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
//
// Created by hanshan on 19-6-11.
//
#include
#include
#include
using namespace std;
class Solution {
public:
int videoStitching(vector>& clips, int T) {
queue> q;
q.push({T, 0});
while (!q.empty()){
auto tmp = q.front();
q.pop();
for(int i=0; i= tmp.first){
if(clips[i][0] == 0)
return tmp.second+1;
q.push({clips[i][0], tmp.second+1});
}
}
}
return -1;
}
};
int videoStitching(vector>& clips, int T) {
sort(begin(clips), end(clips));
int res = 0;
for (auto i = 0, st = 0, end = 0; st < T; st = end, ++res) {
for (; i < clips.size() && clips[i][0] <= st; ++i)
end = max(end, clips[i][1]);
if (st == end) return -1;
}
return res;
}
class Solution {
public:
int videoStitching(vector>& clips, int T) {
int n = clips.size();
int i, j;
/*
dp[i] refers to the furthest valid duration that could be
gained with i clips
dp[i] = max(dp[i-1], clips[j][1] which clips[j][0] <= dp[i])
*/
vector dp = vector(n + 1, -1);
dp[0] = 0;
for (i = 1; i <= n; i++) {
dp[i] = dp[i-1];
for (j = 0; j < n; j++) {
if (clips[j][0] <= dp[i-1]) {
dp[i] = max(dp[i], clips[j][1]);
}
}
}
/*
we get the smallest i such that dp[i] >= T
*/
for (i = 1; i <= n; i++) {
if (dp[i] >= T) {
return i;
}
}
return -1;
}
};