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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

Solution 1: Greedy

Note that if there are multiple sub-intervals with the same starting point, it is optimal to choose the one with the largest right endpoint.

Therefore, we can preprocess all sub-intervals. For each position $i$, calculate the largest right endpoint among all sub-intervals starting at $i$, and record it in the array $last[i]$.

We define a variable mx to represent the farthest position that can currently be reached, a variable ans to represent the current minimum number of sub-intervals needed, and a variable pre to represent the right endpoint of the last used sub-interval.

Next, we start enumerating all positions $i$ from $0$, using $last[i]$ to update mx. If after updating, $mx = i$, it means that the next position cannot be covered, so the task cannot be completed, return $-1$.

At the same time, we record the right endpoint pre of the last used sub-interval. If $pre = i$, it means that a new sub-interval needs to be used, so we add $1$ to ans and update pre to mx.

After the traversal is over, return ans.

The time complexity is $O(n+m)$, and the space complexity is $O(m)$. Where $n$ and $m$ are the lengths of the clips array and the value of time, respectively.

Similar problems:

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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

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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;
    }
}

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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;
    }
};

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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
}

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