What is the primary approach to solving the Remove Adjacent Almost-Equal Characters problem?

The primary approach involves using dynamic programming to track and minimize operations required to remove adjacent almost-equal character pairs in the string.

How do greedy techniques apply in the Remove Adjacent Almost-Equal Characters problem?

Greedy techniques are used to select the optimal character for each position, ensuring that adjacent almost-equal pairs are removed while minimizing operations.

What does 'almost-equal' mean in the context of this problem?

'Almost-equal' refers to two characters that are adjacent in the alphabet, such as 'a' and 'b' or 'y' and 'z'. These pairs need to be broken up to solve the problem.

How can space complexity be optimized in this problem?

Space complexity can be optimized by using a limited state-space for dynamic programming, tracking only necessary transitions and minimizing memory usage.

What are the time and space complexities of the optimal solution?

The time complexity of the optimal solution is generally O(n), and the space complexity is also O(n), where n is the length of the string.

#2957

Medium

auto_awesome状态·转移·动态规划

LeetCode 题解工作台

消除相邻近似相等字符

给你一个下标从 0 开始的字符串 word 。一次操作中，你可以选择 word 中任意一个下标 i ，将 word[i] 修改成任意一个小写英文字母。请你返回消除 word 中所有相邻近似相等字符的最少操作次数。两个字符 a 和 b 如果满足 a == b 或者 a 和 b 在字母表中…

字符串动态规划贪心

题目描述

给你一个下标从 0 开始的字符串 word 。

一次操作中，你可以选择 word 中任意一个下标 i ，将 word[i] 修改成任意一个小写英文字母。

请你返回消除 word 中所有相邻 近似相等 字符的最少操作次数。

两个字符 a 和 b 如果满足 a == b 或者 a 和 b 在字母表中是相邻的，那么我们称它们是 近似相等 字符。

示例 1：

输入：word = "aaaaa"
输出：2
解释：我们将 word 变为 "acaca" ，该字符串没有相邻近似相等字符。
消除 word 中所有相邻近似相等字符最少需要 2 次操作。

示例 2：

输入：word = "abddez"
输出：2
解释：我们将 word 变为 "ybdoez" ，该字符串没有相邻近似相等字符。
消除 word 中所有相邻近似相等字符最少需要 2 次操作。

示例 3：

输入：word = "zyxyxyz"
输出：3
解释：我们将 word 变为 "zaxaxaz" ，该字符串没有相邻近似相等字符。
消除 word 中所有相邻近似相等字符最少需要 3 次操作

提示：

1 <= word.length <= 100
word 只包含小写英文字母。

lightbulb

解题思路

方法一：贪心

我们从下标 $1$ 开始遍历字符串 $word$ ，如果 $word[i]$ 和 $word[i - 1]$ 相邻近似相等，那么我们就贪心地将 $word[i]$ 替换成一个与 $word[i - 1]$ 和 $word[i + 1]$ 都不相等的字符（可以不执行替换操作，记录操作次数即可）。然后，我们跳过 $word[i + 1]$ ，继续遍历字符串 $word$ 。

最后，我们返回记录的操作次数。

时间复杂度 $O(n)$ ，其中 $n$ 是字符串 $word$ 的长度。空间复杂度 $O(1)$ 。

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class Solution:
    def removeAlmostEqualCharacters(self, word: str) -> int:
        ans = 0
        i, n = 1, len(word)
        while i < n:
            if abs(ord(word[i]) - ord(word[i - 1])) < 2:
                ans += 1
                i += 2
            else:
                i += 1
        return ans

speed

复杂度分析

指标	值
时间	Depends on the final approach
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
Strong understanding of dynamic programming is required to identify and manage state transitions.
question_mark
Greedy strategies must be applied carefully to ensure optimal character transformations.
question_mark
The ability to minimize space complexity while solving the problem shows an efficient solution approach.

warning

常见陷阱

外企场景

error
Failing to properly account for all adjacent almost-equal character pairs during transformations can lead to suboptimal solutions.
error
Incorrect greedy character selection might result in more operations than necessary, especially in longer strings.
error
Excessive space complexity due to improper state tracking or inefficient memory management can negatively impact performance.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Consider variations where the allowed transformations could involve changing characters to any set of characters, not just lowercase English letters.
arrow_right_alt
Explore the scenario where adjacent characters have varying levels of 'almost-equality,' requiring more complex state transitions.
arrow_right_alt
Analyze the problem with longer strings, where the dynamic programming approach's time complexity becomes more critical.

help

常见问题

外企场景

继续练习

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