LeetCode 1024. Video Stitching Solution in Java, C++, Python & Go | Explanation + Code

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

Solutions

Solution 1: Greedy

Note that if there are multiple sub-intervals with the same starting point, it is optimal to choose the one with the largest right endpoint.

Therefore, we can preprocess all sub-intervals. For each position i, calculate the largest right endpoint among all sub-intervals starting at i, and record it in the array last[i].

We define a variable mx to represent the farthest position that can currently be reached, a variable ans to represent the current minimum number of sub-intervals needed, and a variable pre to represent the right endpoint of the last used sub-interval.

Next, we start enumerating all positions i from 0, using last[i] to update mx. If after updating, mx = i, it means that the next position cannot be covered, so the task cannot be completed, return -1.

At the same time, we record the right endpoint pre of the last used sub-interval. If pre = i, it means that a new sub-interval needs to be used, so we add 1 to ans and update pre to mx.

After the traversal is over, return ans.

The time complexity is O(n+m), and the space complexity is O(m). Where n and m are the lengths of the clips array and the value of time, respectively.

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PythonJavaC++Go

class Solution: def videoStitching(self, clips: List[List[int]], time: int) -> int: last = [0] * time for a, b in clips: if a < time: last[a] = max(last[a], b) ans = mx = pre = 0 for i, v in enumerate(last): mx = max(mx, v) if mx <= i: return -1 if pre == i: ans += 1 pre = mx return ans(code-box)

class Solution { public int videoStitching(int[][] clips, int time) { int[] last = new int[time]; for (var e : clips) { int a = e[0], b = e[1]; if (a < time) { last[a] = Math.max(last[a], b); } } int ans = 0, mx = 0, pre = 0; for (int i = 0; i < time; ++i) { mx = Math.max(mx, last[i]); if (mx <= i) { return -1; } if (pre == i) { ++ans; pre = mx; } } return ans; } }(code-box)

class Solution { public: int videoStitching(vector<vector<int>>& clips, int time) { vector<int> last(time); for (auto& v : clips) { int a = v[0], b = v[1]; if (a < time) { last[a] = max(last[a], b); } } int mx = 0, ans = 0; int pre = 0; for (int i = 0; i < time; ++i) { mx = max(mx, last[i]); if (mx <= i) { return -1; } if (pre == i) { ++ans; pre = mx; } } return ans; } };(code-box)

func videoStitching(clips [][]int, time int) int { last := make([]int, time) for _, v := range clips { a, b := v[0], v[1] if a < time { last[a] = max(last[a], b) } } ans, mx, pre := 0, 0, 0 for i, v := range last { mx = max(mx, v) if mx <= i { return -1 } if pre == i { ans++ pre = mx } } return ans }(code-box)