How do I approach the Minimum Number of Flips to Make the Binary String Alternating problem?

Start by understanding the alternating string pattern and use dynamic programming to track flips needed at odd and even positions.

What dynamic programming technique is used in this problem?

State transition dynamic programming is used to track the minimum flips for each possible alternating pattern based on the string's current state.

What are common mistakes when solving the Minimum Number of Flips to Make the Binary String Alternating problem?

Failing to optimize for large input sizes and not efficiently tracking the number of flips for odd and even indexed positions can lead to inefficient solutions.

How can I optimize my solution for large strings in this problem?

Focus on tracking only essential values like the number of flips needed at odd and even indices, and avoid unnecessary operations or additional data structures.

Can the problem be adapted to other types of strings?

Yes, you can adapt the problem to non-binary strings or different operations, but the underlying pattern of alternating characters remains the same.

#1888

Medium

auto_awesome状态·转移·动态规划

LeetCode 题解工作台

使二进制字符串字符交替的最少反转次数

给你一个二进制字符串 s 。你可以按任意顺序执行以下两种操作任意次：类型 1 ：删除字符串 s 的第一个字符并将它添加到字符串结尾。类型 2 ：选择字符串 s 中任意一个字符并将该字符反转，也就是如果值为 '0' ，则反转得到 '1' ，反之亦然。请你返回使 s 变成交替字符串…

字符串动态规划滑动窗口

题目描述

给你一个二进制字符串 s 。你可以按任意顺序执行以下两种操作任意次：

类型 1 ：删除 字符串 s 的第一个字符并将它添加到字符串结尾。
类型 2 ：选择 字符串 s 中任意一个字符并将该字符反转，也就是如果值为 '0' ，则反转得到 '1' ，反之亦然。

请你返回使 s 变成交替字符串的前提下， 类型 2 的最少操作次数。

我们称一个字符串是交替的，需要满足任意相邻字符都不同。

比方说，字符串 "010" 和 "1010" 都是交替的，但是字符串 "0100" 不是。

示例 1：

输入：s = "111000"
输出：2
解释：执行第一种操作两次，得到 s = "100011" 。
然后对第三个和第六个字符执行第二种操作，得到 s = "101010" 。

示例 2：

输入：s = "010"
输出：0
解释：字符串已经是交替的。

示例 3：

输入：s = "1110"
输出：1
解释：对第二个字符执行第二种操作，得到 s = "1010" 。

提示：

1 <= s.length <= 10⁵
s[i] 要么是 '0' ，要么是 '1' 。

lightbulb

解题思路

方法一：滑动窗口

我们注意到，操作 $1$ 的作用实际上是让字符串成为一个环，而操作 $2$ 是使得环中的一段长度为 $n$ 的子串变成交替二进制串。

因此，我们只需要枚举每个长度为 $n$ 的子串，计算将其变成交替二进制串的代价，取最小值即可。

我们可以预先计算出字符串 $s$ 与两种交替二进制串的差异数量，记为 $\textit{cnt}$ ，则将 $s$ 变成第一种交替二进制串的代价为 $\textit{cnt}$ ，将 $s$ 变成第二种交替二进制串的代价为 $n - \textit{cnt}$ 。我们初始化 $\textit{ans} = \min(\textit{cnt}, n - \textit{cnt})$ 。

接下来，我们枚举每个长度为 $n$ 的子串，更新 $\textit{cnt}$ 的值。对于每个位置 $i$ ，我们将 $s[i]$ 从第一种交替二进制串的差异数量中减去，并将 $s[i]$ 加入到第二种交替二进制串的差异数量中。更新 $\textit{ans} = \min(\textit{ans}, \textit{cnt}, n - \textit{cnt})$ 。

最后返回 $\textit{ans}$ 即可。

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

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class Solution:
    def minFlips(self, s: str) -> int:
        n = len(s)
        target = "01"
        cnt = sum(c != target[i & 1] for i, c in enumerate(s))
        ans = min(cnt, n - cnt)
        for i in range(n):
            cnt -= s[i] != target[i & 1]
            cnt += s[i] != target[(i + n) & 1]
            ans = min(ans, cnt, n - cnt)
        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
Look for understanding of state transitions and their optimization in a string manipulation context.
question_mark
Check if the candidate can connect dynamic programming principles with practical application, especially in minimizing operations.
question_mark
Assess whether the candidate focuses on the efficiency of the algorithm, especially for large input sizes.

warning

常见陷阱

外企场景

error
Failing to recognize the importance of tracking odd and even positions separately can lead to inefficient solutions.
error
Not optimizing for large strings may result in time complexity that doesn't meet the problem's constraints.
error
Overcomplicating the solution with unnecessary operations can lead to excessive space or time complexity.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Consider variants where the string contains characters other than '0' and '1', such as binary sequences with additional symbols.
arrow_right_alt
Adapt the problem to consider non-binary strings where adjacent characters must still differ, like alternating colors in a grid.
arrow_right_alt
Change the problem to allow more complex operations, such as flipping multiple bits at once or limiting the number of allowed flips.

help

常见问题

外企场景

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