[leetcode] 1024. Video Stitching

Description

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 [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]], 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.      

Constraints:

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

分析

題目的意思是：給定clips數組，表示的是視訊的區間，求出能夠組成T的最小視訊區間的個數。這道題求最小，用動态規劃的方法就是初始化為inf，用dp[i]表示視訊片段[0,i)區間的最小長度，這樣從1，周遊到T，得到dp[T]就是所求了。

對于任意區間c1,c2,如果滿足i>c1 i<=c2,那麼它可以覆寫[0,i)的後半部分，前半部分則用dp[c1]來計算就行了，是以遞推公式為：

dp[i]=min(dp[i],dp[c1]+1)      

代碼

class Solution:
    def videoStitching(self, clips: List[List[int]], T: int) -> int:
        dp=[float('inf')]*(T+1)
        dp[0]=0
        for i in range(1,T+1):
            for c1,c2 in clips:
                if(i>c1 and i<=c2):
                    dp[i]=min(dp[i],dp[c1]+1)
        if(dp[T]==float('inf')):
            return -1
        return dp[T]      

參考文獻