1024. Video Stitching.md

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1024. Video Stitching
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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

Solution

• greedy and sort
• keep appending the interval to a list [interval 1, interval 2] and global_end means the rightmost distance I can reach
• then keep searching later interval and find the current farther reach and compare with global_end

class Solution {
    public int videoStitching(int[][] clips, int T) {
        //sort them by left point 
        //scan through the inteval in order
        //      1. mark the right point as the 
        int i = 0;
        int n = clips.length;
        int global_end = 0; //globalend 
        int farReach = 0;
        int count = 0;
        Arrays.sort(clips, (a, b) -> (a[0] - b[0])); // sort by start time
        for( ; global_end < T; ){
            
            farReach = 0;
            //far distance I can reach under such end(righttest can reach)
            while(i < n && global_end >= clips[i][0]){
                farReach = Math.max(farReach, clips[i++][1]);
            }
            
            //break condition
            if(farReach<=global_end) return -1;
            
            //update the global end
            global_end = farReach;
            count++;
        }
        return count;
    }
}

Great reference

• https://leetcode.com/problems/minimum-number-of-taps-to-open-to-water-a-garden/discuss/506853/Java-A-general-greedy-solution-to-process-similar-problems