CODEWITHSUNDEEP
python
loretoparisi
loretoparisi

Reputation: 16271

Python Longest Increasing Subsequence of Indexes in a Array

This algorithm (originally implemented in unl-aligner) calculates the longest list of increasing numbers with correspondingly increasing indices in the sequence, so given

seq = [0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15]

it will return

[0, 2, 6, 9, 11, 15]

the implementation looks like

def subseq(seq, keyfn=lambda value: value):
    if not seq: return seq
    tail = []
    prev = []
    for i in range(len(seq)):
        for k in range(len(tail)-1, -1, -1):
            if keyfn(seq[tail[k]]) < keyfn(seq[i]):
                if len(tail) == k+1:
                    tail.append(i)
                elif keyfn(seq[tail[k+1]]) > keyfn(seq[i]):
                    tail[k+1] = i
                prev.append(tail[k])                    
                break
        else:
            tail.append(i)
            prev.append(None)

    i = tail[-1]
    subseq = [seq[i]]
    while prev[i] is not None:
        i = prev[i]
        subseq.append(seq[i])
    subseq.reverse()
    return subseq

The algorithm performs a linear scan, while a bisect (binary) search should be preferred. Which is the best approach to refactor it to perform a binary search?

Upvotes: 0

Views: 136

Answers (1)

Işık Kaplan
Işık Kaplan

Reputation: 2982

With this answer:

bisect = "longest_subsequence([1,2,3,4,5,6,7,2,2,2,2,2,5,1,7,8])"
_subseq = "subseq([1,2,3,4,5,6,7,2,2,2,2,2,5,1,7,8])"

from timeit import timeit

print(timeit(bisect, globals=globals(), number=10000))  # 0.2994734
print(timeit(_subseq, globals=globals(), number=10000))  # 0.32428109999999993

This is the result on a totally random test, for your example they seem almost exact time-wise

Upvotes: 1

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