How does the greedy approach help solve the "Split a String in Balanced Strings" problem?

The greedy approach allows you to keep track of the balance between 'L' and 'R' as you iterate through the string. Every time the balance is zero, it indicates a balanced substring can be split.

What is the time complexity of the solution to "Split a String in Balanced Strings"?

The time complexity is O(n), where n is the length of the string, because the algorithm processes the string in one pass.

Can the approach handle strings of any length?

Yes, the solution is scalable for strings up to 1000 characters, as the time complexity is linear.

What is the role of the balance variable in the greedy approach?

The balance variable tracks the difference between the number of 'L' and 'R' characters encountered so far. A balance of zero indicates a balanced substring.

What is the space complexity of the solution to "Split a String in Balanced Strings"?

The space complexity is O(1), as the solution only uses a few variables to track the balance and number of substrings.

#1221

Easy

auto_awesome贪心·invariant

LeetCode 题解工作台

分割平衡字符串

平衡字符串中， 'L' 和 'R' 字符的数量是相同的。给你一个平衡字符串 s ，请你将它分割成尽可能多的子字符串，并满足：每个子字符串都是平衡字符串。返回可以通过分割得到的平衡字符串的最大数量。示例 1：输入： s = "RLRRLLRLRL" 输出： 4 解释： s 可以分割为 …

字符串贪心计数

题目描述

平衡字符串 中，'L' 和 'R' 字符的数量是相同的。

给你一个平衡字符串 s，请你将它分割成尽可能多的子字符串，并满足：

每个子字符串都是平衡字符串。

返回可以通过分割得到的平衡字符串的 最大数量 。

示例 1：

输入：s = "RLRRLLRLRL"
输出：4
解释：s 可以分割为 "RL"、"RRLL"、"RL"、"RL" ，每个子字符串中都包含相同数量的 'L' 和 'R' 。

示例 2：

输入：s = "RLRRRLLRLL"
输出：2
解释：s 可以分割为 "RL"、"RRRLLRLL"，每个子字符串中都包含相同数量的 'L' 和 'R' 。
注意，s 无法分割为 "RL"、"RR"、"RL"、"LR"、"LL" 因为第 2 个和第 5 个子字符串不是平衡字符串。

示例 3：

输入：s = "LLLLRRRR"
输出：1
解释：s 只能保持原样 "LLLLRRRR" 。

提示：

2 <= s.length <= 1000
s[i] = 'L' 或 'R'
s 是一个平衡字符串

lightbulb

解题思路

方法一：贪心

我们用变量 $l$ 维护当前字符串的平衡度，即 $l$ 的值为当前字符串中 $L$ 的数量减去 $R$ 的数量。当 $l$ 的值为 0 时，我们就找到了一个平衡字符串。

遍历字符串 $s$ ，当遍历到第 $i$ 个字符时，如果 $s[i] = L$ ，则 $l$ 的值加 1，否则 $l$ 的值减 1。当 $l$ 的值为 0 时，我们将答案加 1。

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

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class Solution:
    def balancedStringSplit(self, s: str) -> int:
        ans = l = 0
        for c in s:
            if c == 'L':
                l += 1
            else:
                l -= 1
            if l == 0:
                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
Look for candidates who demonstrate a clear understanding of greedy algorithms and balancing.
question_mark
Candidates should explain how the balance variable directly relates to substring splitting.
question_mark
The solution should be efficient, ideally with linear time complexity.

warning

常见陷阱

外企场景

error
Candidates may attempt to split the string based on arbitrary substrings without maintaining the balance correctly.
error
Mistakes can occur when the balance is not reset correctly or if splits are made without checking the balance.
error
Some may overcomplicate the problem by trying to split substrings explicitly instead of using a simple balance tracking method.

swap_horiz

进阶变体

外企场景

arrow_right_alt
What if the string contains additional characters besides 'L' and 'R'? This would change the problem entirely, making it unsuitable for this approach.
arrow_right_alt
Consider the case where the string consists of only 'L's or only 'R's. This would lead to an invalid problem definition, as a balanced string requires both characters.
arrow_right_alt
What happens when the input string is already split into balanced substrings? In this case, the algorithm should still work efficiently without any extra operations.

help

常见问题

外企场景

继续练习

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