LeetCode-in-Java

1024. Video Stitching

Medium

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

Solution

import java.util.Arrays;

public class Solution {
    public int videoStitching(int[][] clips, int time) {
        Arrays.sort(clips, (a, b) -> a[0] == b[0] ? a[1] - b[1] : a[0] - b[0]);
        int count = 0;
        int covered = 0;
        int i = 0;
        int start = 0;
        while (start < time) {
            while (i < clips.length && clips[i][0] <= start) {
                covered = Math.max(covered, clips[i][1]);
                i++;
            }
            if (start == covered) {
                return -1;
            }
            count++;
            start = covered;
        }
        return count;
    }
}

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