Introduction

In this blog post, we will explore how to solve the LeetCode problem "2957" 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 0-indexed string word. In one operation, you can pick any index i of word and change word[i] to any lowercase English letter. Return the minimum number of operations needed to remove all adjacent almost-equal characters from word. Two characters a and b are almost-equal if a == b or a and b are adjacent in the alphabet. Example 1: Input: word = "aaaaa" Output: 2 Explanation: We can change word into "acaca" which does not have any adjacent almost-equal characters. It can be shown that the minimum number of operations needed to remove all adjacent almost-equal characters from word is 2. Example 2: Input: word = "abddez" Output: 2 Explanation: We can change word into "ybdoez" which does not have any adjacent almost-equal characters. It can be shown that the minimum number of operations needed to remove all adjacent almost-equal characters from word is 2. Example 3: Input: word = "zyxyxyz" Output: 3 Explanation: We can change word into "zaxaxaz" which does not have any adjacent almost-equal characters. It can be shown that the minimum number of operations needed to remove all adjacent almost-equal characters from word is 3. Constraints: 1 <= word.length <= 100 word consists only of lowercase English letters.

Explanation

Here's an efficient solution to the problem:

• Greedy Approach: Iterate through the string. If a character is almost-equal to the previous character, increment the operation count and effectively skip the current character by not considering it as the 'previous' for the next comparison. The key is that modifying the current character is always as good or better than modifying the previous one.

• Almost-Equal Check: Implement a helper function to determine if two characters are almost-equal based on the given criteria (equal or adjacent).

• Handle Edge Cases: Start with an empty 'previous' character and handle the first character comparison appropriately.

• Complexity:

  • Runtime Complexity: O(n), where n is the length of the word.
  • Storage Complexity: O(1)

Code

    def solve():
    word = input()
    n = len(word)
    ans = 0
    prev = ''

    for i in range(n):
        if prev == '':
            prev = word[i]
        elif is_almost_equal(word[i], prev):
            ans += 1
        else:
            prev = word[i]
    return ans

def is_almost_equal(a, b):
    if a == b:
        return True
    return abs(ord(a) - ord(b)) == 1

print(solve())