LeetCode 1024 Video Stitching (dp，jump game)

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

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

The main idea of the topic ： Yes n Intervals , Each section can be cut arbitrarily , Find at least a few intervals to cover [0, time]

Topic analysis ： Create a new one nums Array , The length is max(clips[i].r),nums[clips[i].l] = clips[i].r - clips[i].l, After the conversion, the question becomes Jump Game and Jump Game Ⅱ Combination of two questions , Through the first Jump Game Judge whether there is a solution , Re pass Jump Game Ⅱ Solve it

0ms, Time beats 100%

class Solution {
    
    private boolean ok(int[] nums, int time) {
        int ma = 0;
        for (int i = 0; i < time && ma >= i; i++) {
            ma = Math.max(ma, i + nums[i]);
        }
        return ma >= time;
    }
    
    public int videoStitching(int[][] clips, int time) {
        int n = 0;
        for (int i = 0; i < clips.length; i++) {
            n = Math.max(n, clips[i][1]);
        }
        n++;
        int[] nums = new int[n];
        for (int i = 0; i < clips.length; i++) {
            int l = clips[i][0], r = clips[i][1];
            nums[l] = Math.max(nums[l], r - l);
        }
        if (!ok(nums, time)) {
            return -1;
        }
        int cur = 0, ma = 0, ans = 0;
        for (int i = 0; i < time && cur < time; i++) {
            ma = Math.max(ma, i + nums[i]);
            if (cur == i) {
                cur = ma;
                ans++;
            }
        }
        return ans;
    }
}