Analysis

The first phase of patience sort, the card game simulation, can be implemented to take O(n log n) comparisons in the worst case for an n-element input array: there will be at most n piles, and by construction, the top cards of the piles form an increasing sequence from left to right, so the desired pile can be found by binary search.^[1] The second phase, the merging of piles, can be done in ${\displaystyle O(n\log n)}$ time as well using a priority queue.^[1]

When the input data contain natural "runs", i.e., non-decreasing subarrays, then performance can be strictly better. In fact, when the input array is already sorted, all values form a single pile and both phases run in O(n) time. The average-case complexity is still O(n log n): any uniformly random sequence of values will produce an expected number of ${\displaystyle O({\sqrt {n}})}$ piles,^[3] which take ${\displaystyle O(n\log {\sqrt {n}})=O(n\log n)}$ time to produce and merge.^[1]

An evaluation of the practical performance of patience sort is given by Chandramouli and Goldstein, who show that a naive version is about ten to twenty times slower than a state-of-the-art quicksort on their benchmark problem. They attribute this to the relatively small amount of research put into patience sort, and develop several optimizations that bring its performance to within a factor of two of that of quicksort.^[1]

If values of cards are in the range 1, . . . , n, there is an efficient implementation with ${\displaystyle O(n\log n)}$ worst-case running time for putting the cards into piles, relying on a Van Emde Boas tree.^[3]

Algorithm for finding a longest increasing subsequence

First, execute the sorting algorithm as described above. The number of piles is the length of a longest subsequence. Whenever a card is placed on top of a pile, put a back-pointer to the top card in the previous pile (that, by assumption, has a lower value than the new card has). In the end, follow the back-pointers from the top card in the last pile to recover a decreasing subsequence of the longest length; its reverse is an answer to the longest increasing subsequence algorithm.

S. Bespamyatnikh and M. Segal^[3] give a description of an efficient implementation of the algorithm, incurring no additional asymptotic cost over the sorting one (as the back-pointers storage, creation and traversal require linear time and space). They further show how to report all the longest increasing subsequences from the same resulting data structures.