Introduction

In this blog post, we will explore how to solve the LeetCode problem "1638" 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 two strings s and t, find the number of ways you can choose a non-empty substring of s and replace a single character by a different character such that the resulting substring is a substring of t. In other words, find the number of substrings in s that differ from some substring in t by exactly one character. For example, the underlined substrings in "computer" and "computation" only differ by the 'e'/'a', so this is a valid way. Return the number of substrings that satisfy the condition above. A substring is a contiguous sequence of characters within a string. Example 1: Input: s = "aba", t = "baba" Output: 6 Explanation: The following are the pairs of substrings from s and t that differ by exactly 1 character: ("aba", "baba") ("aba", "baba") ("aba", "baba") ("aba", "baba") ("aba", "baba") ("aba", "baba") The underlined portions are the substrings that are chosen from s and t. ​​Example 2: Input: s = "ab", t = "bb" Output: 3 Explanation: The following are the pairs of substrings from s and t that differ by 1 character: ("ab", "bb") ("ab", "bb") ("ab", "bb") ​​​​The underlined portions are the substrings that are chosen from s and t. Constraints: 1 <= s.length, t.length <= 100 s and t consist of lowercase English letters only.

Explanation

Here's the breakdown of the solution:

• Iterate through all possible substrings of s: For each substring, we'll attempt to find a corresponding substring in t that differs by only one character.
• Iterate through all possible substrings of t with the same length: For each substring of t having the same length as the current substring of s, we compare the two.

• Compare substrings and count differences: If the two substrings differ by exactly one character, we increment the count.

• Runtime Complexity: O(m*n*len(s)*len(t)), where m and n are the lengths of s and t, respectively. Storage Complexity: O(1).

Code

    def countSubstrings(s: str, t: str) -> int:
    """
    Given two strings s and t, find the number of ways you can choose a non-empty substring of s and replace a single character by a different character such that the resulting substring is a substring of t. In other words, find the number of substrings in s that differ from some substring in t by exactly one character.

    Args:
        s (str): The first string.
        t (str): The second string.

    Returns:
        int: The number of substrings that satisfy the condition.
    """
    count = 0
    for i in range(len(s)):
        for j in range(i, len(s)):
            sub_s = s[i:j+1]
            for k in range(len(t) - len(sub_s) + 1):
                sub_t = t[k:k+len(sub_s)]
                diff = 0
                for l in range(len(sub_s)):
                    if sub_s[l] != sub_t[l]:
                        diff += 1
                if diff == 1:
                    count += 1
    return count