Leetcode – Split a String in Balanced Strings Solution

Spread the love

The “Split a String in Balanced Strings” problem is an engaging exercise that tests your ability to work with strings and basic logic. It’s an excellent task for those just starting with algorithmic challenges, but it also presents opportunities for optimization and diving deeper into string manipulation techniques. In this article, we will explore multiple ways to solve this problem, ranging from brute-force methods to optimized solutions. We will discuss their time and space complexities and demonstrate the solutions using Python code examples.

Problem Statement


The problem requires you to split a given string s into as many balanced strings as possible. A balanced string is a string that has an equal number of ‘L’s and ‘R’s. You can return the maximum number of balanced strings you can make.


  • 1 <= s.length <= 1000
  • s[i] is either ‘L’ or ‘R’

Understanding the Problem

The problem is about cutting a given string into substrings, such that each substring is balanced. To be balanced, a substring must contain an equal number of ‘L’s and ‘R’s. Your task is to find the maximum number of such substrings.


Approach 1: Using a Counter

The simplest way to approach this problem is to iterate through the string from left to right and maintain a counter that keeps track of the balance between ‘L’s and ‘R’s at any given point. Whenever the counter reaches zero, we know that we’ve found a balanced string, and we can reset the counter.


  1. Initialize a counter variable balance to 0.
  2. Initialize a variable count to store the number of balanced strings found, also set to 0.
  3. Iterate through the string character by character.
    • Increment balance for every ‘R’ and decrement it for every ‘L’.
    • Whenever balance becomes zero, increment count and continue.

Python Code Implementation

def balancedStringSplit(s):
    balance = 0
    count = 0
    for char in s:
        if char == 'R':
            balance += 1
            balance -= 1
        if balance == 0:
            count += 1
    return count

# Test the function
print(balancedStringSplit("RLRRLLRLRL"))  # Output should be 4
print(balancedStringSplit("RLLLLRRRLR"))  # Output should be 3

Time Complexity

This approach runs in O(n) time, where n is the length of the string s.

Space Complexity

The space complexity is O(1), as we only use a couple of extra variables.

Approach 2: Optimizing with Early Returns

In certain scenarios, especially when dealing with very long strings, it might be beneficial to stop the loop as soon as you find a balanced string. This could be done if the problem were to ask for any balanced string rather than the maximum number. However, given that the problem wants us to find the maximum number of balanced strings, an early return wouldn’t be appropriate here.

Approach 3: Using Pythonic Techniques

Python offers a range of techniques to make code more compact without sacrificing readability. While this may not improve the algorithmic complexity, it often results in more maintainable code. For instance, we can use the Counter class from Python’s standard library to create a more Pythonic solution. However, given the simplicity of this problem, using Counter might be an overkill, but it’s an option worth considering for more complex scenarios.

Code Implementation Using Pythonic Techniques

For this problem, using Pythonic techniques may not make a significant difference, but here’s a one-liner using list comprehension.

def balancedStringSplit(s):
    return sum([(balance := balance + 1 if char == 'R' else balance - 1) == 0 for balance, char in enumerate(s, start=0)])

# Test the function
print(balancedStringSplit("RLRRLLRLRL"))  # Output should be 4
print(balancedStringSplit("RLLLLRRRLR"))  # Output should be 3


The “Split a String in Balanced Strings” problem may appear trivial, but it’s a wonderful introduction to the power of a simple counter variable and the ability to keep track of a particular state (in this case, balance) while iterating through a data structure.

Leave a Reply