Problem Description

You are given a string s. Your task is to determine if there exists any substring of length 2 that appears in both the original string s and its reverse.

A substring of length 2 consists of two consecutive characters from the string. For example, if s = "abc", the substrings of length 2 are "ab" and "bc".

The reverse of a string is obtained by reading the characters from right to left. For example, if s = "abc", its reverse is "cba".

You need to check if any 2-character substring from s also appears as a substring in the reversed version of s. If such a substring exists, return true; otherwise, return false.

For example:

• If s = "abcd", the reverse is "dcba". The substring "ab" from the original does not appear in "dcba", "bc" does not appear in "dcba", and "cd" does not appear in "dcba". So the answer would be false.
• If s = "aba", the reverse is "aba". The substring "ab" appears in both the original and reverse, so the answer would be true.

The solution uses a hash table to store all 2-character substrings from the reversed string, then checks if any 2-character substring from the original string exists in this set. The pairwise function is used to generate consecutive pairs of characters from the string.

Intuition

The key insight is that we need to find if any 2-character substring appears in both the original string and its reverse. Instead of checking every possible combination, we can think of this as a set intersection problem.

First, let's understand what we're looking for. If a substring "xy" appears in the original string s, and we want it to also appear in the reverse of s, then somewhere in the reversed string there must be the same pattern "xy".

The straightforward approach would be to:

1. Extract all 2-character substrings from the original string
2. Extract all 2-character substrings from the reversed string
3. Check if there's any common substring between these two sets

Since we're looking for any match (not all matches), we can optimize this by:

1. First collecting all 2-character substrings from the reversed string into a set for O(1) lookup
2. Then iterating through the original string's 2-character substrings and checking if any exists in our set

The use of pairwise function elegantly generates consecutive character pairs. For a string like "abc", pairwise would give us ('a','b') and ('b','c'). By storing these pairs as tuples in a set from the reversed string, we can quickly check membership when iterating through pairs from the original string.

This approach is efficient because:

• We only need to traverse each string once
• Set operations (insertion and lookup) are O(1) on average
• We can return immediately upon finding the first match, avoiding unnecessary computation

Solution Approach

The solution implements a hash table-based approach to efficiently find common 2-character substrings between a string and its reverse.

Step 1: Extract substrings from the reversed string

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

• s[::-1] reverses the string s
• pairwise() generates consecutive character pairs from the reversed string
• We use a set comprehension to store all these pairs as tuples in a hash set st
• For example, if s = "abcd", then s[::-1] = "dcba", and st would contain {('d','c'), ('c','b'), ('b','a')}

Step 2: Check for matching substrings in the original string

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

• pairwise(s) generates consecutive character pairs from the original string
• For each pair (a, b), we check if it exists in the set st using the in operator (O(1) lookup)
• The any() function returns True as soon as it finds the first matching pair, providing early termination
• If no matching pair is found after checking all pairs, it returns False

Data Structures Used:

• Set (Hash Table): Stores the 2-character substrings from the reversed string as tuples, enabling O(1) average-case lookup time
• Tuple: Each 2-character substring is represented as a tuple (char1, char2) for immutability and hashability

Time Complexity: O(n) where n is the length of the string, as we traverse the string twice (once for reverse, once for original)

Space Complexity: O(n) for storing the set of 2-character substrings from the reversed string

Example Walkthrough

Let's walk through the solution with s = "abcba":

Step 1: Reverse the string and extract 2-character substrings

• Original string: "abcba"
• Reversed string: "abcba" (it's a palindrome in this case)
• Apply pairwise() to reversed string to get consecutive pairs:
  • Position 0-1: ('a', 'b')
  • Position 1-2: ('b', 'c')
  • Position 2-3: ('c', 'b')
  • Position 3-4: ('b', 'a')
• Store these in set st = {('a','b'), ('b','c'), ('c','b'), ('b','a')}

Step 2: Check original string's 2-character substrings

• Apply pairwise() to original string "abcba":
  • Position 0-1: ('a', 'b') → Check: Is ('a','b') in st? YES! ✓
• Since we found a match, any() immediately returns True

The function returns True because the substring "ab" appears in both the original and reversed string.

