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.

Note:

1 <= clips.length <= 100
0 <= clips[i][0], clips[i][1] <= 100
0 <= T <= 100

思路：

DP，dp[i]表示从0到i位置都有的情况下，选取的最小值

然后每来一个新的Interval【s,t】，更新dp[s]..dp[t]，复杂度正好10^6

为了保证interval.s小的先算，先排下序就完了

class Solution(object):
    def videoStitching(self, clips, T):
        """
        :type clips: List[List[int]]
        :type T: int
        :rtype: int
        """
        clips.sort(key=lambda a:a[0])
        dp=[float('inf')]*(T+1)
        dp[0] = 0
        for s,t in clips:
            for i in range(s,min(t,T)+1):
                for j in range(s,i):
                    dp[i]=min(dp[i], dp[j]+1)
        return dp[T] if dp[T]<float('inf') else -1

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