1024. Video Stitching
Medium
You are given a series of video clips from a sporting event that lasted T
seconds. These video clips can be overlapping with each other and have varied lengths.
Each video clip clips[i]
is an interval: it starts at time clips[i][0]
and ends at time clips[i][1]
. 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, T]
). If the task is impossible, return -1
.
Example 1:
Input: clips = [[0,2],[4,6],[8,10],[1,9],[1,5],[5,9]], T = 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]], T = 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]], T = 9 Output: 3 Explanation: We can take clips [0,4], [4,7], and [6,9].
Example 4:
Input: clips = [[0,4],[2,8]], T = 5 Output: 2 Explanation: Notice you can have extra video after the event ends.
Constraints:
1 <= clips.length <= 100
0 <= clips[i][0] <= clips[i][1] <= 100
0 <= T <= 100
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 | class Solution { public int videoStitching(int[][] clips, int T) { int[] dp = new int[T+1]; Arrays.fill(dp, T+2); dp[0] = 0; Map<Integer, Integer> intevals = new HashMap<>(); for(int[] clip : clips){ int l = clip[0], r = clip[1]; intevals.put(l, Math.min(Math.max(intevals.getOrDefault(l, 0), r), T)); } for(int i=0; i<=T; ++i){ if(intevals.containsKey(i)){ int l = i, r = intevals.get(i); for(int j=l+1; j<=r; ++j){ dp[j] = Math.min(dp[j], dp[l]+1); } } } return dp[T]==T+2?-1:dp[T]; } } |
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