# LC 1221. Split a String in Balanced Strings

### [Problem link](https://leetcode.com/problems/split-a-string-in-balanced-strings/)

###### tags: `leedcode` `python` `c++` `easy` `Greedy`

**Balanced** strings are those that have an equal quantity of <code>'L'</code> and <code>'R'</code> characters.

Given a **balanced** string <code>s</code>, 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 2<sup>nd</sup> and 5<sup>th</sup> substrings are not balanced.

**Example 3:**

Input: s = "LLLLRRRR"
Output: 1
Explanation: s can be split into "LLLLRRRR".

**Constraints:**

- <code>2 <= s.length <= 1000</code>
- <code>s[i]</code> is either <code>'L'</code> or <code>'R'</code>.
- <code>s</code> is a **balanced** string.

## Solution 1 - Greedy
#### Python

class Solution:
    def balancedStringSplit(self, s: str) -> int:
        cnt = 0
        res = 0

        for c in s:
            if c == 'R':
                cnt += 1
            else:
                cnt -= 1
            if cnt == 0:
                res += 1

        return res

#### C++

class Solution {
public:
    int balancedStringSplit(string s) {
        int res = 0;
        int cnt = 0;
        for (char& c: s) {
            if (c == 'R') {
                cnt++;
            } else {
                cnt--;
            }
            if (cnt == 0) {
                res++;
            }
        }
        return res;
    }
};

>### Complexity
>| | Time Complexity | Space Complexity |
>| ----------- | --------------- | ---------------- |
>| Solution 1 | O(n) | O(1) |

## Note
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