Merge branch 'master' of github.com:BaffinLee/leetcode-javascript

Author: BaffinLee

Diff for: 1701-1800/1750. Minimum Length of String After Deleting Similar Ends.md

@@ -0,0 +1,95 @@
+# 1750. Minimum Length of String After Deleting Similar Ends
+
+- Difficulty: Medium.
+- Related Topics: Two Pointers, String.
+- Similar Questions: .
+
+## Problem
+
+Given a string `s` consisting only of characters `'a'`, `'b'`, and `'c'`. You are asked to apply the following algorithm on the string any number of times:
+
+
+	
+- Pick a **non-empty** prefix from the string `s` where all the characters in the prefix are equal.
+	
+- Pick a **non-empty** suffix from the string `s` where all the characters in this suffix are equal.
+	
+- The prefix and the suffix should not intersect at any index.
+	
+- The characters from the prefix and suffix must be the same.
+	
+- Delete both the prefix and the suffix.
+
+
+Return **the **minimum length** of **`s` **after performing the above operation any number of times (possibly zero times)**.
+
+ 
+Example 1:
+
+```
+Input: s = "ca"
+Output: 2
+Explanation: You can't remove any characters, so the string stays as is.
+```
+
+Example 2:
+
+```
+Input: s = "cabaabac"
+Output: 0
+Explanation: An optimal sequence of operations is:
+- Take prefix = "c" and suffix = "c" and remove them, s = "abaaba".
+- Take prefix = "a" and suffix = "a" and remove them, s = "baab".
+- Take prefix = "b" and suffix = "b" and remove them, s = "aa".
+- Take prefix = "a" and suffix = "a" and remove them, s = "".
+```
+
+Example 3:
+
+```
+Input: s = "aabccabba"
+Output: 3
+Explanation: An optimal sequence of operations is:
+- Take prefix = "aa" and suffix = "a" and remove them, s = "bccabb".
+- Take prefix = "b" and suffix = "bb" and remove them, s = "cca".
+```
+
+ 
+**Constraints:**
+
+
+	
+- `1 <= s.length <= 105`
+	
+- `s` only consists of characters `'a'`, `'b'`, and `'c'`.
+
+
+
+## Solution
+
+```javascript
+/**
+ * @param {string} s
+ * @return {number}
+ */
+var minimumLength = function(s) {
+    var left = 0;
+    var right = s.length - 1;
+    while (left < right && s[left] === s[right]) {
+        while (s[left] === s[left + 1] && left + 1 < right) left++;
+        while (s[right] === s[right - 1] && right - 1 > left) right--;
+        left++;
+        right--;
+    }
+    return right - left + 1;
+};
+```
+
+**Explain:**
+
+nope.
+
+**Complexity:**
+
+* Time complexity : O(n).
+* Space complexity : O(0).