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, 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

    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