Split Concatenated Strings Python Solution

LeetCode 555, Medium. Topics: Greedy, Array, String. View on LeetCode. Generate this problem as a practice environment: tested reference solution, 13 parametrized pytest cases, and a playground notebook:

lcpy gen -n 555   # by problem number
lcpy gen -s split_concatenated_strings   # by problem name

Problem

You are given an array of strings strs. You could concatenate these strings together into a loop, where for each string, you could choose to reverse it or not. Among all the possible loops Return the lexicographically largest string after cutting the loop, which will make the looped string into a regular one. Specifically, to find the lexicographically largest string, you need to experience two phases:

Concatenate all the strings into a loop, where you can reverse some strings or not and connect them in the same order as given.
Cut and make one breakpoint in any place of the loop, which will make the looped string into a regular one starting from the character at the cutpoint.

And your job is to find the lexicographically largest one among all the possible regular strings.

Examples

Input: strs = ["abc","xyz"]
Output: "zyxcba"
Explanation: You can get the looped string "-abcxyz-", "-abczyx-", "-cbaxyz-", "-cbazyx-", where '-' represents the looped status. The answer string came from the fourth looped one, where you could cut from the middle character 'a' and get "zyxcba".

Input: strs = ["abc"]
Output: "cba"

Constraints

1 <= strs.length <= 1000
1 <= strs[i].length <= 1000
1 <= sum(strs[i].length) <= 1000
strs[i] consists of lowercase English letters.

Solution

Reference implementation from solution.py on GitHub, full suite in test_solution.py:

class Solution:
    # Time: O(total length^2) worst case from cut-point candidates
    # Space: O(total length)
    def split_looping_string(self, strs: list[str]) -> str:
        n = len(strs)
        best_parts = [max(s, s[::-1]) for s in strs]
        best = ""
        for i in range(n):
            left = "".join(best_parts[:i])
            right = "".join(best_parts[i + 1 :])
            for t in (strs[i], strs[i][::-1]):
                for k in range(len(t)):
                    cand = t[k:] + right + left + t[:k]
                    if cand > best:
                        best = cand
        return best

Complexity

Time	Space
O(total length^2) worst case from cut-point candidates	O(total length)

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Split Concatenated Strings Python Solution

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