# Solving Leetcode Interviews in Seconds with AI: Split a String in Balanced Strings


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1221" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> Balanced strings are those that have an equal quantity of 'L' and 'R' characters. Given a balanced string s, split it into some number of substrings such that:  Each substring is balanced.  Return the maximum number of balanced strings you can obtain.   Example 1:  Input: s = "RLRRLLRLRL" Output: 4 Explanation: 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 Explanation: 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 Explanation: s can be split into "LLLLRRRR".    Constraints:  2 <= s.length <= 1000 s[i] is either 'L' or 'R'. s is a balanced string.  

	# Explanation
	Here's a solution to the balanced string splitting problem:

*   **High-Level Approach:** The core idea is to iterate through the string and keep track of the counts of 'L' and 'R' characters. Whenever the counts become equal, it signifies a balanced substring, and we increment the count of balanced substrings.

*   **Efficiency:** The solution iterates through the string only once, making it efficient.

*   **Complexity:**
    *   Runtime Complexity: O(n), where n is the length of the input string `s`.
    *   Storage Complexity: O(1), as we only use a few constant extra variables.

	
	# Code
	```python
	def balancedStringSplit(s: str) -> int:
    """
    Splits a balanced string into the maximum number of balanced substrings.

    Args:
        s: The balanced string consisting of 'L' and 'R' characters.

    Returns:
        The maximum number of balanced substrings.
    """
    balance = 0
    count = 0
    l_count = 0
    r_count = 0

    for char in s:
        if char == 'L':
            l_count += 1
        else:
            r_count += 1

        if l_count == r_count:
            count += 1
            l_count = 0
            r_count = 0

    return count
	```