Python: length of the longest common subsequence of lists

Question

Is there a built-in function in python that returns the length of the longest common subsequence of two lists?

a=[1,2,6,5,4,8]
b=[2,1,6,5,4,4]

print a.llcs(b)

>>> 3

I tried to find the longest common subsequence and then get its length, but I think there should be a better solution.

Answer 1

You can easily reconfigure LCS to LLCS:

def lcs_length(a, b):
    table = [[0] * (len(b) + 1) for _ in xrange(len(a) + 1)]
    for i, ca in enumerate(a, 1):
        for j, cb in enumerate(b, 1):
            table[i][j] = (
                table[i - 1][j - 1] + 1 if ca == cb else
                max(table[i][j - 1], table[i - 1][j]))
    return table[-1][-1]

Demo:

>>> a=[1,2,6,5,4,8]
>>> b=[2,1,6,5,4,4]
>>> lcs_length(a, b)
4

If you want the longest common substring (a different but related issue where the subsequence is adjacent), use:

def lcsubstring_length(a, b):
    table = [[0] * (len(b) + 1) for _ in xrange(len(a) + 1)]
    l = 0
    for i, ca in enumerate(a, 1):
        for j, cb in enumerate(b, 1):
            if ca == cb:
                table[i][j] = table[i - 1][j - 1] + 1
                if table[i][j] > l:
                    l = table[i][j]
    return l

lcs_length, , ( , ).

3:

>>> lcsubstring_length(a, b)
3

0 s:

def lcsubstring_length(a, b):
    table = {}
    l = 0
    for i, ca in enumerate(a, 1):
        for j, cb in enumerate(b, 1):
            if ca == cb:
                table[i, j] = table.get((i - 1, j - 1), 0) + 1
                if table[i, j] > l:
                    l = table[i, j]
    return l