Problem Description

Given a string s, find any substring of length 2 which is also present in the reverse of s. Return true if such a substring exists, and false otherwise.

Intuition

To tackle this problem, consider the potential match between substrings of s and its reverse. The solution leverages a set to store all possible contiguous pairs (substrings of length 2) from the reverse of the string. By reversing s, all pairs are examined in opposite order, facilitating the matching of each character from s with its corresponding character in the reverse.

Next, iterate through the string and extract each pair (substring of length 2). For each extracted pair, check against the previously stored pairs from the reversed string. If a match is found, it indicates the presence of a reverse order substring in the original string, and therefore, you return true. If no matches are found by the end of the traversal, then return false. This efficient approach ensures all necessary substrings are compared without redundancy.

Solution Approach

The solution utilizes a hash table to efficiently store and retrieve substrings. Here’s a detailed breakdown of the implementation:

1. Reverse the String: Begin by reversing the input string s. This is necessary to capture the substrings to be compared against substrings of the original s.

2. Extract Substrings from Reversed String: Using Python's set comprehension, create a set st that contains pairs (substrings of length 2) from the reversed string. This set ensures that only unique substrings are captured and allows for O(1) average-time complexity for membership checks.

   st = {(a, b) for a, b in pairwise(s[::-1])}

3. Check Substrings in the Original String: Traverse the original string s. For each contiguous pair (substring of length 2), check if this pair exists in the set st.

   return any((a, b) in st for a, b in pairwise(s))

4. Return Result: If any such substring is found in both the original and reversed string’s substring set, return true. Otherwise, return false when the iteration completes without finding any such pairs.

This approach is efficient due to the use of a hash set and minimizes the complexity by ensuring each pair is only checked a single time.

Example Walkthrough

Consider the string s = "abcda". Our task is to determine if there exists any substring of length 2 in s that is also present in the reverse of s.

1. Reverse the String:

   • Reverse s to get s[::-1] = "adcba".

2. Extract Substrings from Reversed String:

   • From s[::-1] = "adcba", collect all substrings of length 2:
     • "ad", "dc", "cb", "ba".
   • Store these in a set to avoid duplicates:
     • st = {"ad", "dc", "cb", "ba"}.

3. Check Substrings in the Original String:

   • Traverse the original string s to find all substrings of length 2 and check if they exist in st.
   • Substrings in s = "abcda" are:
     • "ab", "bc", "cd", "da".
   • Check each of these substrings against the set st:
     • "ab" is not in st.
     • "bc" is not in st.
     • "cd" is not in st.
     • "da" is in st.

4. Return Result:

   • Since "da" is found in both the original string s and the set st from the reversed string, the function returns true.

This demonstrates that "da" exists as a substring in both s and its reverse, satisfying the problem's conditions.

Solution Implementation

Time and Space Complexity

• Time Complexity: The pairwise function, if implemented optimally, will iterate over the string s with a linear pass, making it O(n). Creating a set of pairs from the reversed string and checking existence in the set both require n operations, thus maintaining the O(n) complexity.

• Space Complexity: The space complexity results from storing all pairs of adjacent characters in a set, which could theoretically grow to O(|\Sigma|^2) where \Sigma represents the character set. For lowercase English letters, |\Sigma| = 26, leading to a maximal space of storing 26^2 distinct pairs.

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