Split a String in Balanced Strings
Easy
You are given a balanced string s consisting of only 'L' and 'R' characters. A balanced string is defined as a string with an equal number of 'L' and 'R' characters. Your task is to split the given string s into the maximum number of balanced substrings.
Example 1:
Input: s = "RLRRLLRLRL"
Output: 4
Explanation: The string can be split into "RL", "RRLL", "RL", "RL", where each substring is balanced.
Example 2:
Input: s = "RLRRRLLRLL"
Output: 2
Explanation: The string can be split into "RL", "RRRLLRLL", where each substring is balanced.
Example 3:
Input: s = "LLLLRRRR"
Output: 1
Explanation: The string can only be split into the entire string "LLLLRRRR", which is balanced.
Constraints:
- The length of the string
sis between 2 and 1000, inclusive. - Each character in the string
sis either 'L' or 'R'. - The string
sis guaranteed to be a balanced string.
Write a function that takes a balanced string s as input and returns the maximum number of balanced substrings you can obtain by splitting the string.
Solution
Naive Solution
A brute force approach would be to iterate through all possible substrings and check if each substring is balanced. We can keep track of the number of balanced substrings found and return the maximum possible number.
Code (Python)
def balanced_string_split_naive(s):
count = 0
for i in range(len(s)):
for j in range(i + 1, len(s) + 1):
sub = s[i:j]
l_count = sub.count('L')
r_count = sub.count('R')
if l_count == r_count and l_count > 0:
count += 1
return count
Time Complexity
O(n^3), where n is the length of the string. This is due to the nested loops and the count method within the inner loop, each potentially taking O(n) time.
Space Complexity
O(n), where n is the length of the string due to substring extraction.
Optimal Solution
The optimal solution involves iterating through the string once, keeping track of the counts of 'L' and 'R'. Whenever the counts are equal, we increment a counter, signaling a balanced substring.
Code (Python)
def balanced_string_split(s):
count = 0
balance = 0
for char in s:
if char == 'L':
balance += 1
else:
balance -= 1
if balance == 0:
count += 1
return count
Time Complexity
O(n), where n is the length of the string. This is because we iterate through the string once.
Space Complexity
O(1), as we only use a constant amount of extra space for the count and balance variables.
Edge Cases
- Empty String: The problem statement specifies that the string will not be empty, and it will be a balanced string. Therefore, we do not need to handle the empty string case.
- Single Balanced String: If the entire string is balanced (e.g., "LLRR"), the algorithm correctly returns 1.
- Multiple Balanced Substrings: The algorithm correctly identifies and counts multiple balanced substrings (e.g., "RLRLRL" returns 3).
- Large Input String: The algorithm can handle large strings up to the constraint limits (1000) without performance issues due to O(n) time complexity.
