查找在列表中的单词中重复的一组字母

Question

I have a list of words:

list1 = ['technology','technician','technical','technicality']

I want to check which phrase is repeated in each of the word. In this case, it is 'tech'. I have tried converting all the characters to ascii values, but I am stuck there as I am unable to think of any logic. Can somebody please help me with this?

Answer 1

This is generally called the Longest common substring/subsequence problem.

---

A very basic (but slow) strategy:

longest_substring = ""
curr_substring = ""

# Loop over a particular word (ideally, shortest).
for start_idx in range(shortest_word):

    # Select a substring from that word.
    for length in range(1, len(shortest_word) - start_idx):
        curr_substring = shortest_word[start_idx : start_idx + length]

        # Check if substring is present in all words,
        # and exit loop or update depending on outcome.

        if "curr_substring not in all words":
            break

        if "new string is longer":
            longest_substring = curr_substring

Answer 2

Iterate over first word, increase length of prefix if there is only one prefix in all words checked by set, when difference in prefix is found return last result

list1 = ['technology', 'technician', 'technical', 'technicality']


def common_prefix(li):
    s = set()
    word = li[0]
    while(len(s) < 2):
        old_s = s
        for i in range(1, len(word)):
            s.add(word[:i])
    return old_s.pop()


print(common_prefix(list1))

output: techn

Answer 3

Find the shortest word. Iterate over increasingly small chunks of the first word, starting with a chunk equal in length to the shortest word, checking that each is contained in all of the other strings. If it is, return that substring.

list1 = ['technology', 'technician', 'technical', 'technicality']

def shortest_common_substring(lst):
    shortest_len = min(map(len, lst))
    shortest_word = next((w for w in lst if len(w) == shortest_len), None)
    
    for i in range(shortest_len, 1, -1):
        for j in range(0, shortest_len - i):
            substr = lst[0][j:i]
            
            if all(substr in w for w in lst[1:]):
                return substr

And just for fun, let's replace that loop with a generator expression, and just take the first thing it gives us (or None ).

def shortest_common_substring(lst):
    shortest_len = min(map(len, lst))
    shortest_word = next((w for w in lst if len(w) == shortest_len), 0)
    
    return next((lst[0][j:i] for i in range(shortest_len, 1, -1)
                             for j in range(0, shortest_len - i)
                             if all(lst[0][j:i] in w for w in lst[1:])),
                None)