Solution Explanation: Split Concatenated Strings

This problem asks to find the lexicographically largest string that can be formed by concatenating a given array of strings, allowing for reversal of individual strings, and then cutting the resulting loop to create a linear string.

Approach: Greedy String Manipulation

The solution employs a greedy approach. The core idea is to maximize the lexicographical order at each step.

1. Maximize Individual Strings: The algorithm first iterates through each string in the input array strs. For each string, it compares the string with its reversed version. If the reversed string is lexicographically larger, the original string is replaced with its reversed version. This ensures that we start with the lexicographically largest possible version of each individual string.

2. Iterate Through Split Points: The algorithm then iterates through each possible split point in the array strs. This means considering each string as the potential starting point of the final concatenated string after the loop is broken.

3. Explore Rotations and Reversals: For each split point, the algorithm considers two scenarios for the string at that split point:

   • Unreversed: The string at the split point is used as-is. The algorithm then concatenates the remaining strings (after the split point) followed by the strings before the split point.
   • Reversed: The string at the split point is reversed. Again, the remaining strings (after the split point) are concatenated followed by strings before the split point.

4. Internal Substring Rotations: Within each of the above scenarios (unreversed or reversed at the split point), the algorithm iterates through all possible positions within the string at the split point, effectively rotating that string. Each rotation is concatenated with the rest of the string as calculated earlier. This ensures that we find the largest lexicographical ordering even with internal rotation of the strings.

5. Lexicographical Comparison and Update: After each concatenation, the resulting string is compared lexicographically to the current maximum (ans). If the new string is larger, ans is updated.

Time and Space Complexity

• Time Complexity: O(n*m^2), where 'n' is the number of strings in the input array and 'm' is the maximum length of a string in the array. The nested loops for split points, rotations, and string manipulations contribute to this complexity.

• Space Complexity: O(nm). This is because we might need to store concatenated strings of maximum length nm in the worst case. The space used to store the reversed strings in the first step is also proportional to the total length of strings.

Code Explanation (Python)

The Python code directly implements the steps described above. The max() function is used for lexicographical comparison. String slicing ([::-1]) is used for efficient reversal.

class Solution:
    def splitLoopedString(self, strs: List[str]) -> str:
        strs = [s[::-1] if s[::-1] > s else s for s in strs] #Maximize individual strings
        ans = ''.join(strs) # Initialize with the concatenation of all maximal strings as a baseline
        for i, s in enumerate(strs):
            t = ''.join(strs[i + 1 :]) + ''.join(strs[:i]) # Strings after and before the split point
            for j in range(len(s)):
                a = s[j:]
                b = s[:j]
                ans = max(ans, a + t + b) # Unreversed string at the split point
                ans = max(ans, b[::-1] + t + a[::-1]) # Reversed string at the split point
        return ans
 

The other languages (Java, C++, Go) implement the same algorithm with similar logic and time/space complexity. They use their respective string manipulation and comparison functions.