LeetCode 1024. Video Stitching

原题链接在这里：https://leetcode.com/problems/video-stitching/

题目：

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

题解：

Could sort the clips based on starting time.

For each step, go through all the clips having starting time <= current start time, get to the furtherest.

if furtherest doesn't change, start == end, that means there is a gap. return -1.

Otherwise, update start to end and increment res.

End as soon as current furtherest reaches T.

Time Complexity: O(nlogn). n = clips.length.

Space: O(1).

AC Java:

 1 class Solution {
 2     public int videoStitching(int[][] clips, int T) {
 3         if(clips == null || clips.length == 0 || clips[0].length != 2 || T <= 0){
 4             return -1;
 5         }
 6
 7         Arrays.sort(clips, (a, b) -> a[0] - b[0]);
 8
 9         int start = 0;
10         int end = 0;
11         int res = 0;
12         int i = 0;
13         while(end < T){
14             while(i<clips.length && clips[i][0] <= start){
15                 end = Math.max(end, clips[i][1]);
16                 i++;
17             }
18
19             if(start == end){
20                 return -1;
21             }
22
23             start = end;
24             res++;
25         }
26
27         return res;
28     }
29 }

Let dp[i] denotes minimum number of clips needed to cover up to i.

For each clip, if clip[0] <= i <= clip[1], then i could be covered using minimum clips covered up to clip[0] plus this clip.

Update dp[i].

Check if dp[T] is never updated. If it is never updated, then return -1.

Time Complexity: O(nT). n = clips.length.

Space: O(T).

AC Java:

 1 class Solution {
 2     public int videoStitching(int[][] clips, int T) {
 3         int [] dp = new int[T+1];
 4         dp[0] = 0;
 5
 6         for(int i = 1; i<=T; i++){
 7             dp[i] = T+1;
 8             for(int [] clip : clips){
 9                 if(clip[0]<=i && clip[1]>=i){
10                     dp[i] = Math.min(dp[i], dp[clip[0]]+1);
11                 }
12             }
13         }
14
15         return dp[T] == T+1 ? -1 : dp[T];
16     }
17 }