Video Stitching
April 30, 2019
Introduction
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 [0,2].
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.
Note:
1 <= clips.length <= 100
0 <= clips[i][0], clips[i][1] <= 100
0 <= T <= 100
Solution
Recursive solution with caching:
func use(clips [][]int, skipClip, useTime int) [][]int {
ret := [][]int{}
for i, clip := range clips {
if i == skipClip {
continue
}
start := max(0, clip[0] - useTime)
end := max(0, clip[1] - useTime)
if start != end {
ret = append(ret, []int{start, end})
}
}
return ret
}
var m map[string]int
func videoStitching(clips [][]int, T int) int {
if m == nil {
m = make(map[string]int)
}
key := fmt.Sprintf("%v,%v", clips, T)
if v, ok := m[key]; ok {
return v
}
minSoFar := -1
for i, clip := range clips {
if clip[0] == 0 {
if clip[1] >= T {
return 1
}
cnt := videoStitching(use(clips, i, clip[1]), T-clip[1])
if cnt != -1 {
if minSoFar == -1 || minSoFar > cnt + 1 {
minSoFar = cnt + 1
}
}
}
}
m[key] = minSoFar
return minSoFar
}
func max(a, b int) int {
if a > b {
return a
} else {
return b
}
}
Explanation
This is a good example of recursive solution with caching.
Solutions
Another way to solve this problem is to use Dynamic Programming to transform it to the “Jump Game II”.
func videoStitching(clips [][]int, T int) int {
dp := make([]int, T+1)
for _, clip := range clips {
j := clip[0]
if j < T {
dp[j] = max(dp[j], clip[1])
}
}
// make relative numbers
for i := 0; i < T; i++ {
dp[i] = max(0, dp[i] - i)
}
// now this is a jump game
return jump(dp)
}
// this solution is from Jump Game II
func jump(nums []int) int {
n := len(nums)
if n <= 1 {
return 0
}
steps := 1
i, m := 0, nums[0]
for m > 0 {
if m >= n-1 {
return steps
}
steps++
maxSoFar := 0
for ;i <= m; i++ {
maxSoFar = max(maxSoFar, i + nums[i])
if maxSoFar >= n-1 {
return steps
}
}
m = maxSoFar
}
return -1
}
func max(a, b int) int {
if a > b {
return a
} else {
return b
}
}
Explanation
This is the best performance solution with O(n) complexity. The part of the code took from “Jump Game II” post.