feat: finish 1341

Author: BaffinLee

Diff for: 1301-1400/1341. Split a String in Balanced Strings.md

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+# 1341. Split a String in Balanced Strings
+
+- Difficulty: Easy.
+- Related Topics: String, Greedy, Counting.
+- Similar Questions: .
+
+## Problem
+
+**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.
+
+
+
+## Solution
+
+```javascript
+/**
+ * @param {string} s
+ * @return {number}
+ */
+var balancedStringSplit = function(s) {
+    var num = 0;
+    var S = 0;
+    var L = 0;
+    for (var i = 0; i < s.length; i++) {
+        if (s[i] === 'L') {
+            L++;
+        } else {
+            S++;
+        }
+        if (S === L) {
+            num++;
+            S = 0;
+            L = 0;
+        }
+    }
+    return num;
+};
+```
+
+**Explain:**
+
+Every time you meet a balanced string, split it.
+
+**Complexity:**
+
+* Time complexity : O(n).
+* Space complexity : O(1).