What is the main idea behind Time Needed to Rearrange a Binary String?

The core concept is flipping all 01 pairs simultaneously each second until no 01 remains, tracking total iterations.

Can I optimize beyond simple simulation?

Yes, by tracking zero positions or using dynamic programming, you can reduce redundant operations and compute the time efficiently.

How does state transition dynamic programming apply here?

Each 0's movement past 1s represents a state transition; DP records the time each zero takes, allowing quick computation of total seconds.

What are common mistakes to avoid?

Not performing simultaneous flips, mishandling initial strings with no 01, or inefficiently updating the string can lead to wrong answers or slow solutions.

What constraints should I remember?

The string length is between 1 and 1000, and each character is either 0 or 1, which affects performance considerations for different approaches.

#2380

Medium

auto_awesome状态·转移·动态规划

LeetCode 题解工作台

二进制字符串重新安排顺序需要的时间

给你一个二进制字符串 s 。在一秒之中，所有子字符串 "01" 同时被替换成 "10" 。这个过程持续进行到没有 "01" 存在。请你返回完成这个过程所需要的秒数。示例 1：输入： s = "0110101" 输出： 4 解释：一秒后，s 变成 "1011010" 。再过 1 秒后，…

字符串动态规划模拟

题目描述

给你一个二进制字符串 s 。在一秒之中，所有子字符串 "01" 同时被替换成 "10" 。这个过程持续进行到没有 "01" 存在。

请你返回完成这个过程所需要的秒数。

示例 1：

输入：s = "0110101"
输出：4
解释：
一秒后，s 变成 "1011010" 。
再过 1 秒后，s 变成 "1101100" 。
第三秒过后，s 变成 "1110100" 。
第四秒后，s 变成 "1111000" 。
此时没有 "01" 存在，整个过程花费 4 秒。
所以我们返回 4 。

示例 2：

输入：s = "11100"
输出：0
解释：
s 中没有 "01" 存在，整个过程花费 0 秒。
所以我们返回 0 。

提示：

1 <= s.length <= 1000
s[i] 要么是 '0' ，要么是 '1' 。

进阶：

你能以 O(n) 的时间复杂度解决这个问题吗？

lightbulb

解题思路

方法一：暴力模拟

由于本题数据范围不大，所以可以暴力模拟，每一轮将字符串中所有 “01” 替换为 “10”，统计轮次作为答案。

时间复杂度 $O(n^2)$ 。每一轮时间复杂度 $O(n)$ ，最多进行 $n$ 轮操作。

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class Solution:
    def secondsToRemoveOccurrences(self, s: str) -> int:
        ans = 0
        while s.count('01'):
            s = s.replace('01', '10')
            ans += 1
        return ans

speed

复杂度分析

指标	值
时间	complexity depends on the approach: direct simulation can reach O(n^2) in worst cases, while tracking positions or DP can achieve O(n). Space complexity ranges from O(n) for storing positions or DP arrays.
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
The candidate may attempt a naive loop-based simulation first.
question_mark
Look for candidates identifying overlapping flips and simultaneous state changes.
question_mark
Efficient solutions often emerge by tracking 0 positions or using DP to model time steps.

warning

常见陷阱

外企场景

error
Failing to simulate flips simultaneously, leading to incorrect step counts.
error
Ignoring edge cases where the string has no 01 pairs from the start.
error
Using inefficient string concatenations repeatedly, causing performance issues.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Determine time for similar transformations in ternary strings with multiple patterns.
arrow_right_alt
Compute minimal steps if only a single 01 can flip per second instead of all simultaneously.
arrow_right_alt
Count the total number of flips instead of seconds until stabilization.

help

常见问题

外企场景

继续练习

#2370 最长理想子序列 #2390 从字符串中移除星号 #2430 对字母串可执行的最大删除数