1024. Video Stitching

Description

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.

• 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, 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

Solutions

• class Solution {
public int videoStitching(int[][] clips, int time) {
int[] last = new int[time];
for (var e : clips) {
int a = e[0], b = e[1];
if (a < time) {
last[a] = Math.max(last[a], b);
}
}
int ans = 0, mx = 0, pre = 0;
for (int i = 0; i < time; ++i) {
mx = Math.max(mx, last[i]);
if (mx <= i) {
return -1;
}
if (pre == i) {
++ans;
pre = mx;
}
}
return ans;
}
}

• class Solution {
public:
int videoStitching(vector<vector<int>>& clips, int time) {
vector<int> last(time);
for (auto& v : clips) {
int a = v[0], b = v[1];
if (a < time) {
last[a] = max(last[a], b);
}
}
int mx = 0, ans = 0;
int pre = 0;
for (int i = 0; i < time; ++i) {
mx = max(mx, last[i]);
if (mx <= i) {
return -1;
}
if (pre == i) {
++ans;
pre = mx;
}
}
return ans;
}
};

• class Solution:
def videoStitching(self, clips: List[List[int]], time: int) -> int:
last = [0] * time
for a, b in clips:
if a < time:
last[a] = max(last[a], b)
ans = mx = pre = 0
for i, v in enumerate(last):
mx = max(mx, v)
if mx <= i:
return -1
if pre == i:
ans += 1
pre = mx
return ans


• func videoStitching(clips [][]int, time int) int {
last := make([]int, time)
for _, v := range clips {
a, b := v[0], v[1]
if a < time {
last[a] = max(last[a], b)
}
}
ans, mx, pre := 0, 0, 0
for i, v := range last {
mx = max(mx, v)
if mx <= i {
return -1
}
if pre == i {
ans++
pre = mx
}
}
return ans
}