Python: Length of longest common subsequence of lists

Question

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

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

print a.llcs(b)

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Answer 1

You can easily retool a LCS into a 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 wanted the longest common substring (a different, but related problem, where the subsequence is contiguous), 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

This is very similar to the lcs_length dynamic programming approach, but we track the maximum length found so far (since it is no longer guaranteed the last element in the table is the maximum).

This returns 3:

>>> lcsubstring_length(a, b)
3

A sparse table variant to not have to track all the 0s:

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