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Split Concatenated Strings

Problem Description​

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:

1- Concatenate all the strings into a loop, where you can reverse some strings or not and connect them in the same order as given. 2- 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​

Example 1:

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".

Example 2:

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 for Assign Cookies​

Approach:​

1- For each string in the input list, store the lexicographically larger string between the original and its reverse. 2- Initialize an empty string to store the result. 3- For each string, concatenate the parts from the list excluding the current string. 4- For each rotation of the current string, form possible concatenated results by appending the rotated string before and after the concatenated parts. 5- Compare and update the result with the lexicographically largest string formed.

Code in Different Languages​

C++​

class Solution {
public:
    string splitLoopedString(vector<string>& strs) {
        for (auto& s : strs) {
            string t{s.rbegin(), s.rend()};
            s = max(s, t);
        }
        int n = strs.size();
        string ans = "";
        for (int i = 0; i < strs.size(); ++i) {
            auto& s = strs[i];
            string t;
            for (int j = i + 1; j < n; ++j) {
                t += strs[j];
            }
            for (int j = 0; j < i; ++j) {
                t += strs[j];
            }
            for (int j = 0; j < s.size(); ++j) {
                auto a = s.substr(j);
                auto b = s.substr(0, j);
                auto cur = a + t + b;
                if (ans < cur) {
                    ans = cur;
                }
                reverse(a.begin(), a.end());
                reverse(b.begin(), b.end());
                cur = b + t + a;
                if (ans < cur) {
                    ans = cur;
                }
            }
        }
        return ans;
    }
};

Java​

class Solution {
    public String splitLoopedString(String[] strs) {
        int n = strs.length;
        for (int i = 0; i < n; ++i) {
            String s = strs[i];
            String t = new StringBuilder(s).reverse().toString();
            if (s.compareTo(t) < 0) {
                strs[i] = t;
            }
        }
        String ans = "";
        for (int i = 0; i < n; ++i) {
            String s = strs[i];
            StringBuilder sb = new StringBuilder();
            for (int j = i + 1; j < n; ++j) {
                sb.append(strs[j]);
            }
            for (int j = 0; j < i; ++j) {
                sb.append(strs[j]);
            }
            String t = sb.toString();
            for (int j = 0; j < s.length(); ++j) {
                String a = s.substring(j);
                String b = s.substring(0, j);
                String cur = a + t + b;
                if (ans.compareTo(cur) < 0) {
                    ans = cur;
                }
                cur = new StringBuilder(b)
                          .reverse()
                          .append(t)
                          .append(new StringBuilder(a).reverse().toString())
                          .toString();
                if (ans.compareTo(cur) < 0) {
                    ans = cur;
                }
            }
        }
        return ans;
    }
}

Python​

class Solution:
    def splitLoopedString(self, strs: List[str]) -> str:
        strs = [s[::-1] if s[::-1] > s else s for s in strs]
        ans = ''.join(strs)
        for i, s in enumerate(strs):
            t = ''.join(strs[i + 1 :]) + ''.join(strs[:i])
            for j in range(len(s)):
                a = s[j:]
                b = s[:j]
                ans = max(ans, a + t + b)
                ans = max(ans, b[::-1] + t + a[::-1])
        return ans

Complexity Analysis​

Time Complexity: O(n × l^2)​

Reason: where n is the number of strings in the vector and l is the average length of the strings.

Space Complexity: O(n × l )​

Reason: Due to the space required for storing the modified strings and temporary concatenations.

References​