Video Stitching
<p>You are given a series of video clips from a sporting event that lasted <code>time</code> seconds. These video clips can be overlapping with each other and have varying lengths.</p>

<p>Each video clip is described by an array <code>clips</code> where <code>clips[i] = [start<sub>i</sub>, end<sub>i</sub>]</code> indicates that the ith clip started at <code>start<sub>i</sub></code> and ended at <code>end<sub>i</sub></code>.</p>

<p>We can cut these clips into segments freely.</p>

<ul>
	<li>For example, a clip <code>[0, 7]</code> can be cut into segments <code>[0, 1] + [1, 3] + [3, 7]</code>.</li>
</ul>

<p>Return <em>the minimum number of clips needed so that we can cut the clips into segments that cover the entire sporting event</em> <code>[0, time]</code>. If the task is impossible, return <code>-1</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> clips = [[0,2],[4,6],[8,10],[1,9],[1,5],[5,9]], time = 10
<strong>Output:</strong> 3
<strong>Explanation:</strong> 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].
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> clips = [[0,1],[1,2]], time = 5
<strong>Output:</strong> -1
<strong>Explanation:</strong> We cannot cover [0,5] with only [0,1] and [1,2].
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> 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
<strong>Output:</strong> 3
<strong>Explanation:</strong> We can take clips [0,4], [4,7], and [6,9].
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= clips.length &lt;= 100</code></li>
	<li><code>0 &lt;= start<sub>i</sub> &lt;= end<sub>i</sub> &lt;= 100</code></li>
	<li><code>1 &lt;= time &lt;= 100</code></li>
</ul>
Greedy; Array; Dynamic Programming
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;
    }
}