Video Stitching

You are given a series of video clips from a sporting event that lastedTseconds. These video clips can be overlapping with each other and have varied lengths.

Each video clipclips[i] is an interval: it starts at timeclips[i][0]and ends at timeclips[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

分析

DP存最小个数：一loop所有长度，内loop所有range,长度在range内的话的dp取最小。

class Solution:
    def videoStitching(self, clips: List[List[int]], T: int) -> int:
       # clips.sort(key = lambda x:(x[1],x[0])) 可有可无
        dp = [float('inf')]* (T+1)
        dp[0]=0
        for t in range(T+1):           
            for s,e in clips:
                if s <= t<=e:
                    dp[t] = min(dp[t],dp[s]+1)

            if dp[t] == float('inf'):
                return -1
        return dp[-1]

dp存最远可达距离

class Solution:
    def videoStitching(self, clips: List[List[int]], T: int) -> int:
        n = len(clips)
        dp = [-1]*(n+1) #最远能到达的距离
        dp[0] = 0
        for i in range(1,n+1):
            for s,e in clips:               
                if dp[i-1] >= s:
                    dp[i] = max(dp[i],e)
            if dp[i] >= T:
                return i
        return -1

贪心

class Solution:
    def videoStitching(self, clips: List[List[int]], T: int) -> int:
        preend,end,res=-1,0,0
        for s,e in sorted(clips):
            if s > end or end>= T:
                break
            if preend < s <= end:
                res,preend = res+1,end
            end = max(end,e)
        return res if end >= T else -1