Split a String in Balanced Strings

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".

Balanced String Split

Problem: Given a balanced string s, split it into the maximum number of balanced substrings, where a balanced string has an equal number of 'L' and 'R' characters.

Example:

• Input: s = "RLRRLLRLRL"
• Output: 4
• Explanation: The string can be split into "RL", "RRLL", "RL", "RL".

Naive Approach

A brute-force approach would be to iterate through all possible substrings and check if each substring is balanced. If a substring is balanced, increment the count and start searching for the next balanced substring from the end of the current substring. This approach is inefficient due to the nested loops required to generate all substrings.

Optimal Approach

A more efficient approach involves iterating through the string once, keeping track of the balance between 'L' and 'R' characters. Whenever the balance reaches zero, it indicates that a balanced substring has been found. Increment the count and continue iterating.

def balanced_string_split(s):
    balance = 0
    count = 0
    for char in s:
        if char == 'L':
            balance += 1
        else:
            balance -= 1
        if balance == 0:
            count += 1
    return count

# Example usage
s = "RLRRLLRLRL"
result = balanced_string_split(s)
print(f"The maximum number of balanced strings is: {result}") # Output: 4

s = "RLRRRLLRLL"
result = balanced_string_split(s)
print(f"The maximum number of balanced strings is: {result}") # Output: 2

s = "LLLLRRRR"
result = balanced_string_split(s)
print(f"The maximum number of balanced strings is: {result}") # Output: 1

Big O Runtime Analysis

The algorithm iterates through the string s once. Therefore, the time complexity is O(n), where n is the length of the string.

Big O Space Usage Analysis

The algorithm uses a constant amount of extra space to store the balance and count variables. Therefore, the space complexity is O(1).

Edge Cases

• Empty String: If the input string is empty, the algorithm should return 0.
• Invalid Input: If the input string contains characters other than 'L' and 'R', the algorithm might produce incorrect results. However, the problem statement specifies that the string will only contain 'L' and 'R'.
• Unbalanced String: The problem statement specifies that the input string is balanced, so we don't need to handle the case where the number of 'L's and 'R's are not equal.

def balanced_string_split(s):
    if not s:
        return 0

    balance = 0
    count = 0
    for char in s:
        if char == 'L':
            balance += 1
        elif char == 'R':
            balance -= 1
        else:
          #Consider this an edge case, but problem statement specifies only L and R
          continue #or raise an exception if input validation is required

        if balance == 0:
            count += 1
    return count