Add solution #1544

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,357 LeetCode solutions in JavaScript
+# 1,358 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1179,6 +1179,7 @@
 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|
 1542|[Find Longest Awesome Substring](./solutions/1542-find-longest-awesome-substring.js)|Hard|
+1544|[Make The String Great](./solutions/1544-make-the-string-great.js)|Easy|
 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/1544-make-the-string-great.js

@@ -0,0 +1,38 @@
+/**
+ * 1544. Make The String Great
+ * https://leetcode.com/problems/make-the-string-great/
+ * Difficulty: Easy
+ *
+ * Given a string s of lower and upper case English letters.
+ *
+ * A good string is a string which doesn't have two adjacent characters s[i] and s[i + 1] where:
+ * - 0 <= i <= s.length - 2
+ * - s[i] is a lower-case letter and s[i + 1] is the same letter but in upper-case or vice-versa.
+ *
+ * To make the string good, you can choose two adjacent characters that make the string bad and
+ * remove them. You can keep doing this until the string becomes good.
+ *
+ * Return the string after making it good. The answer is guaranteed to be unique under the given
+ * constraints.
+ *
+ * Notice that an empty string is also good.
+ */
+
+/**
+ * @param {string} s
+ * @return {string}
+ */
+var makeGood = function(s) {
+  const stack = [];
+
+  for (const char of s) {
+    const lastChar = stack[stack.length - 1];
+    if (lastChar && ((char.toLowerCase() === lastChar.toLowerCase()) && (char !== lastChar))) {
+      stack.pop();
+    } else {
+      stack.push(char);
+    }
+  }
+
+  return stack.join('');
+};