Let's also trace through a negative example with s = "abcd":

Step 1: Reverse and extract substrings

• Original: "abcd"
• Reversed: "dcba"
• Pairs from reversed: ('d','c'), ('c','b'), ('b','a')
• Set st = {('d','c'), ('c','b'), ('b','a')}

Step 2: Check original string's substrings

• Pairs from original:
  • ('a','b') → Not in st ✗
  • ('b','c') → Not in st ✗
  • ('c','d') → Not in st ✗
• No matches found, return False

The key insight is that we're looking for exact 2-character matches, and the set-based approach lets us check this efficiently in linear time.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n)

The algorithm consists of two main operations:

• Creating a set st from consecutive pairs in the reversed string s[::-1], which takes O(n) time to reverse the string and O(n) time to iterate through all consecutive pairs.
• Checking if any consecutive pair from the original string exists in the set, which takes O(n) time to iterate through pairs and O(1) average time for each set lookup.

Therefore, the overall time complexity is O(n) where n is the length of the string.

Space Complexity: O(|Σ|²)

The space is primarily consumed by the set st which stores tuples of consecutive character pairs. In the worst case, the set can contain all possible combinations of two characters from the alphabet. Since the string consists of lowercase English letters where |Σ| = 26, the maximum number of possible pairs is 26 × 26 = 676. This gives us a space complexity of O(|Σ|²) which is O(26²) = O(676) = O(1) in terms of constant space, but more precisely expressed as O(|Σ|²) to show its dependency on the alphabet size.

The reversed string s[::-1] also takes O(n) space temporarily, but since n can be arbitrarily large while |Σ|² is bounded by the alphabet size, the dominant factor in the worst case is O(min(n, |Σ|²)), which simplifies to O(|Σ|²) as stated in the reference answer.

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

Common Pitfalls

1. Not Handling Edge Cases for Short Strings

The pairwise() function returns an empty iterator when the input string has length less than 2. While the current solution handles this correctly (returning False for empty iterations), developers might incorrectly assume they need special handling for these cases.

Pitfall Example:

def isSubstringPresent(self, s: str) -> bool:
    # Unnecessary edge case handling
    if len(s) < 2:
        return False  # This check is redundant
  
    reversed_pairs = {(a, b) for a, b in pairwise(s[::-1])}
    return any((a, b) in reversed_pairs for a, b in pairwise(s))

Why it's unnecessary: The pairwise() function naturally handles strings of length 0 or 1 by returning no pairs, causing both the set comprehension and any() to work correctly without explicit checks.

2. Comparing Strings Instead of Character Tuples

A common mistake is trying to store and compare 2-character substrings as strings instead of tuples, which can lead to subtle bugs or inefficiencies.

Pitfall Example:

def isSubstringPresent(self, s: str) -> bool:
    # Using string concatenation instead of tuples
    reversed_pairs = {a + b for a, b in pairwise(s[::-1])}
    return any(a + b in reversed_pairs for a, b in pairwise(s))

Issue: While this works, string concatenation is less efficient than tuple creation and comparison. Tuples are immutable and have better hash performance.

3. Manually Implementing Sliding Window Instead of Using pairwise()

Without knowledge of itertools.pairwise() (introduced in Python 3.10), developers might write error-prone manual implementations.

Pitfall Example:

def isSubstringPresent(self, s: str) -> bool:
    reversed_s = s[::-1]
    reversed_pairs = set()
  
    # Manual sliding window - prone to off-by-one errors
    for i in range(len(reversed_s) - 1):  # Easy to forget the -1
        reversed_pairs.add((reversed_s[i], reversed_s[i + 1]))
  
    for i in range(len(s) - 1):
        if (s[i], s[i + 1]) in reversed_pairs:
            return True
    return False

Better Alternative for Python < 3.10:

def isSubstringPresent(self, s: str) -> bool:
    # Define pairwise locally if not available
    def pairwise(iterable):
        a = iter(iterable)
        b = iter(iterable)
        next(b, None)
        return zip(a, b)
  
    reversed_pairs = {(a, b) for a, b in pairwise(s[::-1])}
    return any((a, b) in reversed_pairs for a, b in pairwise(s))

4. Incorrect Understanding of the Problem

Some might misinterpret the problem as finding palindromic substrings or checking if the same 2-character substring appears twice in the original string.

Pitfall Example:

def isSubstringPresent(self, s: str) -> bool:
    # Wrong: Checking for palindromic substrings
    for i in range(len(s) - 1):
        if s[i] == s[i + 1]:  # This checks for repeated characters
            return True
    return False

Correct Understanding: We need to check if any 2-character substring from s appears anywhere in reverse(s), not just at the corresponding reversed position.