LeetCode Q 1024 - 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 [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:

Solution 1: Sorting

Code:

public int videoStitching(int[][] clips, int T) {
  Arrays.sort(clips, (a, b) -> {
    if (a[0] == b[0]) return a[1] - b[1];
    return a[0] - b[0];
  });

  int len = clips.length;
  if (clips[0][0] != 0 || clips[len - 1][1] < T) return -1;
  
  int curEnd = 0, count = 0;
  for (int i = 0; i < len; ) {
    if (curEnd < clips[i][0]) return -1;
    int maxEnd = curMax;
    while (i < len && clips[i][0] <= curEnd) {
      maxEnd = Math.max(maxEnd, clips[i++][1]);
    }
    curEnd = maxEnd;
    count++;
    if (curEnd >= T) return count;
  }

  return -1;
}