Question

1024. Video Stitching

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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

Solution 1

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

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;
  }
};

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
}