Problem Description

You are tasked with the challenge of creating a highlight reel from a series of overlapping video clips taken during a sporting event. The entire event has a duration of time seconds. The video clips are provided in the form of a list, where each clip is represented by a pair of numbers: the start time (start_i) and the end time (end_i). The clips may overlap and can have different lengths. Interestingly, you have the freedom to cut these clips into any number of segments you wish.

The main objective is to find the smallest number of clips necessary to cover the entire duration of the event from 0 to time seconds. If it is not possible to cover the entire event with the given clips, the function should return -1.

Intuition

To solve this problem, we first need to think about how to use the given clips to cover the entire time span of the event. A brute force approach might try to consider all possible combinations of clips, but this would be highly inefficient. Instead, we need a smarter strategy that can quickly determine the optimal set of clips without redundant computations.

The intuition behind the solution is to first figure out, for each second in the event's duration, what is the furthest point we can reach if we were to start a clip at that second. We begin by initializing an array that will hold this information. As we iterate through each clip, we record the farthest end time we can achieve, starting from each start time that is less than time.

Next, we iterate through each second in the event, extending our reach as we find greater end times. The goal is to keep extending our current segment's end time (pre) until we can no longer do so or until we reach the end of the event. If at any point, the farthest reach (mx) is less than or equal to the current second, it means there is a gap in the coverage and hence the task is impossible.

On the other hand, when the current second reaches the previous maximum reach (pre), it indicates that we have utilized a clip to its fullest extent, and we need to select another clip to extend our coverage. This selection increments our answer to the minimum number of clips needed. Eventually, if we are able to cover the whole event, we return the number of clips used; otherwise, if we encounter a coverage gap, we return -1.

This strategy is efficient because it takes advantage of the greedy approach, always picking the clip that extends coverage the furthest at each step, and ensuring that a minimum number of clips are used. The algorithm will sweep through the time units only once, maintaining the maximum reach and updating the number of clips used as needed.

Learn more about Greedy and Dynamic Programming patterns.

Solution Approach

The solution involves a greedy algorithm that aims to extend the reach of the current clip segment as far as possible within the given time. Let's walk through the implementation:

• First, we initialize a list called last with a length equal to the time. This list is used to track the furthest end time (last[a]) that can be reached from each starting second a.

• The solution iterates over each clip [a, b] from the provided clips list. For each clip, it checks if the start time a is within the event duration (a < time). If so, it updates last[a] with the maximum of its current value and the end time b of the current clip. This step effectively finds the furthest end time achievable from each start time.

last = [0] * time
for a, b in clips:
    if a < time:
        last[a] = max(last[a], b)

• We then initialize three variables: ans to keep track of the number of clips used, mx to store the furthest end time reachable at any point, and pre which represents the end of the current clip segment we are creating.

• The next loop goes through each second (index i) in the last list, updating mx to the maximum of its current value and last[i] which represents the furthest time we can reach from second i. Here are two important conditions:

  • If at any point mx (the maximum reach so far) is less than or equal to the current second i, this means that there is a gap in our clip coverage and we cannot create a continuous highlight reel up to this second. Hence, we return -1.

  • If the current second i equals pre (the previous segment's farthest reach), it means that we need to start a new segment by selecting another clip. We increment ans (the number of clips used), and update pre to mx to reflect the new segment's farthest reach.

ans = mx = pre = 0
for i, v in enumerate(last):
    mx = max(mx, v)
    if mx <= i:
        return -1
    if pre == i:
        ans += 1
        pre = mx

• After the loop is completed, if we are able to reach the end of the event's duration (time), the variable ans will hold the minimum number of clips needed. The function returns this value.

This approach uses a greedy algorithm for selecting clips, as well as dynamic programming techniques to efficiently calculate the maximum coverage at each point in time. By continuously updating the furthest reach and tracking the number of clips, the algorithm ensures that it always progresses towards the goal and only uses the necessary clips to cover the entire event duration.

Example Walkthrough

Let's illustrate the solution approach with a small example. Suppose we have an event that lasts for 10 seconds (time = 10), and we have the following video clips provided in the form of (start_i, end_i) pairs:

Clips = [(0, 3), (1, 5), (4, 8), (7, 10)]

Following the solution approach:

1. Initialize the last list with a length equal to time (which is 10 in this case), to track the furthest end time we can reach from each starting second.

   last = [0]*10  # [0, 0, 0, 0, 0, 0, 0, 0, 0, 0]

2. Iterate through each clip to update the last list.

   • For clip (0, 3), since 0 < 10, update last[0] to 3.
   • For clip (1, 5), since 1 < 10, update last[1] to 5.
   • For clip (4, 8), since 4 < 10, update last[4] to 8.
   • For clip (7, 10), since 7 < 10, update last[7] to 10.

   After iterating, last looks like this:

   last = [3, 5, 0, 0, 8, 0, 0, 10, 0, 0]

3. Initialize ans (the number of clips used), mx (the farthest end time reachable at any point), and pre (the end of the current clip segment) to 0.

4. Iterate through the last list to find the minimum number of clips:

   • For i = 0, mx becomes max(0, 3), so mx = 3.
   • For i = 1, mx becomes max(3, 5), so mx = 5.
   • Since i has not reached pre yet, we continue without incrementing ans.
   • For i = 2, mx remains 5 as last[2] is 0 and 5 > 2.
   • ...
   • When i = 3 and pre = 3, we find our first segment covering from second 0 to second 5. Increment ans to 1, and update pre to mx (5).
   • Continue this process until i=7.
   • For i = 7, mx becomes max(5, 10), so mx = 10.
   • i = 7 is equal to pre which is 5, so increment ans to 2, and update pre to mx (10).
   • We now have covered the duration from 0 to 10 seconds with 2 segments.

5. Finally, since the last list has been fully traversed and we've been able to cover every second up to the end of the event (10), we conclude that 2 is the minimum number of clips needed to cover the event.

Following this example, the function would return 2 since we can cover the entire event with the [(1, 5), (7, 10)] clips.

Solution Implementation

Time and Space Complexity

Time Complexity

The provided Python code has a for loop that iterates through the clips list, which has a length that we can refer to as n. Inside this loop, we have constant time operations (if comparison and max function), meaning that this part of the algorithm has a time complexity of O(n).

Following that, there is another for loop which iterates through the last list. Since last has a length equal to the time parameter, this loop iterates time times. Once again, we perform constant time operations within the loop (max function, if comparisons, and simple assignments).

Therefore, the second loop has a time complexity of O(time). Since these two loops are not nested but executed in sequence, the overall time complexity of the code combines both complexities, resulting in O(n + time).

Space Complexity

For space complexity, the code allocates an array last of length equal to time, which requires O(time) space. The rest of the variables used in the code (ans, mx, and pre) require constant space, O(1).

Therefore, the overall space complexity of the function is O(time) because this is the largest space requirement that does not change with different inputs of clips.

Learn more about how to find time and space complexity quickly using problem constraints.