Split a String in Balanced Strings - Video Solutions

Problem Overview: You receive a string consisting only of characters L and R. A substring is balanced when the number of L and R characters are equal. The goal is to split the string into the maximum number of balanced substrings.

Approach 1: Greedy Approach Using Counter (Time: O(n), Space: O(1))

This problem fits naturally with a greedy strategy. Traverse the string once while maintaining a single counter. Increment the counter when you see L and decrement it when you see R (or vice versa). Whenever the counter returns to zero, the current segment contains equal numbers of both characters, meaning a balanced substring has been found. Increment the result and continue scanning the rest of the string.

The key insight: the earliest point where the running difference between L and R becomes zero forms a valid balanced substring. Splitting immediately maximizes the number of possible splits later. Since the algorithm processes each character exactly once and stores only a few integers, the time complexity is O(n) and the space complexity is O(1). This approach relies on simple counting logic and works efficiently even for large strings.

Approach 2: Two Counter Method (Time: O(n), Space: O(1))

Another way to implement the same idea is by explicitly maintaining two counters: one for L and one for R. Iterate through the string and update the appropriate counter for each character. After each update, compare the two counters. When they become equal, a balanced substring is complete. Increase the split count and continue scanning the remaining characters.

This approach is slightly more verbose but often easier to reason about for beginners because it directly tracks both character counts. The algorithm still performs a single pass through the string, giving O(n) time complexity with O(1) auxiliary space. The trade‑off is simply maintaining two integers instead of one difference counter.

Recommended for interviews: Interviewers typically expect the greedy counter approach. It shows you recognize that balanced segments can be finalized the moment the running difference reaches zero. The two‑counter method demonstrates the same concept but with more explicit state tracking. Showing both approaches communicates a clear understanding of the underlying counting principle and why the greedy split guarantees the maximum number of balanced substrings.