unreasonably effective

link

At an index, if all strings have the same character, set of these characters will have length one.

Say there are strings and the shortest string has length . There are iterations. In each iteration both collecting characters at a given index and creating the set takes time .

Time: , space: .

class Solution:
    def longestCommonPrefix(self, strs: List[str]) -> str:
        common_prefix = []
        for heads in zip(*strs):
            if len(set(heads)) != 1:
                break
            common_prefix.append(heads[0])

        return "".join(common_prefix)