Introduction

In this blog post, we will explore how to solve the LeetCode problem "3083" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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. Example 1: Input: s = "leetcode" Output: true Explanation: Substring "ee" is of length 2 which is also present in reverse(s) == "edocteel". Example 2: Input: s = "abcba" Output: true Explanation: All of the substrings of length 2 "ab", "bc", "cb", "ba" are also present in reverse(s) == "abcba". Example 3: Input: s = "abcd" Output: false Explanation: There is no substring of length 2 in s, which is also present in the reverse of s. Constraints: 1 <= s.length <= 100 s consists only of lowercase English letters.

Explanation

Here's a solution to the problem:

• High-Level Approach: The solution iterates through all possible substrings of length 2 in the input string s. For each substring, it checks if the substring exists in the reversed string. If a match is found, the function immediately returns True. If the loop completes without finding any matches, the function returns False.

• Complexity:

  • Runtime Complexity: O(n^2), where n is the length of the string s.
  • Storage Complexity: O(n), due to the reversed string.

Code

    def has_substring_in_reverse(s: str) -> bool:
    """
    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.

    Example 1:
    Input: s = "leetcode"
    Output: true
    Explanation: Substring "ee" is of length 2 which is also present in reverse(s) == "edocteel".

    Example 2:
    Input: s = "abcba"
    Output: true
    Explanation: All of the substrings of length 2 "ab", "bc", "cb", "ba" are also present in reverse(s) == "abcba".

    Example 3:
    Input: s = "abcd"
    Output: false
    Explanation: There is no substring of length 2 in s, which is also present in the reverse of s.

    Constraints:
    1 <= s.length <= 100
    s consists only of lowercase English letters.
    """
    reversed_s = s[::-1]
    for i in range(len(s) - 1):
        substring = s[i:i+2]
        if substring in reversed_s:
            return True
    return False