This is another good one, I think. Strings and arrays are commonly-requested topics.

1. We can observe that “the first string must contain the longest common prefix”. That’s an interesting tool that we can leverage.
2. So we know that we can iterate over that length. We want to “scan” all the strings in parallel, and that’s how.
3. We have an interesting “return from a nested loop” potential. Helper functions may make this clearer. Potential sequel homework.

class Solution(object):
    def longestCommonPrefix(self, strs):
        if len(strs) == 0:
            return ""
        root = strs[0]
        if len(root) == 0:
            return ""
        
        prefixLen = 0
        for i in range(len(root)):
            for s in strs[1:]:
                if i >= len(s):
                    return root[:prefixLen]
                if s[i] != root[i]:
                    return root[:prefixLen]
            prefixLen += 1
        return root