---
###### tags: `Leetcode`
---
# Leetcode 1221. Split a String in Balanced Strings

[link](https://leetcode.com/problems/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".

#### Constraints:
- 2 <= s.length <= 1000
- s[i] is either 'L' or 'R'.
- s is a balanced string.

---

題意: Split the string `s` into multiple substrings which are balanced strings as many as possible.

---

The balanced string means the number of `'L'` and `'R'` have equal quantity so I think we could have `'l'` and `'r'` to store the quantity of both characters. If `'l'` and `'r'` are the same, we know we have 1 substring. Iterating the string `'s'` and then returning the numbers of substrings we count.

#### Solution 1

class Solution:
    def balancedStringSplit(self, s: str) -> int:
        count = 0
        l = r = 0
        for c in s:
            if c == 'R': r += 1
            else: l += 1

            if l == r:
                count += 1
                l = r = 0
                
        return count
        

O(T): O(n)
O(S): O(1)