Find all possible varients of max pair of 2

Question

Given a string of numbers like 123456, I want to find all the possibilities they can be paired in by 2 or by itself. For example, from the string 123456 I would like to get the following:

12 3 4 5 6, 12 34 5 6, 1 23 4 56, etc.

The nearest I was able to come to was this:

strr = list("123456")
x = list("123456")

for i in range(int(len(strr)/2)):
    newlist = []
    for j in range(i):
        newlist.append(x[j])
    newlist.append(x[i] + x[i+1])
    for j in range(len(x))[i+2:]:
        newlist.append(x[j])
    x = newlist.copy()
    b = x.copy()
    for f in range(len(b))[i:]:
        if f == i:
            print(b)
            continue
        b[f] = b[f - 1][1] + b[f]
        b[f - 1] = b[f - 1][0]
        print(b)

This code gives the output:

output

Asked By: Rasmus

||

Source

Answer 1

It’s easy to solve this problem with a recursive generator. This is similar to how you solve change-making problems, just here we have only two "coins", either two characters together, or one character at a time. The total change we’re trying to make is the length of the input string. The fact that the characters are digits in a numeric string is irrelevant.

def singles_and_pairs(string):
    if len(string) <= 1:             # base case
        yield list(string)           # yield either [] or [string] and then quit
        return

    for result in singles_and_pairs(string[:-1]): # first recursion
        result.append(string[-1:])
        yield result

    for result in singles_and_pairs(string[:-2]): # second recursion
        result.append(string[-2:])
        yield result

If you plan on running this on large input strings, you might want to add memoization, since the recursive calls recalculate the same results quite often.

Answered By: Blckknght

Answer 2

Pheew, this one took me some time to get right, but it seems to finally work (edited for prettier ordering):

def max_2_partitions(my_string):
    if not my_string:
        return [[]]
    if len(my_string) == 1:
        return [[my_string]]
    ret = []
    for i in range(len(my_string)):
        for l in max_2_partitions(my_string[:i] + my_string[i + 1:]):
            li = sorted([my_string[i]]+l, key = lambda x: (len(x),x))
            if li not in ret:
                ret.append(li)
        for j in range(i+1,len(my_string)):
            for l in max_2_partitions(my_string[:i]+my_string[i+1:j]+my_string[j+1:]):
                li = sorted([my_string[i] + my_string[j]] + l, key = lambda x: (len(x),x))
                if li not in ret:
                    ret.append(li)
    return sorted(ret, key=lambda x: (-len(x),x))

Example:

print(max_2_partitions("1234"))
# [['1', '2', '3', '4'], ['1', '2', '34'], ['1', '3', '24'], ['1', '4', '23'], ['2', '3', '14'], ['2', '4', '13'], ['3', '4', '12'], ['12', '34'], ['13', '24'], ['14', '23']]

Answered By: Swifty

Answer 3

12 lines of code, full permutations:

You can first create permutations of the string, and then add spacing:

from itertools import permutations

def solution(A):
    result = []
    def dfs(A,B):
        if not B:
            result.append(A)
        else:
            for i in range(1,min(2,len(B))+1):
                dfs(A+[B[:i]],B[i:])
    
    for x in permutations(A):
        dfs([],''.join(x))
    return result

print(f"{solution('123') = }")
# solution('123') = [['1', '2', '3'], ['1', '23'], ['12', '3'], ['1', '3', '2'], ['1', '32'], ['13', '2'], ['2', '1', '3'], ['2', '13'], ['21', '3'], ['2', '3', '1'], ['2', '31'], ['23', '1'], ['3', '1', '2'], ['3', '12'], ['31', '2'], ['3', '2', '1'], ['3', '21'], ['32', '1']]

Answered By: C-3PO

Find all possible varients of max pair of 2

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