1024. Video Stitching

Posted 2021-03-14 wentiliangkaihua

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. 1 <= clips.length <= 100
2. 0 <= clips[i][0], clips[i][1] <= 100
3. 0 <= T <= 100

class Solution {
    public int videoStitching(int[][] clips, int T) {
        Arrays.sort(clips, (a, b) -> a[0] - b[0]);
        int res = 0;
        for(int i = 0, st = 0, end = 0; st < T; res++, st = end){
            for(; i < clips.length && clips[i][0] <= st; i++){
                end = Math.max(end, clips[i][1]);
            }
            if(st == end) return -1;
        }
        return res;
    }
}

 

 

 https://leetcode.com/problems/video-stitching/discuss/269988/C%2B%2BJava-6-lines-O(n-log-n)