Problem Description

The problem presents a challenge where you are given an array of strings strs. The task is to concatenate these strings to form a loop, with the option of reversing any of the strings. Once the loop is created, you must cut the loop at any point and unravel it into a regular string that begins with the character at the cut point. The goal is to find the lexicographically largest string (i.e., the string that would appear last if the strings were sorted alphabetically) that can be obtained following these steps.

Intuition

The intuition behind the solution involves looking for the optimal way to arrange and possibly reverse the strings in order to maximize the lexicographical order of the final string.

Firstly, we observe that reversing a string can only be beneficial if the reversed string is lexicographically larger than the original one. With this in mind, we iterate through each string in the array and replace it with its reverse if the reversed string would come later in lexicographical order.

Next, we want to try various breakpoints to determine the largest possible string we can create. We loop through each string and consider every possible position within that string to break the loop. Once we've chosen where to break a string, there are two possible strings we can form: one that starts from the breakpoint and goes to the end of the string, then continues with the rest of the loop, followed by the start of the string to the breakpoint; and the other that is the reverse - starting from the character before the breakpoint, going backwards to the beginning, continuing with the end of the loop, and finishing with the reverse of the substring from the breakpoint to the end.

By comparing all possible breakpoints and choosing the one that yields the lexicographically largest string, we can find the solution to the problem.

The key here is to understand that the loop allows for flexible starting points and the option to include reversed versions of strings, which can affect the final result.

Learn more about Greedy patterns.

Solution Approach

The implementation of the solution uses a step-by-step strategy to ensure that we analyze all possible strings that can be created through reversing and concatenation.

Step 1: Optimize the Individual Strings

As the first step, we iterate through the given array strs and decide for each individual string if it should be reversed. This decision is based on a simple comparison: if the reversed string is lexicographically larger than the original string, then we replace the original with its reversed version. The comparison s[::-1] > s is used for this, and list comprehension in Python makes this step concise and efficient.

strs = [s[::-1] if s[::-1] > s else s for s in strs]

Step 2: Find the Lexicographically Largest String

Now that we have an optimized version of the array strs, we want to concatenate the strings to form loops and check every possible breakpoint to find the largest string. The loop has two parts:

• A part before the breakpoint (a)
• A part after the breakpoint (b)

To explore all options, we iterate over the array and for each string, we again iterate over each character to consider all possible breakpoints within that string.

For each breakpoint, we create two potential strings:

1. One that starts from the breakpoint to the end of the string and continues with the rest of the strings in the array (a + t + b)
2. The reverse scenario, which starts with the reverse of the string from the beginning to the breakpoint, then continues with the rest, and finishes with the reverse of the substring from the breakpoint to the end (b[::-1] + t + a[::-1]).

In the code, t represents the concatenation of the strings in the array except for the current string being processed. It is obtained by concatenating the strings following the current string and those before it.

for i, s in enumerate(strs):
    t = ''.join(strs[i + 1 :]) + ''.join(strs[:i])
    for j in range(len(s)):
        a = s[j:]    # Part after the breakpoint
        b = s[:j]    # Part before the breakpoint
        ans = max(ans, a + t + b)
        ans = max(ans, b[::-1] + t + a[::-1])

We use the max() function to keep track of the lexicographically largest string encountered so far. By the end of these nested loops, ans will hold the lexicographically largest string that can be formed based on the given rules.

Data Structures and Patterns

• String and List manipulations are heavily used in this problem.
• The inherent ordered nature of strings in Python is leveraged to make lexicographical comparisons.
• List comprehension is used for its readability and efficiency in replacing elements based on a condition.
• Nested loops are utilized to consider all possible strings resulting from different breakpoints.

Example Walkthrough

Consider the following example with an array of strings strs = ["bca", "ada", "acb"]. The task is to find the lexicographically largest string that can be obtained by reversing any of the strings, forming a loop, cutting the loop at any point, and unraveling it into a string.

Optimizing Individual Strings

Firstly, we determine if reversing any of the strings would produce a larger lexicographical string:

• Original: "bca", Reversed: "acb" -> Reversed is larger, so we use "acb".
• Original: "ada", Reversed: "ada" -> Both are the same, so no change is made.
• Original: "acb", Reversed: "bca" -> Original is larger, so no change is made.

Now, our optimized array strs looks like this: ["acb", "ada", "acb"].

Finding the Lexicographically Largest String

We now construct loops and check every possible breakpoint:

Starting with the first string "acb" and breaking at each point, we get:

1. Breakpoint after "a": String = "cb" + "adaacb" + "a" = "cbadaacba" (Reverse = "abcaadabc")
2. Breakpoint after "c": String = "b" + "adaacb" + "ac" = "badaacb" (Reverse = "caadaacdab")
3. Breakpoint after "b": String = "" + "adaacb" + "acb" = "adaacbacb"

We do this for each string in our optimized array. Then, we compare all these strings along with their reverse versions to find the maximum:

First String ("acb"):

• "cbadaacba" and "abcaadabc" -> Maximum is "cbadaacba"
• "badaacb" and "caadaacdab" -> Maximum is "caadaacdab"
• "adaacbacb" -> Already the entire string, no reverse needed.

Second String ("ada"): Similar tests are conducted with "ada" as the breaking point, but none will be larger than "caadaacdab", so for brevity's sake, we will not list them.

Third String ("acb"): Similar tests are conducted with "acb" as the breaking point, but again, none will be larger than "caadaacdab".

After comparing all possible breakpoints and their reversed counterparts from each string, we find that "caadaacdab" is the lexicographically largest string. This string will be kept as the value of ans.

Result

The solution returns that the lexicographically largest string possible by following the given method on the array ["bca", "ada", "acb"] is "caadaacdab".

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the initial string reversal in the list comprehension is O(n * m), where n is the number of strings in the list and m is the average length of the strings, as reversal takes O(m) and it is done for each of the n strings.

Then, we iterate over each of the n strings, and for each string, we perform a series of string concatenations. The concatenation t = ''.join(strs[i + 1 :]) + ''.join(strs[:i]) occurs n times and each time it has a cost of O(k), where k is the total length of the strings in the array strs. This step contributes O(n * k) to the time complexity.

Inside the outer loop, we further loop within each string having an average length m, performing multiple operations. The generation of substrings a = s[j:] and b = s[:j] is done mn times and each takes O(m). This contributes another O(m^2 * n).

Finally, max operations and reversals are done within this inner loop, which again contributes a factor of m for each string operation.

Adding all up, we have: Total Time Complexity = O(n * m) + O(n * k) + O(m^2 * n) In the worst case, where k is large, the term O(n * k) will dominate, so the final time complexity can be approximated as O(n * k).

Space Complexity

The space complexity of the solution is determined by the storage needed for the modified strings list and the temporary variables used for concatenation and comparison. String reversal and list comprehension create a new list of the same size, O(n). The temporary variables t, a, and b can together hold up to O(k) space, where k is the total length of all strings. The max function does not use additional space proportional to the input size.

Thus, the space complexity is O(n + k).

Learn more about how to find time and space complexity quickly using problem constraints.