lc 1024. video stitching

You are given a series of video clips from a sporting event that lasted T seconds. These video clips can be overlapping with each other and have varied lengths.
Each video clip clips[i] is an interval: it starts at time clips[i][0] and ends at time clips[i][1]. 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, T]). If the task is impossible, return -1.

Example 1:
Input: clips = [[0,2],[4,6],[8,10],[1,9],[1,5],[5,9]], T = 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]], T = 5
Output: -1
Explanation:
We can’t cover [0,5] with only [0,1] and [0,2].

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]], T = 9
Output: 3
Explanation:
We can take clips [0,4], [4,7], and [6,9].

Example 4:
Input: clips = [[0,4],[2,8]], T = 5
Output: 2
Explanation:
Notice you can have extra video after the event ends.

This is a greedy problem, we should sort this array ascending by clips[i][0], and then for each overlapping(clips[i][0] <= start(‘end’ in last round)), we should update the end; once end == start, meaning that we can not advance end, we can not cover all of T, thus return -1

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class :
    def videoStitching(self, clips: List[List[int]], T: int) -> int:
        res, start, end, i = 0, 0, 0, 0
        clips.sort()
        while start < T:
            while i < len(clips) and clips[i][0] <= start:
                end = max(end, clips[i][1])
                i += 1
            if end == start: return -1
            start = end
            res += 1
        return res