Introduction

In this blog post, we will explore how to solve the LeetCode problem "3335" 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 and an integer t, representing the number of transformations to perform. In one transformation, every character in s is replaced according to the following rules: If the character is 'z', replace it with the string "ab". Otherwise, replace it with the next character in the alphabet. For example, 'a' is replaced with 'b', 'b' is replaced with 'c', and so on. Return the length of the resulting string after exactly t transformations. Since the answer may be very large, return it modulo 109 + 7. Example 1: Input: s = "abcyy", t = 2 Output: 7 Explanation: First Transformation (t = 1): 'a' becomes 'b' 'b' becomes 'c' 'c' becomes 'd' 'y' becomes 'z' 'y' becomes 'z' String after the first transformation: "bcdzz" Second Transformation (t = 2): 'b' becomes 'c' 'c' becomes 'd' 'd' becomes 'e' 'z' becomes "ab" 'z' becomes "ab" String after the second transformation: "cdeabab" Final Length of the string: The string is "cdeabab", which has 7 characters. Example 2: Input: s = "azbk", t = 1 Output: 5 Explanation: First Transformation (t = 1): 'a' becomes 'b' 'z' becomes "ab" 'b' becomes 'c' 'k' becomes 'l' String after the first transformation: "babcl" Final Length of the string: The string is "babcl", which has 5 characters. Constraints: 1 <= s.length <= 105 s consists only of lowercase English letters. 1 <= t <= 105

Explanation

Here's the breakdown of the solution:

• Dynamic Programming for 'z' Expansion: The core idea is to precompute how much each 'z' will expand into after t transformations. We can store this in a DP table, where dp[i] represents the length after i transformations. dp[0] = 1, and dp[i] = dp[i-1] * 2 % MOD.

• Character Transformation Logic: For each character in the input string, calculate how many steps away from 'a' it is. If the total steps (original distance + t) is less than 26, the character will simply shift, contributing 1 to the length.

• 'z' Case: If the total steps exceed 26, it means the character will turn into a 'z' after some transformations. The contribution will be dp[t - (26 - original_distance)], which represents the expansion of that 'z'.

• Runtime Complexity: O(t + n) & Storage Complexity: O(t) where t is the number of transformations and n is the length of the input string s.

Code

    def solve():
    s = input()
    t = int(input())
    MOD = 10**9 + 7
    n = len(s)
    dp = [0] * (t + 1)
    dp[0] = 1
    for i in range(1, t + 1):
        dp[i] = (dp[i - 1] * 2) % MOD

    ans = 0
    for char in s:
        dist = ord(char) - ord('a')
        total_dist = dist + t
        if total_dist < 26:
            ans = (ans + 1) % MOD
        else:
            remaining_transformations = t - (25 - dist)
            ans = (ans + dp[remaining_transformations]) % MOD

    return ans

print(solve())