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:
class Solution {
public int videoStitching(int[][] clips, int T) {
Arrays.sort(clips, (a, b) -> a[0] - b[0] != 0 ? a[0] - b[0] : b[1] - a[1]);
if (clips[0][0] != 0) return -1;
if (clips[0][1] >= T) return 1;
List<int[]> arr = new ArrayList();
arr.add(clips[0]);
// drop clips that have same start but lower end because we would never pick them
for (int i = 1; i < clips.length; i ++) {
if (clips[i][0] != arr.get(arr.size() - 1)[0]) {
arr.add(clips[i]);
}
}
int curr = 0;
int count = 1;
while (arr.get(curr)[1] < T) {
int next = -1;
for (int i = curr + 1; i < arr.size(); i ++) {
if (arr.get(i)[0] > arr.get(curr)[0] && arr.get(i)[0] <= arr.get(curr)[1]) {
// find the next clip, which can connect to current clip with largest end
if (arr.get(i)[1] > arr.get(curr)[1] && (next == -1 || arr.get(i)[1] > arr.get(next)[1])) {
next = i;
}
} else {
break;
}
}
if (next == -1) return -1;
if (arr.get(next)[1] >= T) return count + 1;
count ++;
curr = next;
}
return -1;
}
}