Days Without Meetings

Medium·LeetCode #3169 Pro OnlyPattern Transfer

LeetCode

Problem Statement

An employee works on days 1 to days. You get a list of meetings, where meetings[i] = [start, end] means a meeting runs on every day from start to end, including both. Meetings can overlap.

Return how many work days have no meeting at all.

Input

An integer days, and a list meetings of [start, end] pairs.

Output

One integer: the number of days in 1 … days that no meeting touches.

Example 1

Input: days = 10, meetings = [[5,7],[1,3],[9,10]]

Output: 2

Days 4 and 8 have no meeting.

Example 2

Input: days = 5, meetings = [[2,4],[1,3]]

Output: 1

The meetings together cover days 1 to 4. Only day 5 is free.

Example 3

Input: days = 6, meetings = [[1,6]]

Output: 0

One meeting fills every day.

Constraints

• 1 ≤ days ≤ 10^9
• 1 ≤ meetings.length ≤ 10^5
• meetings[i].length == 2
• 1 ≤ start ≤ end ≤ days

Pattern Recognition Clues

Recognising these key signals in the problem description immediately identifies the pattern:

“meetings may overlap”

→ Shared days must be counted once → merge overlapping ranges first.

“days up to 10^9”

→ A per-day array is impossible. Work with the ranges, not the days.

“start and end are both included”

→ A block holds `end − start + 1` days, not `end − start`.

Approach Complexity Ladder

Brute Force — one flag per dayO(D + N · L)O(D)

Mark every day of every meeting in a boolean array, then count unmarked days

Optimal — sort, merge on the fly, subtract covered daysO(N log N)O(1)

Merge the meetings into disjoint blocks and add each block's length once

Brute Force Intuition

Make an array busy of size days + 1. For each meeting, mark every day from start to end. Then count the days from 1 to days that are still unmarked.

Recognition rule for this sub-pattern:

Sort by start. If next.start ≤ current.end, they overlap — extend current.end. Otherwise start a new interval.

Remember this · Intervals

Trigger

Input is [start, end] pairs; overlap, merge, insert, meeting rooms, free time, minimum removals.

Think

After sorting, do I only ever need to compare against the most recent interval?

Key question

Sort by start or by end — which one makes the greedy choice obvious?

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