How do you solve Split a String in Balanced Strings on LeetCode?

Split a String in Balanced Strings (LeetCode #1221) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The balance counter effectively tracks the number of 'L' and 'R' characters.

What is the brute force solution for Split a String in Balanced Strings?

The brute force approach for Split a String in Balanced Strings is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible substring to see if it is balanced. This is straightforward but inefficient because it requires examining many combinations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Split a String in Balanced Strings?

Split a String in Balanced Strings uses the following concepts: String, Greedy, Counting. The recommended patterns to study are: Two Pointers, Greedy.

What are the interview tips for Split a String in Balanced Strings?

Always consider edge cases, such as strings that are already balanced or consist of only one character type. Explain your thought process clearly while coding; it helps interviewers understand your approach. Practice writing clean and efficient code, as readability is often as important as functionality.

What are common mistakes when solving Split a String in Balanced Strings?

Failing to account for the balance correctly, leading to incorrect counts. Not recognizing that the problem can be solved in a single pass.

#1221

Split a String in Balanced Strings

Easy

StringGreedyCountingTwo PointersGreedy

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal approach uses a single pass through the string while maintaining a balance counter. This allows us to efficiently count balanced substrings without checking every possible substring.

⚙️

Algorithm

3 steps

1Step 1: Initialize a balance counter and a result counter to zero.
2Step 2: Loop through each character in the string, adjusting the balance counter: +1 for 'L' and -1 for 'R'.
3Step 3: Whenever the balance counter reaches zero, increment the result counter.

solution.py8 lines

1def balanced_string_split(s):
2    balance = 0
3    count = 0
4    for char in s:
5        balance += 1 if char == 'L' else -1
6        if balance == 0:
7            count += 1
8    return count

ℹ

Complexity note: The time complexity is O(n) since we only loop through the string once. The space complexity is O(1) because we use a fixed amount of extra space for the counters.

1The balance counter effectively tracks the number of 'L' and 'R' characters.
2Each time the balance returns to zero, it indicates a complete balanced substring.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.