Add solution #1541

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,355 LeetCode solutions in JavaScript
+# 1,356 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1177,6 +1177,7 @@
 1537|[Get the Maximum Score](./solutions/1537-get-the-maximum-score.js)|Hard|
 1539|[Kth Missing Positive Number](./solutions/1539-kth-missing-positive-number.js)|Easy|
 1540|[Can Convert String in K Moves](./solutions/1540-can-convert-string-in-k-moves.js)|Medium|
+1541|[Minimum Insertions to Balance a Parentheses String](./solutions/1541-minimum-insertions-to-balance-a-parentheses-string.js)|Medium|
 1550|[Three Consecutive Odds](./solutions/1550-three-consecutive-odds.js)|Easy|
 1551|[Minimum Operations to Make Array Equal](./solutions/1551-minimum-operations-to-make-array-equal.js)|Medium|
 1566|[Detect Pattern of Length M Repeated K or More Times](./solutions/1566-detect-pattern-of-length-m-repeated-k-or-more-times.js)|Easy|

solutions/1541-minimum-insertions-to-balance-a-parentheses-string.js

@@ -0,0 +1,48 @@
+/**
+ * 1541. Minimum Insertions to Balance a Parentheses String
+ * https://leetcode.com/problems/minimum-insertions-to-balance-a-parentheses-string/
+ * Difficulty: Medium
+ *
+ * Given a parentheses string s containing only the characters '(' and ')'. A parentheses string
+ * is balanced if:
+ * - Any left parenthesis '(' must have a corresponding two consecutive right parenthesis '))'.
+ * - Left parenthesis '(' must go before the corresponding two consecutive right parenthesis '))'.
+ *
+ * In other words, we treat '(' as an opening parenthesis and '))' as a closing parenthesis.
+ * - For example, "())", "())(())))" and "(())())))" are balanced, ")()", "()))" and "(()))" are
+ *   not balanced.
+ *
+ * You can insert the characters '(' and ')' at any position of the string to balance it if needed.
+ *
+ * Return the minimum number of insertions needed to make s balanced.
+ */
+
+/**
+ * @param {string} s
+ * @return {number}
+ */
+var minInsertions = function(s) {
+  let result = 0;
+  let openCount = 0;
+
+  for (let i = 0; i < s.length; i++) {
+    if (s[i] === '(') {
+      openCount++;
+    } else {
+      if (i + 1 < s.length && s[i + 1] === ')') {
+        i++;
+      } else {
+        result++;
+      }
+      if (openCount > 0) {
+        openCount--;
+      } else {
+        result++;
+      }
+    }
+  }
+
+  result += openCount * 2;
+
+  return result;
+};