# 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 [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(vector<vector<int>>& clips, int T) {
int res = 0, n = clips.size(), i = 0, st = 0, end = 0;
sort(clips.begin(), clips.end());
while (st < T) {
while (i < n && clips[i][0] <= st) {
end = max(end, clips[i++][1]);
}
if (st == end) return -1;
st = end;
++res;
}
return res;
}
};
``````

``````class Solution {
public:
int videoStitching(vector<vector<int>>& clips, int T) {
sort(clips.begin(), clips.end());
vector<int> dp(101, 101);
dp[0] = 0;
for (auto &clip : clips) {
for (int i = clip[0] + 1; i <= clip[1]; ++i) {
dp[i] = min(dp[i], dp[clip[0]] + 1);
}
}
return dp[T] == 101 ? -1 : dp[T];
}
};
``````

``````class Solution {
public:
int videoStitching(vector<vector<int>>& clips, int T) {
vector<int> dp(T + 1, T + 1);
dp[0] = 0;
for (int i = 1; i <= T; ++i) {
for (auto &clip : clips) {
if (i >= clip[0] && i <= clip[1]) {
dp[i] = min(dp[i], dp[clip[0]] + 1);
}
}
}
return dp[T] == T + 1 ? -1 : dp[T];
}
};
``````

Github 同步地址:

https://github.com/grandyang/leetcode/issues/1024

https://leetcode.com/problems/video-stitching/

https://leetcode.com/problems/video-stitching/discuss/269988/C%2B%2BJava-6-lines-O(n-log-n)

https://leetcode.com/problems/video-stitching/discuss/270036/JavaC%2B%2BPython-Greedy-Solution-O(1)-Space

LeetCode All in One 题目讲解汇总(持续更新中…)

 微信打赏 Venmo 打赏
（欢迎加入博主的知识星球，博主将及时答疑解惑，并分享刷题经验与总结，试运营期间前五十位可享受半价优惠～）

×

Help us with donation