15. Video Stitching

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 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]], time = 9
Output: 3
Explanation: We can take clips [0,4], [4,7], and [6,9].

Example 4:

Input: clips = [[0,4],[2,8]], time = 5
Output: 2
Explanation: Notice you can have extra video after the event ends.

Solution: (DP)

class Solution
{
public:
    int videoStitching(vector<vector<int>> &clips, int time)
    {

        int n = clips.size();
        vector<int> dp(time + 1, n + 1);

        sort(clips.begin(), clips.end());
        dp[0] = 0;

        for (int i = 0; i < clips.size(); i++)
        {

            int start = clips[i][0];
            int end = clips[i][1];

            for (int j = start; j <= min(end, time); j++)
            {
                dp[j] = min(dp[j], dp[start] + 1);
            }
        }

        if (dp[time] == n + 1)
        {
            return -1;
        }

        return dp[time];
    }
};

Time Complexity: O(n * time + nlog n)

Solution: (Greedy)

Approach: (Similar to Jump Game) Tracking the max jump that can be taken from current position

class Solution
{
public:
    int videoStitching(vector<vector<int>> &clips, int time)
    {

        sort(clips.begin(), clips.end());

        int maxEnd = 0;

        int count = 0;

        int i = 0;
        while (maxEnd < time)
        {
            int curEnd = 0;
            while (i < clips.size() && clips[i][0] <= maxEnd)
            {
                curEnd = max(curEnd, clips[i][1]);
                i++;
            }

            if (curEnd <= maxEnd)
            {
                return -1;
            }

            count++;
            maxEnd = curEnd;
        }

        return count;
    }
};

Time Complexity: O(n log n)