Cfrgtkky

I have a new idea, as the figure shows:

We can construct a directed graph by word[0] == word[-1], then the problem is converted to find the maximum length path.

As mentioned by others, the problem is to find the longest path in a directed acyclic graph.

For anything graph related in Python, networkx is your friend.

You just need to initialize the graph, add the nodes, add the edges and launch dag_longest_path:

import networkx as nx

import matplotlib.pyplot as plt



words = ['giraffe', 'elephant', 'ant', 'tiger', 'racoon', 'cat',

         'hedgehog', 'mouse']



G = nx.DiGraph()

G.add_nodes_from(words)



for word1 in words:

    for word2 in words:

        if word1 != word2 and word1[-1] == word2[0]:

            G.add_edge(word1, word2)

nx.draw_networkx(G)

plt.show()

print(nx.algorithms.dag.dag_longest_path(G))

It outputs:

['hedgehog', 'giraffe', 'elephant', 'tiger', 'racoon']

Note : this algorithm only works if there are no cycles (loops) in the graph. It means it will fail with ['ab', 'ba'] because there would be a path of infinite length: ['ab', 'ba', 'ab', 'ba', 'ab', 'ba', ...]

In the spirit of brute force solutions, you can check all permutations of the words list and choose the best continuous starting sequence:

from itertools import permutations



def continuous_starting_sequence(words):

    chain = [words[0]]

    for i in range(1, len(words)):

        if not words[i].startswith(words[i - 1][-1]):

            break

        chain.append(words[i])

    return chain



words = ['giraffe', 'elephant', 'ant', 'tiger', 'racoon', 'cat', 'hedgehog', 'mouse']

best = max((continuous_starting_sequence(seq) for seq in permutations(words)), key=len)



print(best)

# ['hedgehog', 'giraffe', 'elephant', 'tiger', 'racoon']

Since we're considering all permutations, we know that there must be a permutation that starts with the largest word chain.

This, of course, has O(n n!) time complexity :D

This function creates a type of iterator called a generator (see: What does the "yield" keyword do?). It recursively creates further instances of the same generator to explore all possible tail sequences:

words = ['giraffe', 'elephant', 'ant', 'tiger', 'racoon', 'cat', 'hedgehog', 'mouse']



def chains(words, previous_word=None):

    # Consider an empty sequence to be valid (as a "tail" or on its own):

    yield 

    # Remove the previous word, if any, from consideration, both here and in any subcalls:

    words = [word for word in words if word != previous_word]

    # Take each remaining word...

    for each_word in words:

        # ...provided it obeys the chaining rule

        if not previous_word or each_word.startswith(previous_word[-1]):

            # and recurse to consider all possible tail sequences that can follow this particular word:

            for tail in chains(words, previous_word=each_word):

                # Concatenate the word we're considering with each possible tail:

                yield [each_word] + tail  



all_legal_sequences = list(chains(words))  # convert the output (an iterator) to a list

all_legal_sequences.sort(key=len) # sort the list of chains in increasing order of chain length

for seq in all_legal_sequences: print(seq)

# The last line (and hence longest chain) prints as follows:

# ['hedgehog', 'giraffe', 'elephant', 'tiger', 'racoon']

Or, to get straight to the longest chain more efficiently:

print(max(chains(words), key=len)

Finally, here is an alternative version that allows repeated words in the input (i.e. if you include a word N times, you may use it up to N times in the chain):

def chains(words, previous_word_index=None):

    yield 

    if previous_word_index is not None:

        previous_letter = words[previous_word_index][-1]

        words = words[:previous_word_index] + words[previous_word_index + 1:]

    for i, each_word in enumerate( words ):

        if previous_word_index is None or each_word.startswith(previous_letter):

            for tail in chains(words, previous_word_index=i):

                yield [each_word] + tail  

Here is a working recursive brute-force approach:

def brute_force(pool, last=None, so_far=None):

    so_far = so_far or 

    if not pool:

        return so_far

    candidates = 

    for w in pool:

        if not last or w.startswith(last):

            c_so_far, c_pool = list(so_far) + [w], set(pool) - set([w])

            candidates.append(brute_force(c_pool, w[-1], c_so_far))

    return max(candidates, key=len, default=so_far)



>>> brute_force(words)

['hedgehog', 'giraffe', 'elephant', 'tiger', 'racoon']

At every recursive call, this tries to continue the chain with every eligible word form the remaining pool. It then chooses the longest such continuation.

I have a tree-based approach for this question which might be faster. I am still working on implementation of the code but here is what I would do:

    1. Form a tree with the root node as first word. 

    2. Form the branches if there is any word or words that starts 

with the alphabet with which this current word ends.

    3. Exhaust the entire given list based on the ending alphabet

 of current word and form the entire tree.

    4. Now just find the longest path of this tree and store it.

    5. Repeat steps 1 to 4 for each of the words given in the list

 and print the longest path among the longest paths we got above.

I hope this might give a better solution in case there is a large list of words given. I will update this with the actual code implementation.

Another answer using a recursive approach:

def word_list(w_list, remaining_list):

    max_result_len=0

    res = w_list

    for word_index in range(len(remaining_list)):

        # if the last letter of the word list is equal to the first letter of the word

        if w_list[-1][-1] == remaining_list[word_index][0]:

            # make copies of the lists to not alter it in the caller function

            w_list_copy = w_list.copy()

            remaining_list_copy = remaining_list.copy()

            # removes the used word from the remaining list

            remaining_list_copy.pop(word_index)

            # append the matching word to the new word list

            w_list_copy.append(remaining_list[word_index])

            res_aux = word_list(w_list_copy, remaining_list_copy)

            # Keep only the longest list

            res = res_aux if len(res_aux) > max_result_len else res 

    return res



words = ['giraffe', 'elephant', 'ant', 'tiger', 'racoon', 'cat', 'hedgehog', 'mouse']

word_list(['dog'], words)

output:

['dog', 'giraffe', 'elephant', 'tiger', 'racoon']

Hopefully, a more intuitive way of doing it without recursion. Iterate through the list and let Python's sort and list comprehension do the work for you:

words = ['giraffe', 'elephant', 'ant', 'tiger', 'racoon', 'cat', 'hedgehog', 'mouse']



def chain_longest(pivot, words):

    new_words = 

    new_words.append(pivot)

    for word in words:

        potential_words = [i for i in words if i.startswith(pivot[-1]) and i not in new_words]

        if potential_words:

            next_word = sorted(potential_words, key = lambda x: len)[0]

            new_words.append(next_word)

            pivot = next_word

        else:

            pass

    return new_words



max([chain_longest(i, words) for i in words], key = len)

>>

['hedgehog', 'giraffe', 'elephant', 'tiger', 'racoon']

Set a pivot and check for potential_words if they start with your pivot word and do not appear in your new list of words. If found then just sort them by length and take the first element.

The list comprehension goes through every word as a pivot and returns you the longest chain.