Video Stitching

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

class Solution {
    public int videoStitching(int[][] clips, int T) {
        Arrays.sort(clips, (a, b) -> a[0] == b[0] ? b[1] - a[1] : a[0] - b[0]);
        if (clips[0][0] != 0)
            return -1;
        int count = 1;
        int ending = clips[0][1];
        for (int i = 1; i < clips.length; i++) {
            if (ending >= T)
                break;
            int maxEnd = ending;
            while (i < clips.length && clips[i][0] <= ending) {
                maxEnd = Math.max(maxEnd, clips[i][1]);
                i++;
            }
            if (ending != maxEnd) {
                i--;
                ending = maxEnd;
                count++;
            } else
                break;
        }
        return ending >= T ? count : -1;
    }
}