Introduction

In this blog post, we will explore how to solve the LeetCode problem "1024" 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 series of video clips from a sporting event that lasted time seconds. These video clips can be overlapping with each other and have varying lengths. Each video clip is described by an array clips where clips[i] = [starti, endi] indicates that the ith clip started at starti and ended at endi. We can cut these clips into segments freely. For example, a clip [0, 7] can be cut into segments [0, 1] + [1, 3] + [3, 7]. Return the minimum number of clips needed so that we can cut the clips into segments that cover the entire sporting event [0, time]. If the task is impossible, return -1. Example 1: Input: clips = [[0,2],[4,6],[8,10],[1,9],[1,5],[5,9]], time = 10 Output: 3 Explanation: We take the clips [0,2], [8,10], [1,9]; a total of 3 clips. Then, we can reconstruct the sporting event as follows: We cut [1,9] into segments [1,2] + [2,8] + [8,9]. Now we have segments [0,2] + [2,8] + [8,10] which cover the sporting event [0, 10]. Example 2: Input: clips = [[0,1],[1,2]], time = 5 Output: -1 Explanation: We cannot cover [0,5] with only [0,1] and [1,2]. Example 3: Input: clips = [[0,1],[6,8],[0,2],[5,6],[0,4],[0,3],[6,7],[1,3],[4,7],[1,4],[2,5],[2,6],[3,4],[4,5],[5,7],[6,9]], time = 9 Output: 3 Explanation: We can take clips [0,4], [4,7], and [6,9]. Constraints: 1 <= clips.length <= 100 0 <= starti <= endi <= 100 1 <= time <= 100

Explanation

Here's a breakdown of the approach, complexities, and the Python code:

• Greedy Approach: Sort the clips based on their start times. Iterate through the sorted clips, greedily selecting the clip that extends the current coverage the furthest.
• Coverage Tracking: Maintain a variable to track the current reachable time. In each step, choose the clip that maximizes the reachable time while starting no later than the current reachable time.

• Impossibility Check: If at any point, we cannot extend the coverage, it means the task is impossible.

• Runtime Complexity: O(n log n), dominated by sorting the clips. Storage Complexity: O(n), where n is the number of clips (due to sorting in place, if not possible it will be O(n) for storing the sorted list).

Code

    def videoStitching(clips, time):
    """
    Finds the minimum number of clips needed to cover the time interval [0, time].

    Args:
        clips: A list of video clips, where each clip is a list [start, end].
        time: The total time of the sporting event.

    Returns:
        The minimum number of clips needed, or -1 if impossible.
    """

    clips.sort()  # Sort clips by start time
    count = 0
    reachable = 0
    i = 0
    max_reach = 0

    while reachable < time:
        max_reach = reachable  # Initialize max_reach for this iteration

        while i < len(clips) and clips[i][0] <= reachable:
            max_reach = max(max_reach, clips[i][1])
            i += 1

        if max_reach == reachable:  # No progress, impossible to cover
            return -1

        reachable = max_reach
        count += 1

    return count