To address common interval-related programming problems (since the original problem statement was an image), here are solutions to three frequent ones:

1. Merge Overlapping Intervals

Problem: Given an array of intervals, merge all overlapping intervals and return the merged list.

Approach:

• Sort intervals by start time.
• Iterate and merge adjacent overlapping intervals.

Solution Code:

def merge(intervals):
    if not intervals:
        return []
    intervals.sort()
    merged = [intervals[0]]
    for current in intervals[1:]:
        last = merged[-1]
        if current[0] <= last[1]:
            last[1] = max(last[1], current[1])  # Merge
        else:
            merged.append(current)  # Add non-overlapping
    return merged

Explanation:

• Sorting ensures overlapping intervals are adjacent.
• Merge by updating the end time of the last interval if current starts before/at its end.
• Time complexity: O(n log n) (sorting), Space: O(n) (result list).

2. Maximum Non-Overlapping Intervals

Problem: Select the maximum number of non-overlapping intervals.

Approach:

• Sort intervals by end time (to prioritize earliest ending intervals).
• Pick intervals that don’t overlap with the last selected one.

Solution Code:

def max_non_overlapping(intervals):
    if not intervals:
        return 0
    intervals.sort(key=lambda x: x[1])
    count = 1
    last_end = intervals[0][1]
    for s, e in intervals[1:]:
        if s >= last_end:
            count +=1
            last_end = e
    return count

Explanation:

• Choosing earliest ending intervals maximizes room for subsequent intervals.
• Time complexity: O(n log n), Space: O(1) (excluding input).

3. Minimum Meeting Rooms

Problem: Find the minimum number of conference rooms needed for meetings.

Approach:

• Separate start/end times and sort both.
• Use two pointers to track active rooms (increment if start < end, else free a room).

Solution Code:

def min_meeting_rooms(intervals):
    starts = sorted([i[0] for i in intervals])
    ends = sorted([i[1] for i in intervals])
    rooms = 0
    end_ptr = 0
    for s in starts:
        if s < ends[end_ptr]:
            rooms +=1
        else:
            end_ptr +=1
    return rooms

Explanation:

• Sorting start/end times allows efficient tracking of room usage.
• Time complexity: O(n log n), Space: O(n) (for start/end lists).

Choose the solution that matches your problem! If the original problem differs, share the text statement for a precise answer.
Let me know if you need further clarification.

作者声明:本文包含人工智能生成内容。