The core insight is sorting by width ascending and then running longest increasing subsequence on heights.
This is a classic dynamic programming problem that can be solved in O(n log n) by sorting envelopes by width ascending and height descending, then finding the longest strictly increasing subsequence of heights. Explain the sorting trick to avoid same-width nesting, then apply LIS using binary search for efficiency.
Pro tip: Mention the edge case of equal widths and how sorting height descending prevents invalid nesting, and discuss the trade-off between O(n^2) DP and O(n log n) LIS to show depth.
Restate the problem: we need the maximum chain where each envelope is strictly smaller in both dimensions. Clarify that envelopes can be rotated? (No, typically not, but confirm).
Sort by width ascending; for equal widths, sort by height descending. This ensures that when we look for increasing heights, we don't pick two envelopes with the same width.
Extract the heights array from the sorted envelopes. The problem becomes finding the longest strictly increasing subsequence (LIS) of this array.
Use the patience sorting algorithm with binary search to compute LIS in O(n log n). Maintain a tails array where tails[i] is the smallest tail of an increasing subsequence of length i+1.
The length of the tails array is the maximum number of envelopes that can be nested. Return that length.
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