Naive Solution

A brute-force solution would involve iterating through all possible substrings and checking if each substring is balanced. We would keep track of the maximum number of balanced substrings found. However, this approach is inefficient.

Code (Python)

def max_balanced_substrings_naive(s):
    n = len(s)
    max_count = 0
    for i in range(1 << (n - 1)):
        count = 0
        current_substring = ""
        balanced = True
        l_count = 0
        r_count = 0
        for j in range(n):
            current_substring += s[j]
            if s[j] == 'L':
                l_count += 1
            else:
                r_count += 1

            if (i >> j) & 1 or j == n - 1:
                if l_count == r_count:
                    count += 1
                    l_count = 0
                    r_count = 0
                else:
                    balanced = False
                    break

        if balanced:
            max_count = max(max_count, count)
    return max_count

Time Complexity: O(2^n * n)

This is because we iterate through all possible combinations of substrings (2^(n-1)) and, for each combination, we iterate through the string (n).

Space Complexity: O(n)

In the worst case, the recursion depth can go up to n.

Optimal Solution

A more efficient approach involves iterating through the string once, keeping track of the counts of 'L' and 'R' characters. Whenever the counts are equal, we increment the number of balanced substrings.

Code (Python)

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

Time Complexity: O(n)

We iterate through the string once.

Space Complexity: O(1)

We only use a constant amount of extra space.

Edge Cases

• Empty string (not possible according to constraints)
• String with only 'L' or only 'R' (not possible according to constraints)
• Invalid input (not possible according to constraints, input is always balanced)