#1221

Easy

auto_awesomeGreedy choice plus invariant validation

LeetCode Problem Workspace

Split a String in Balanced Strings

Split a string into the maximum number of balanced substrings where each substring has an equal number of 'L' and 'R'.

Problem Statement

You are given a balanced string s consisting of characters 'L' and 'R'. A balanced string has an equal number of 'L' and 'R' characters. Your task is to split the string into as many substrings as possible such that each substring is also balanced, meaning it contains equal numbers of 'L' and 'R'.

Return the maximum number of balanced substrings you can obtain. The string s is guaranteed to be balanced, and its length will be between 2 and 1000 characters, inclusive.

Examples

Example 1

Input: s = "RLRRLLRLRL"

Output: 4

s can be split into "RL", "RRLL", "RL", "RL", each substring contains same number of 'L' and 'R'.

Example 2

Input: s = "RLRRRLLRLL"

Output: 2

s can be split into "RL", "RRRLLRLL", each substring contains same number of 'L' and 'R'. Note that s cannot be split into "RL", "RR", "RL", "LR", "LL", because the 2nd and 5th substrings are not balanced.

Example 3

Input: s = "LLLLRRRR"

Output: 1

s can be split into "LLLLRRRR".

Constraints

2 <= s.length <= 1000
s[i] is either 'L' or 'R'.
s is a balanced string.

Solution Approach

Greedy Approach with Balance Tracking

Loop through the string from left to right, keeping a balance variable that increments for 'L' and decrements for 'R'. Each time the balance reaches zero, a balanced substring is formed, and the split counter is incremented. This approach ensures we maximize the number of substrings with an equal count of 'L' and 'R'.

Invariant Validation

To verify if a substring is balanced, we can check if the balance variable is zero. This ensures that for every 'L' there is a matching 'R' within the substring. The balance approach guarantees correctness without the need to explicitly track individual substring contents.

Time Complexity and Space Complexity

The time complexity depends on the approach used, but generally, it involves a single scan through the string, resulting in O(n) time complexity, where n is the length of the string. The space complexity is O(1) as only a few variables are used to track the balance and the number of splits.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity is O(n) because we traverse the string once. The space complexity is O(1) since we only use a constant amount of extra space for tracking the balance and the split count.

What Interviewers Usually Probe

Look for candidates who demonstrate a clear understanding of greedy algorithms and balancing.
Candidates should explain how the balance variable directly relates to substring splitting.
The solution should be efficient, ideally with linear time complexity.

Common Pitfalls or Variants

Common pitfalls

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

Follow-up variants

What if the string contains additional characters besides 'L' and 'R'? This would change the problem entirely, making it unsuitable for this approach.
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.
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.

FAQ

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.

terminal

Solution

Solution 1: Greedy

We use a variable $l$ to maintain the current balance of the string, i.e., the value of $l$ is the number of 'L's minus the number of 'R's in the current string. When the value of $l$ is 0, we have found a balanced string.

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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

Continue Practicing

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