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Sixian Li

Longest Increasing Subsequence And Patience Game(Part2)


This part covers the actual implementation of the idea mentioned in Part1. I assume you’ve read it, and I’ll use claims proved in that part.

If you understand how the greedy algorithm works for Patience, the implementation is just simulationg the gameplay.


Remember, the algorithm puts the new card on the leftmost pile possible, and the length of the LIS is the total number of piles at the end of the game.

So the problem boils down to find out #piles\#\text{piles}.

Let’s play the game

1. Do we need to store every card in a pile?

When we’re looking for a valid pile, we only compare the new card cc and the top card tt of a pile. If ctc \leq t, we put cc on the pile, and it becomes the new top card. So there’s no need to store every card.

2. How do we find the first valid pile?

Of course, we can look at every pile from left to right, that would be O(n)O(n), and the total complexity will be O(n2)O(n^2). There’s an extra piece of information about those top cards: they form an increasing subsequence, i.e., they’re sorted.

How do we define “leftmost”? It is the first pile such that new cardtop card\text{new card} \leq \text{top card}, so we have new card>top card of the last pile\text{new card} \gt \text{top card of the last pile}.

If new card>some top card\text{new card} \gt \text{some top card}, new card\text{new card} is larger than anything to the left of that pile, so we can skip those elements. Yes, that’s binary search, the lognlogn part in O(nlogn)O(nlogn).


Line 6: right is len(piles) instead of the usual len(piles) - 1 because n can be larger than any top card in current piles, then it should be placed in a new pile.

Line 10-11: If piles[mid] < n, mid is definitely not the pile where n is placed, so we can exclude it.

Line 13: In this branch, n <= piles[mid], mid may be the pile we’re looking for, so we can’t exclude it. But we’re not sure if it’s the leftmost position, so we keep going.

Line 15-16: If the leftmost valid pile doesn’t exist yet, it means n is larger than all top cards, so we create a new pile for it, len(piles) will be incremented by 1 at the same time.

That’s all for this algorithm. Knowing why an algorithm works is always more important than memorizing the implementation. Proving its correctness is a very good way to learn.