Russian Doll Envelopes
<p>You are given a 2D array of integers <code>envelopes</code> where <code>envelopes[i] = [w<sub>i</sub>, h<sub>i</sub>]</code> represents the width and the height of an envelope.</p>

<p>One envelope can fit into another if and only if both the width and height of one envelope are greater than the other envelope&#39;s width and height.</p>

<p>Return <em>the maximum number of envelopes you can Russian doll (i.e., put one inside the other)</em>.</p>

<p><strong>Note:</strong> You cannot rotate an envelope.</p>

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

<pre>
<strong>Input:</strong> envelopes = [[5,4],[6,4],[6,7],[2,3]]
<strong>Output:</strong> 3
<strong>Explanation:</strong> The maximum number of envelopes you can Russian doll is <code>3</code> ([2,3] =&gt; [5,4] =&gt; [6,7]).
</pre>

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

<pre>
<strong>Input:</strong> envelopes = [[1,1],[1,1],[1,1]]
<strong>Output:</strong> 1
</pre>

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

<ul>
	<li><code>1 &lt;= envelopes.length &lt;= 10<sup>5</sup></code></li>
	<li><code>envelopes[i].length == 2</code></li>
	<li><code>1 &lt;= w<sub>i</sub>, h<sub>i</sub> &lt;= 10<sup>5</sup></code></li>
</ul>
Array; Binary Search; Dynamic Programming; Sorting
class Solution {
    public int maxEnvelopes(int[][] envelopes) {
        Arrays.sort(envelopes, (a, b) -> { return a[0] == b[0] ? b[1] - a[1] : a[0] - b[0]; });
        int n = envelopes.length;
        int[] d = new int[n + 1];
        d[1] = envelopes[0][1];
        int size = 1;
        for (int i = 1; i < n; ++i) {
            int x = envelopes[i][1];
            if (x > d[size]) {
                d[++size] = x;
            } else {
                int left = 1, right = size;
                while (left < right) {
                    int mid = (left + right) >> 1;
                    if (d[mid] >= x) {
                        right = mid;
                    } else {
                        left = mid + 1;
                    }
                }
                int p = d[left] >= x ? left : 1;
                d[p] = x;
            }
        }
        return size;
    }
}