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1024. Video Stitching

leetcode-cn Daily Challenge on October 26th, 2020.

Difficulty : Medium

Related Topics : Dynamic Programming

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

Solution

  • mine
    • Java
      • Runtime: 1 ms, faster than 79.91%, Memory Usage: 36.5 MB, less than 13.71% of Java online submissions 
        // O(N * logN)time
// O(N)space
public int videoStitching(int[][] clips, int T) {
    Arrays.sort(clips, (a, b) -> a[0] - b[0]);
    if(clips[0][0] != 0) return -1;
    LinkedList<int[]> list = new LinkedList<>();

    for(int i = 0; i < clips.length; i++){
        if(list.isEmpty()){
            list.add(clips[i]);
            continue;
        }
        int[] t = list.getLast();
        if(clips[i][1] <= t[1]) continue;
        if(clips[i][0] >= T || t[1] >= T) break;

        if(clips[i][0] > t[1]) return -1;

        if(clips[i][0] == t[0]){
            list.removeLast();
            list.add(clips[i]);
            continue;
        }
        int[] last = list.removeLast();
        while(!list.isEmpty() && list.getLast()[1] >= clips[i][0]){
            last = list.removeLast();
        }
        list.add(last);
        list.add(clips[i]);
    }
    int res = list.size();
    if(res == 0 || list.getLast()[1] < T) return -1;
    return res;
}
  • the most votes
  • Runtime: 1 ms, faster than 79.91%, Memory Usage: 36.5 MB, less than 13.71% of Java online submissions 
    // O(N * logN)time
// O(1)space
public int videoStitching(int[][] clips, int T) {
    int res = 0;
    Arrays.sort(clips, (a,b) ->  a[0] - b[0]);
    for (int i = 0, st = 0, end = 0; st < T; st = end, ++res) {
        for (; i < clips.length && clips[i][0] <= st; ++i)
            end = Math.max(end, clips[i][1]);
        if (st == end) return -1;
    }
    return res;
}
  • the fastest code
  • Runtime: 0 ms, faster than 100.00%, Memory Usage: 36.8 MB, less than 13.71% of Java online submissions 
    //O(N* T)time
//O(1)space
public int videoStitching(int[][] clips, int T) {
    //record pre[1]
    int s = 0;
    //record the current max value
    int e = 0;
    int count = 0;
    for (int i = 0; i <= T; i++) {
        e = 0;
        for (int j = 0; j < clips.length; j++) {
            if (clips[j][0] <= s) {
                e = Math.max(e, clips[j][1]);
            }
        }
        count++;
        if (e >= T) {
            return count;
        }
        s = e;
    }
    return -1;
}