//Leetcode 1024: Stiching Video
//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].
public int VideoStitching2(int[][] clips, int T)
{
if (clips.Length == 0)
{
return(-1);
}
IComparer <int[]> myComparer = new ReverseComparer();;
Array.Sort(clips, myComparer);
int start = clips[0][0], end = clips[0][1], i = 1, count = 1;
if (start > 0)
{
return(-1);
}
else if (end >= T)
{
return(1);
}
while (i < clips.Length && end < T && clips[i][0] <= end)
{
int max = end;
while (i < clips.Length && clips[i][0] <= end)
{
max = Math.Max(max, clips[i][1]);
i++;
}
end = max;
count++;
}
return(end >= T? count : -1);
}
public int VideoStitching(int[][] clips, int T)
{
if (clips.Length == 0)
{
return(-1);
}
IComparer <int[]> myComparer = new ReverseComparer();;
Array.Sort(clips, myComparer);
int start = clips[0][0], end = clips[0][1];
int max = end, count = 1;
int current = 1;
if (start > 0)
{
return(-1);
}
else if (end >= T)
{
return(1);
}
while (current < clips.Length && max < T && clips[current][0] <= end)
{
while (current < clips.Length && clips[current][0] <= end) // there is still overlap
{
if (clips[current][1] > max)
{
max = clips[current][1];
}
current++;
}
count++;// we need another clip
end = max;
}
if (max >= T)
{
return(count);
}
else
{
return(-1);
}
}