LeetCode 1024. Video Stitching

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：sorting by first element, for the clip with same first element, save only the largest end one. then use the greedy logic to get the result.

    public int videoStitching(int[][] clips, int T) {
        if (clips.length == 0 || clips[0].length != 2) return -1;
        TreeMap map = new TreeMap<>((o1, o2) -> (o1 - o2));
        for (int i = 0; i < clips.length; i++) {
            int key = clips[i][0];
            if (map.containsKey(key)) {
                map.put(key, Math.max(clips[i][1], map.get(key)));
            } else {
                map.put(key, clips[i][1]);
            }
        }
        int size = map.size();
        int[][] clean = new int[size][2];
        int index = 0;
        for (Integer key : map.keySet()) {
            clean[index][0] = key;
            clean[index++][1] = map.get(key);
        }
        index = 1;
        int[] current = clean[0];
        int count = 1;
        while (index < size) {
            int end = current[1];
            if (end >= T) break;
            while (index < size && clean[index][0] <= current[1]) {
                end = Math.max(end, clean[index++][1]);
            }
            count++;
            current[1] = end;
        }
        if (current[0] != 0 || current[1] < T) return -1;
        return count; 
    } 

this is the better coding version of mine, we are exactly the same logic.

 public int videoStitching(int[][] clips, int T) {
          Arrays.sort(clips, new Comparator() {
              public int compare(int[] a, int[] b) {
                  return a[0]-b[0];
              }
          });    
          int count = 0;
          int curend = 0;
          int laststart = -1;
          
          for(int i = 0; i < clips.length; ) {
              if(clips[i][0] > curend) {
                  return -1;
              }
              int maxend = curend;
              while(i < clips.length && clips[i][0] <= curend) { // while one clip's start is before or equal to current end
                  maxend = Math.max(maxend, clips[i][1]); // find out the one with the max possible end
                  i++;
              }
              count++;
              curend = maxend;
              if(curend >= T) {
                  return count;    
              }
          }
          return -1;        
      }