Maximum Length of Pair Chain
<p>You are given an array of <code>n</code> pairs <code>pairs</code> where <code>pairs[i] = [left<sub>i</sub>, right<sub>i</sub>]</code> and <code>left<sub>i</sub> &lt; right<sub>i</sub></code>.</p>

<p>A pair <code>p2 = [c, d]</code> <strong>follows</strong> a pair <code>p1 = [a, b]</code> if <code>b &lt; c</code>. A <strong>chain</strong> of pairs can be formed in this fashion.</p>

<p>Return <em>the length longest chain which can be formed</em>.</p>

<p>You do not need to use up all the given intervals. You can select pairs in any order.</p>

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

<pre>
<strong>Input:</strong> pairs = [[1,2],[2,3],[3,4]]
<strong>Output:</strong> 2
<strong>Explanation:</strong> The longest chain is [1,2] -&gt; [3,4].
</pre>

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

<pre>
<strong>Input:</strong> pairs = [[1,2],[7,8],[4,5]]
<strong>Output:</strong> 3
<strong>Explanation:</strong> The longest chain is [1,2] -&gt; [4,5] -&gt; [7,8].
</pre>

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

<ul>
	<li><code>n == pairs.length</code></li>
	<li><code>1 &lt;= n &lt;= 1000</code></li>
	<li><code>-1000 &lt;= left<sub>i</sub> &lt; right<sub>i</sub> &lt;= 1000</code></li>
</ul>
Greedy; Array; Dynamic Programming; Sorting
impl Solution {
    pub fn find_longest_chain(mut pairs: Vec<Vec<i32>>) -> i32 {
        pairs.sort_by_key(|pair| pair[1]);
        let mut ans = 0;
        let mut pre = i32::MIN;
        for pair in pairs {
            let (a, b) = (pair[0], pair[1]);
            if pre < a {
                ans += 1;
                pre = b;
            }
        }
        ans
    }
}
