Medium
You are given a series of video clips from a sporting event that lasted time
seconds. These video clips can be overlapping with each other and have varying lengths.
Each video clip is described by an array clips
where clips[i] = [starti, endi]
indicates that the ith clip started at starti
and ended at endi
.
We can cut these clips into segments freely.
[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, time]
. If the task is impossible, return -1
.
Example 1:
Input: clips = [[0,2],[4,6],[8,10],[1,9],[1,5],[5,9]], time = 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]], time = 5
Output: -1
Explanation: We cannot 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]], time = 9
Output: 3
Explanation: We can take clips [0,4], [4,7], and [6,9].
Constraints:
1 <= clips.length <= 100
0 <= starti <= endi <= 100
1 <= time <= 100
class Solution {
fun videoStitching(clips: Array<IntArray>, time: Int): Int {
clips.sortWith { a: IntArray, b: IntArray ->
if (a[0] == b[0]
) {
a[1] - b[1]
} else {
a[0] - b[0]
}
}
var count = 0
var covered = 0
var i = 0
var start = 0
while (start < time) {
while (i < clips.size && clips[i][0] <= start) {
covered = covered.coerceAtLeast(clips[i][1])
i++
}
if (start == covered) {
return -1
}
count++
start = covered
}
return count
}
}