Introduction

In this blog post, we will explore how to solve the LeetCode problem "3474" 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

You are given two strings, str1 and str2, of lengths n and m, respectively. A string word of length n + m - 1 is defined to be generated by str1 and str2 if it satisfies the following conditions for each index 0 <= i <= n - 1: If str1[i] == 'T', the substring of word with size m starting at index i is equal to str2, i.e., word[i..(i + m - 1)] == str2. If str1[i] == 'F', the substring of word with size m starting at index i is not equal to str2, i.e., word[i..(i + m - 1)] != str2. Return the lexicographically smallest possible string that can be generated by str1 and str2. If no string can be generated, return an empty string "". Example 1: Input: str1 = "TFTF", str2 = "ab" Output: "ababa" Explanation: The table below represents the string "ababa" Index T/F Substring of length m 0 'T' "ab" 1 'F' "ba" 2 'T' "ab" 3 'F' "ba" The strings "ababa" and "ababb" can be generated by str1 and str2. Return "ababa" since it is the lexicographically smaller string. Example 2: Input: str1 = "TFTF", str2 = "abc" Output: "" Explanation: No string that satisfies the conditions can be generated. Example 3: Input: str1 = "F", str2 = "d" Output: "a" Constraints: 1 <= n == str1.length <= 104 1 <= m == str2.length <= 500 str1 consists only of 'T' or 'F'. str2 consists only of lowercase English characters.

Explanation

Here's the approach to solve this problem:

• Initialization and Construction: Build a candidate string word of length n + m - 1 initialized with 'a' characters. This minimizes the lexicographical order initially.

• Constraint Enforcement: Iterate through str1. If str1[i] is 'T', enforce word[i:i+m] == str2. If str1[i] is 'F', enforce word[i:i+m] != str2. If a 'F' constraint cannot be met with minimal changes, increment the word and repeat.

• Lexicographical Minimization and Validation: Ensure the final word satisfies all constraints. If not, no such string exists; return "".

• Time Complexity: O(n m 26^(overlap)), where overlap is the length of the overlap between str1 and str2. In worst case, this will be O(n m 26^m). In most realistic cases, overlap should be very small (much less than m) leading to a good runtime.

• Space Complexity: O(n + m) for storing strings.

Code

    def solve():
    str1 = input()
    str2 = input()
    n = len(str1)
    m = len(str2)
    length = n + m - 1

    word = list('a' * length)

    def check(word_list):
        word = "".join(word_list)
        for i in range(n):
            if str1[i] == 'T':
                if i + m > length or word[i:i + m] != str2:
                    return False
            else:
                if i + m <= length and word[i:i + m] == str2:
                    return False
        return True

    def enforce_t(i, word_list):
        if i + m > length:
            return False
        for j in range(m):
            word_list[i + j] = str2[j]
        return True

    def enforce_f(i, word_list):
        if i + m > length:
            return True

        original_char = word_list[i]

        for char_code in range(ord('a'), ord('z') + 1):
            char = chr(char_code)

            temp_word = word_list[:]

            temp_word[i] = char

            is_different = True
            if i + m <= length:

                temp_str = "".join(temp_word[i: i + m])

                if temp_str == str2:
                    is_different = False

            if is_different:

                word_list[i] = char

                return True

        word_list[i] = original_char
        return False

    def increment_word(word_list):
        for i in range(len(word_list) - 1, -1, -1):
            if word_list[i] == 'z':
                word_list[i] = 'a'
            else:
                word_list[i] = chr(ord(word_list[i]) + 1)
                return True
        return False

    possible = True

    for i in range(n):
        if str1[i] == 'T':
            if not enforce_t(i, word):
                possible = False
                break

    if not possible:
        print("")
        return

    for i in range(n):
        if str1[i] == 'F':
            if not enforce_f(i, word):
                possible = False
                break

    if not possible:
        print("")
        return

    if check(word):
      print("".join(word))
    else:
        print("")

solve()