How do you solve Count Days Without Meetings on LeetCode?

Count Days Without Meetings (LeetCode #3169) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Merging intervals is crucial to reduce overlaps.

What is the time complexity of Count Days Without Meetings?

The optimal solution for Count Days Without Meetings runs in O(n log n) time and uses O(n) space. The sorting step takes O(n log n) time, and merging the meetings takes O(n) time. The space complexity is O(n) due to the merged list.

What is the brute force solution for Count Days Without Meetings?

The brute force approach for Count Days Without Meetings is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we will check each day to see if it falls within any of the meeting intervals. If it doesn't, we count it as an available day. This is straightforward but inefficient for large inputs. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Count Days Without Meetings?

Count Days Without Meetings uses the following concepts: Array, Sorting. The recommended patterns to study are: Array, Sorting.

What are the interview tips for Count Days Without Meetings?

Explain your thought process clearly. Consider edge cases before finalizing your solution. Practice merging intervals as it's a common pattern in interviews.

What are common mistakes when solving Count Days Without Meetings?

Not merging overlapping meetings. Forgetting to handle edge cases like no meetings.

#3169

Count Days Without Meetings

Medium

ArraySortingArraySorting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

💡

Intuition

Time O(n log n)Space O(n)

The optimal approach involves merging overlapping meetings first, then counting the gaps between these merged intervals. This is efficient because it reduces the number of checks we need to perform.

⚙️

Algorithm

3 steps

1Step 1: Sort the meetings by their start time.
2Step 2: Merge overlapping meetings into a new list.
3Step 3: Count the available days by calculating gaps between merged meetings and the total days.

solution.py20 lines

1# Full working Python code
2
3def count_days_without_meetings(days, meetings):
4    if not meetings:
5        return days
6    meetings.sort()
7    merged = []
8    for start, end in meetings:
9        if not merged or merged[-1][1] < start:
10            merged.append([start, end])
11        else:
12            merged[-1][1] = max(merged[-1][1], end)
13    available_days = 0
14    last_end = 0
15    for start, end in merged:
16        available_days += start - last_end - 1
17        last_end = end
18    available_days += days - last_end
19    return available_days
20

ℹ

Complexity note: The sorting step takes O(n log n) time, and merging the meetings takes O(n) time. The space complexity is O(n) due to the merged list.

1Merging intervals is crucial to reduce overlaps.
2Counting gaps between meetings efficiently helps find available days.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.