Introduction

In this blog post, we will explore how to solve the LeetCode problem "3106" 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 k. Define a function distance(s1, s2) between two strings s1 and s2 of the same length n as: The sum of the minimum distance between s1[i] and s2[i] when the characters from 'a' to 'z' are placed in a cyclic order, for all i in the range [0, n - 1]. For example, distance("ab", "cd") == 4, and distance("a", "z") == 1. You can change any letter of s to any other lowercase English letter, any number of times. Return a string denoting the lexicographically smallest string t you can get after some changes, such that distance(s, t) <= k. Example 1: Input: s = "zbbz", k = 3 Output: "aaaz" Explanation: Change s to "aaaz". The distance between "zbbz" and "aaaz" is equal to k = 3. Example 2: Input: s = "xaxcd", k = 4 Output: "aawcd" Explanation: The distance between "xaxcd" and "aawcd" is equal to k = 4. Example 3: Input: s = "lol", k = 0 Output: "lol" Explanation: It's impossible to change any character as k = 0. Constraints: 1 <= s.length <= 100 0 <= k <= 2000 s consists only of lowercase English letters.

Explanation

Here's the solution to the problem, with a high-level explanation, complexity analysis, and the complete Python code.

• Greedy Approach: Iterate through the input string s. For each character, try to change it to the smallest possible lexicographical character ('a', 'b', 'c', ...) such that the overall distance between the modified string and the original string does not exceed k.

• Distance Calculation: The distance between two characters c1 and c2 is min(abs(ord(c1) - ord(c2)), 26 - abs(ord(c1) - ord(c2))). Update the remaining k value accordingly.

• Lexicographical Minimization: Prioritize smaller characters during replacement to obtain the lexicographically smallest result.

• Complexity Analysis: The solution has a time complexity of O(n*26) = O(n) and a space complexity of O(n), where n is the length of the string s.

Code

    def get_distance(c1, c2):
    diff = abs(ord(c1) - ord(c2))
    return min(diff, 26 - diff)

def change_string(s, k):
    n = len(s)
    t = list(s)
    for i in range(n):
        for char_code in range(ord('a'), ord('z') + 1):
            new_char = chr(char_code)
            dist = get_distance(s[i], new_char)
            if dist <= k:
                t[i] = new_char
                k -= dist
                break
    return "".join(t)