Add solution #2116

Author: JoshCrozier

README.md

@@ -383,6 +383,7 @@
 2095|[Delete the Middle Node of a Linked List](./2095-delete-the-middle-node-of-a-linked-list.js)|Medium|
 2099|[Find Subsequence of Length K With the Largest Sum](./2099-find-subsequence-of-length-k-with-the-largest-sum.js)|Medium|
 2114|[Maximum Number of Words Found in Sentences](./2114-maximum-number-of-words-found-in-sentences.js)|Easy|
+2116|[Check if a Parentheses String Can Be Valid](./2116-check-if-a-parentheses-string-can-be-valid.js)|Medium|
 2129|[Capitalize the Title](./2129-capitalize-the-title.js)|Easy|
 2154|[Keep Multiplying Found Values by Two](./2154-keep-multiplying-found-values-by-two.js)|Easy|
 2185|[Counting Words With a Given Prefix](./2185-counting-words-with-a-given-prefix.js)|Easy|

solutions/2116-check-if-a-parentheses-string-can-be-valid.js

@@ -0,0 +1,54 @@
+/**
+ * 2116. Check if a Parentheses String Can Be Valid
+ * https://leetcode.com/problems/check-if-a-parentheses-string-can-be-valid/
+ * Difficulty: Medium
+ *
+ * A parentheses string is a non-empty string consisting only of '(' and ')'. It is valid
+ * if any of the following conditions is true:
+ * - It is ().
+ * - It can be written as AB (A concatenated with B), where A and B are valid parentheses
+ *   strings.
+ * - It can be written as (A), where A is a valid parentheses string.
+ *
+ * You are given a parentheses string s and a string locked, both of length n. locked is a
+ * binary string consisting only of '0's and '1's. For each index i of locked,
+ * - If locked[i] is '1', you cannot change s[i].
+ * - But if locked[i] is '0', you can change s[i] to either '(' or ')'.
+ *
+ * Return true if you can make s a valid parentheses string. Otherwise, return false.
+ */
+
+/**
+ * @param {string} s
+ * @param {string} locked
+ * @return {boolean}
+ */
+var canBeValid = function(s, locked) {
+  if (s.length % 2) return false;
+
+  let symmetrical = 0;
+  for (let i = 0; i < s.length; i++) {
+    if (locked[i] === '0' || s[i] === '(') {
+      symmetrical++;
+    } else {
+      symmetrical--;
+    }
+    if (symmetrical < 0) {
+      return false;
+    }
+  }
+
+  symmetrical = 0;
+  for (let i = s.length - 1; i >= 0; i--) {
+    if (locked[i] === '0' || s[i] === ')') {
+      symmetrical++;
+    } else {
+      symmetrical--;
+    }
+    if (symmetrical < 0) {
+      return false;
+    }
+  }
+
+  return true;
+};