Introduction

In this blog post, we will explore how to solve the LeetCode problem "1625" 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 a string s of even length consisting of digits from 0 to 9, and two integers a and b. You can apply either of the following two operations any number of times and in any order on s: Add a to all odd indices of s (0-indexed). Digits post 9 are cycled back to 0. For example, if s = "3456" and a = 5, s becomes "3951". Rotate s to the right by b positions. For example, if s = "3456" and b = 1, s becomes "6345". Return the lexicographically smallest string you can obtain by applying the above operations any number of times on s. A string a is lexicographically smaller than a string b (of the same length) if in the first position where a and b differ, string a has a letter that appears earlier in the alphabet than the corresponding letter in b. For example, "0158" is lexicographically smaller than "0190" because the first position they differ is at the third letter, and '5' comes before '9'. Example 1: Input: s = "5525", a = 9, b = 2 Output: "2050" Explanation: We can apply the following operations: Start: "5525" Rotate: "2555" Add: "2454" Add: "2353" Rotate: "5323" Add: "5222" Add: "5121" Rotate: "2151" Add: "2050"​​​​​ There is no way to obtain a string that is lexicographically smaller than "2050". Example 2: Input: s = "74", a = 5, b = 1 Output: "24" Explanation: We can apply the following operations: Start: "74" Rotate: "47" ​​​​​​​Add: "42" ​​​​​​​Rotate: "24"​​​​​​​​​​​​ There is no way to obtain a string that is lexicographically smaller than "24". Example 3: Input: s = "0011", a = 4, b = 2 Output: "0011" Explanation: There are no sequence of operations that will give us a lexicographically smaller string than "0011". Constraints: 2 <= s.length <= 100 s.length is even. s consists of digits from 0 to 9 only. 1 <= a <= 9 1 <= b <= s.length - 1

Explanation

Here's the breakdown of the solution:

• BFS for Exploration: We use Breadth-First Search (BFS) to explore all possible string configurations reachable from the initial string s by applying the allowed operations. This ensures we find the lexicographically smallest string.
• String Transformations: For each string encountered in the BFS, we generate its neighbors by applying the 'add' operation on odd indices and the 'rotate' operation.

• Lexicographical Minimization: We maintain a visited set to avoid cycles and only process strings that haven't been seen before. The lexicographically smallest string found so far is updated during the BFS.

• Time Complexity: O(n*10*n) where n is the length of the string s. Space Complexity: O(n*10*n) due to the queue and visited set potentially storing all possible string states.

Code

    from collections import deque

def findLexSmallestString(s: str, a: int, b: int) -> str:
    n = len(s)
    visited = set()
    q = deque([s])
    smallest = s

    while q:
        curr = q.popleft()
        if curr in visited:
            continue
        visited.add(curr)

        if curr < smallest:
            smallest = curr

        # Add operation
        add_arr = list(curr)
        for i in range(1, n, 2):
            add_arr[i] = str((int(add_arr[i]) + a) % 10)
        add_str = "".join(add_arr)
        if add_str not in visited:
            q.append(add_str)

        # Rotate operation
        rotate_str = curr[-b:] + curr[:-b]
        if rotate_str not in visited:
            q.append(rotate_str)

    return smallest