<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>

class Solution {
public:
    int maxEnvelopes(vector<vector<int>>& envelopes) {

        const int n = envelopes.size();

        // Sort by width to forget about it and just solve 
        // 300. Longest Increasing Subsequence for heights.
        // The main trick here is to sort descending by height
        // for items with the same width, to avoid  making 
        // chains like [w,1] [w,2] [w,3]
        ranges::sort(envelopes, [](vector<int>& a, vector<int>& b) {
            return (a[0] == b[0]) ? 
                (a[1] > b[1]) : 
                (a[0] < b[0]);
        });

        // Trick to avoid using extra space taken from 
        // 300. Longest Increasing Subsequence

        int len = 0;
        for (int i = 0; i < n; ++i) {
            if (len == 0 || 
                envelopes[i][1] > envelopes[len - 1][1]) {
                envelopes[len][1] = envelopes[i][1];
                len++;
            } else {
                // Main part of the trick - to replace previous values.
                // Search place in [0 .. len - 1] to replace the value.
                // Implement own lower_bound, 
                // because we use original vector of vectors
                int left = 0;
                int right = len - 1;
                int pos = 0;
                while (left <= right) {
                    int mid = left + (right - left) / 2;
                    if (envelopes[mid][1] >= envelopes[i][1]) {
                        pos = mid;
                        right = mid - 1;
                    } else {
                        left = mid + 1;
                    }
                }
                envelopes[pos][1] = envelopes[i][1];
            }
        }

        return len;
    }
};
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