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

Code

similar to Minimum Number of Taps to Open to Water a Garden 、 Jump Game II。

Sorting + DP

Time Complexity: $O(N\log N+NT)$, Space Complexity: $O(T)$

要先 sort 才可以照著 clip 的順序去更新 DP，若沒有 sort，就必須像下面一個解法一樣，對每個 index 都跑過整個 clips。因為當 clip 的順序非 sorted 時，有些 clip 在後，有些在前，他們之間的交互影響必須要反應到 DP 當中。

class Solution {
public:
    int videoStitching(vector<vector<int>>& clips, int time) {
        vector<int> dp(101, 101);
 
        sort(clips.begin(), clips.end());
 
        dp[0] = 0;
 
        for(auto clip: clips) {
            int start = clip[0];
            int end = clip[1];
            for(int i = start; i <= end; i++) {
                dp[i] = min(dp[i], dp[start] + 1);
            }
        }
 
        return dp[time] == 101 ? -1 : dp[time];
    }
};

DP

Time Complexity: $O(NT)$, Space Complexity: $O(T)$

class Solution {
public:
    int videoStitching(vector<vector<int>>& clips, int time) {
        int dp[time + 1];
        for(int i = 0; i < time + 1; i++) {
            dp[i] = time + 1;
        }
 
        dp[0] = 0;
 
        for(int i = 0; i < time + 1; i++) {
            for(auto clip: clips) {
                int start = clip[0];
                int end = clip[1];
                if(i >= start && i <= end) {
                    dp[i] = min(dp[i], dp[start] + 1);
                }
            }
        }
 
        return dp[time] == time + 1 ? -1 : dp[time];
    }
};

Greedy

Time Complexity: $O(T)$, Space Complexity: $O(T)$

不斷更新 end 指標。注意初始值 prev_end = -1, end = 0。

當 i > prev_end，就要更新 prev_end = end，若 i > end 的話就不會進行此更新。

class Solution {
public:
    int videoStitching(vector<vector<int>>& clips, int time) {
        unordered_map<int, int> maxEnd;
        for(auto clip: clips) {
            maxEnd[clip[0]] = max(maxEnd[clip[0]], clip[1]);
        }
 
        int prev_end = -1, end = 0, count = 0;
        for(int i = 0; i <= time; i++) {
            if(end >= time || i > end) break;
            if(maxEnd.count(i)) {
                if(i > prev_end) {
                    count++;
                    prev_end = end;
                }
                end = max(end, maxEnd[i]);
            }
        }
        return end >= time ? count : - 1;
 
    }
};

Source

• Video Stitching - LeetCode