How do you approach solving Minimum Suffix Flips?

The problem can be solved using a greedy strategy, flipping bits only when mismatches occur. This minimizes the number of operations required.

What is the time complexity of solving Minimum Suffix Flips?

The time complexity is O(n), where n is the length of the target string. This is because we only need to iterate through the string once.

What does the greedy choice in Minimum Suffix Flips involve?

The greedy choice involves flipping at positions where mismatches occur, ensuring each flip operation reduces the number of mismatches optimally.

Can you optimize the solution further for Minimum Suffix Flips?

The greedy approach is already optimal for this problem, and no further optimization is needed as it achieves the best time complexity.

What are the common mistakes in solving Minimum Suffix Flips?

Common mistakes include incorrectly tracking the flip count, overcomplicating the solution, and failing to account for the impact of each flip.

#1529

Medium

auto_awesome贪心·invariant

LeetCode 题解工作台

最少的后缀翻转次数

给你一个长度为 n 、下标从 0 开始的二进制字符串 target 。你自己有另一个长度为 n 的二进制字符串 s ，最初每一位上都是 0 。你想要让 s 和 target 相等。在一步操作，你可以选择下标 i （ 0 ）并翻转在闭区间 [i, n - 1] 内的所有位。翻转意味着 '0' 变为…

字符串贪心

题目描述

给你一个长度为 n 、下标从 0 开始的二进制字符串 target 。你自己有另一个长度为 n 的二进制字符串 s ，最初每一位上都是 0 。你想要让 s 和 target 相等。

在一步操作，你可以选择下标 i（0 <= i < n）并翻转在 闭区间 [i, n - 1] 内的所有位。翻转意味着 '0' 变为 '1' ，而 '1' 变为 '0' 。

返回使 s 与 target 相等需要的最少翻转次数。

示例 1：

输入：target = "10111"
输出：3
解释：最初，s = "00000" 。
选择下标 i = 2: "00000" -> "00111"
选择下标 i = 0: "00111" -> "11000"
选择下标 i = 1: "11000" -> "10111"
要达成目标，需要至少 3 次翻转。

示例 2：

输入：target = "101"
输出：3
解释：最初，s = "000" 。
选择下标 i = 0: "000" -> "111"
选择下标 i = 1: "111" -> "100"
选择下标 i = 2: "100" -> "101"
要达成目标，需要至少 3 次翻转。

示例 3：

输入：target = "00000"
输出：0
解释：由于 s 已经等于目标，所以不需要任何操作

提示：

n == target.length
1 <= n <= 10⁵
target[i] 为 '0' 或 '1'

lightbulb

解题思路

方法一：贪心

我们从左到右遍历字符串 $\textit{target}$ ，用变量 $\textit{ans}$ 记录翻转次数。当遍历到下标 $i$ 时，如果当前的翻转次数 $\textit{ans}$ 的奇偶性与 $\textit{target}[i]$ 不同，则需要在下标 $i$ 处进行一次翻转操作，将 $\textit{ans}$ 加 $1$ 。

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, target: str) -> int:
        ans = 0
        for v in target:
            if (ans & 1) ^ int(v):
                ans += 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
Candidate demonstrates an understanding of greedy algorithms by applying it to string manipulation.
question_mark
Candidate proposes a solution with O(n) time complexity, which is optimal for this problem.
question_mark
Candidate identifies and implements the invariant validation effectively, minimizing the number of operations.

warning

常见陷阱

外企场景

error
Failing to track the flip count correctly, leading to unnecessary operations or missed flips.
error
Overcomplicating the solution by trying to optimize beyond the greedy approach, which is already optimal.
error
Misunderstanding the effect of each flip and failing to track its impact correctly on the string.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Consider extending the problem to include different types of strings, such as strings with characters other than '0' and '1'.
arrow_right_alt
Explore the impact of reverse operations, where flips are performed from right to left instead of left to right.
arrow_right_alt
Modify the problem to consider minimizing the number of flips, but with constraints on the number of flips allowed per operation.

help

常见问题

外企场景

继续练习

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