All ways to partition a string

Question

I\'m trying to find a efficient algorithm to get all ways to partition a string

eg for a given string \'abcd\' =>
\'a\' \'bcd\'
\'a\' \'b\' \'cd\'
\'a\'

Answer 1

This is a solution which minimizes developer time by taking advantage of a built-in iterator. It should be reasonably quick for problem sizes for which the answer itself is not infeasibly large.

There is a one-to-one correspondence between partitions of a string and subsets of potential cutpoints. If the length of the string is n then there are n-1 places where you could cut the string. A straightforward way would be to iterate through such subsets, and for each such subset, slice the string in that way. Here is a Python approach which uses the standard modules itertools:

import itertools

def multiSlice(s,cutpoints):
    k = len(cutpoints)
    if k == 0:
        return [s]
    else:
        multislices = [s[:cutpoints[0]]]
        multislices.extend(s[cutpoints[i]:cutpoints[i+1]] for i in range(k-1))
        multislices.append(s[cutpoints[k-1]:])
        return multislices

def allPartitions(s):
    n = len(s)
    cuts = list(range(1,n))
    for k in range(n):
        for cutpoints in itertools.combinations(cuts,k):
            yield multiSlice(s,cutpoints)

For example:

>>> parts = allPartitions('World')
>>> for p in parts: print(p)

['World']
['W', 'orld']
['Wo', 'rld']
['Wor', 'ld']
['Worl', 'd']
['W', 'o', 'rld']
['W', 'or', 'ld']
['W', 'orl', 'd']
['Wo', 'r', 'ld']
['Wo', 'rl', 'd']
['Wor', 'l', 'd']
['W', 'o', 'r', 'ld']
['W', 'o', 'rl', 'd']
['W', 'or', 'l', 'd']
['Wo', 'r', 'l', 'd']
['W', 'o', 'r', 'l', 'd']

Note that this approach produces generates ['World'] as a partition of 'World'. This corresponds to slicing with an empty set of cut points. I regard that as a feature rather than a bug since the standard mathematical definition of partition allows for a partition of a set into one piece. If this in undesirable for your purposes, the fix is easy enough -- just iterate over the nonempty subsets of the cut points. In terms of the above code, this fix amounts to adding two characters to allPartitions: replace

for k in range(n):

by

for k in range(1,n):