# Solving Leetcode Interviews in Seconds with AI: Lexicographically Smallest Generated String


	# 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](https://www.chatmagic.app), 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
	```python
	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()
	```