2839. Check if Strings Can be Made Equal With Operations I

Problem Description

In this problem, we are given two strings s1 and s2, each exactly 4 characters long, consisting of lowercase English letters. The task is to determine if we can transform string s1 into string s2 through a series of operations. The operation allowed in this context is to swap any two characters in either string, given that these characters are exactly two indices apart (that is, there is one character between them).

The goal is to decide if it's possible to make s1 identical to s2 by applying this operation zero or more times on any of the strings and return true if it is possible, or false otherwise.

Intuition

Upon examining the operation allowed, we realize that it only enables us to swap characters that are two indices apart. For a 4-character string, this means we could swap the 1st character with the 3rd, and the 2nd character with the 4th, but not any other pairs. Therefore, the characters that start at even indices (0 and 2 in zero-based indexing) can only be moved within their own set of positions, and the same goes for characters at odd indices (1 and 3).

To decide if it's possible to make the two strings equal, we only need to check if, within these two groups of indices (even and odd), the sets of characters are the same for both s1 and s2.

The reference solution approach leverages precisely this observation. It separately sorts the characters located at even and odd indices for both s1 and s2 and checks if they are equal. If both the even and odd indexed characters match after sorting, that means s1 can be transformed into s2 through the allowed operation.

Here's why:

• The even indexed characters can only be swapped among themselves. The same applies to the odd indexed characters.
• Whether s1 can be transformed into s2 does not depend on the initial order of the characters within the even and odd index groups, just on the presence of the same characters.
• Sorting the characters in a specific group (even or odd) gives their ordered sequence based on the character values, regardless of their initial order.
• If the sequences match after sorting for both even and odd indexed groups, it affirms that one can achieve one string from the other through the swapping operation.

Thus, the canBeEqual function in the solution works efficiently by sorting and comparing sliced strings formed by characters at even and odd positions.

Solution Approach

The implementation of the solution uses Python's list slicing and sorting capabilities. The key to the solution lies in separating the characters based on their index being even or odd and then comparing these subsets from s1 and s2. Here are the main components of this approach:

• List Slicing: In Python, slicing is a feature that allows for accessing parts of sequences like strings, lists, or tuples. In the given solution, s1[::2] and s2[::2] are used to access characters from s1 and s2 that are at even indices, which are 0 and 2 for a 4-character string. Similarly, s1[1::2] and s2[1::2] are used to access the characters at odd indices, which are 1 and 3.
• Sorting: Sorting is used to rearrange the characters in a specific order. This is done to easily compare whether the same set of characters are present in both s1 and s2 for the respective positions (even and odd). The sorted() function in Python returns a new sorted list from the elements of any iterable.
• Equality Check: After sorting, the two lists of characters (each for the even and odd positions) are compared using the equality operator ==. If two lists are equal, it means they contain the same elements in the same order.

Now looking at the code implementation based on this approach:

1class Solution:
2    def canBeEqual(self, s1: str, s2: str) -> bool:
3        # Sort and compare characters at even indices of s1 and s2
4        even_positions_match = sorted(s1[::2]) == sorted(s2[::2])
5        # Sort and compare characters at odd indices of s1 and s2
6        odd_positions_match = sorted(s1[1::2]) == sorted(s2[1::2])
7        # Return True if both matches are True, otherwise False
8        return even_positions_match and odd_positions_match

The two separate boolean variables even_positions_match and odd_positions_match store the result of whether or not the characters at even and odd positions, respectively, match between s1 and s2. The function finally returns True if both sets of positions match after sort, which indicates the operation can be performed to make the strings equal, otherwise, it returns False.

In terms of complexity, since the strings are of constant length, both the space and time complexity of the operations on the strings are also constant, O(1), regardless of the actual implementation detail of the sorted() function.

Example Walkthrough

Let's take an example to understand how the solution approach works. Suppose s1 is abcd and s2 is cbad. We wish to know if s1 can be transformed into s2 by swapping any two characters that are two indices apart in either string.

Now we can slice and sort the characters from s1 and s2 based on their indices being even or odd and compare them:

• For the even indices (0 and 2):

  • s1 has the characters 'a' and 'c' at these indices.
  • s2 has the characters 'c' and 'a'.

  Sort these characters and we get the following:

  • s1 sorted even characters: 'ac' → 'ac'
  • s2 sorted even characters: 'ca' → 'ac'

  Since 'ac' is equal to 'ac', the even-indexed characters match.

• For the odd indices (1 and 3):

  • s1 has the characters 'b' and 'd' at these indices.
  • s2 has the characters 'b' and 'd'.

  Sort these characters as well:

  • s1 sorted odd characters: 'bd' → 'bd'
  • s2 sorted odd characters: 'bd' → 'bd'

  Since 'bd' is equal to 'bd', the odd-indexed characters also match.

Since the sorted subsets for both the even and odd indices from s1 match those in s2, we can confirm that s1 can indeed be transformed into s2 using the allowed operation.

Therefore, when we use the described method on our example, the canBeEqual function in the Solution class would return True. Here's how the class would process our example:

1# Instantiate the solution class
2solution = Solution()
3# Call the canBeEqual function with s1 and s2
4result = solution.canBeEqual('abcd', 'cbad')
5# Output the result
6print(result)  # This would print True to the console

By applying the solution's approach to our example, we are able to walk through the steps to determine that s1 can be transformed into s2 as per the defined operation.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the provided algorithm is driven by the sorting operations performed on sliced strings. Python uses Timsort, an optimized merge sort that, in the general case, has a time complexity of O(n log n).

Since the string s1 is sliced into two halves (s1[::2] and s1[1::2]), and the same is done for s2, these operations can be considered to work on roughly half the length of the input strings each. For a string of length n, each slice would have a size of about n/2.

The sorted() function is then applied to each slice, resulting in a time complexity of O(n/2 * log(n/2)) for each sort operation. There are four such sort operations, two for s1 and two for s2.

Multiplying this by four (since there are four slices), the overall time complexity is 4 * O(n/2 * log(n/2)), which simplifies to O(n log n) because the constants do not change the order of growth.

Space Complexity

The space complexity is driven by the space required to store the sorted slices. Sorting a list in Python is not done in-place (when using the sorted() function); rather, it creates a new list. Therefore, for a slice of n/2 elements, the space required is O(n/2).

Since we sort four slices independently, and assuming only the largest of these affects the overall space complexity, we could argue that the space complexity is O(n). However, all these sorts are not in memory at the same time unless in the case of concurrent execution, which isn't specified here. Therefore, the space complexity is O(n/2), which simplifies to O(n) because dropping constants is standard in Big O notation.

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