Add solution #3039

Author: JoshCrozier

README.md

@@ -2144,6 +2144,7 @@
 3034|[Number of Subarrays That Match a Pattern I](./solutions/3034-number-of-subarrays-that-match-a-pattern-i.js)|Medium|
 3035|[Maximum Palindromes After Operations](./solutions/3035-maximum-palindromes-after-operations.js)|Medium|
 3038|[Maximum Number of Operations With the Same Score I](./solutions/3038-maximum-number-of-operations-with-the-same-score-i.js)|Easy|
+3039|[Apply Operations to Make String Empty](./solutions/3039-apply-operations-to-make-string-empty.js)|Medium|
 3042|[Count Prefix and Suffix Pairs I](./solutions/3042-count-prefix-and-suffix-pairs-i.js)|Easy|
 3066|[Minimum Operations to Exceed Threshold Value II](./solutions/3066-minimum-operations-to-exceed-threshold-value-ii.js)|Medium|
 3068|[Find the Maximum Sum of Node Values](./solutions/3068-find-the-maximum-sum-of-node-values.js)|Hard|

solutions/3039-apply-operations-to-make-string-empty.js

@@ -0,0 +1,45 @@
+/**
+ * 3039. Apply Operations to Make String Empty
+ * https://leetcode.com/problems/apply-operations-to-make-string-empty/
+ * Difficulty: Medium
+ *
+ * You are given a string s.
+ *
+ * Consider performing the following operation until s becomes empty:
+ * - For every alphabet character from 'a' to 'z', remove the first occurrence of that
+ *   character in s (if it exists).
+ *
+ * For example, let initially s = "aabcbbca". We do the following operations:
+ * - Remove the underlined characters s = "aabcbbca". The resulting string is s = "abbca".
+ * - Remove the underlined characters s = "abbca". The resulting string is s = "ba".
+ * - Remove the underlined characters s = "ba". The resulting string is s = "".
+ *
+ * Return the value of the string s right before applying the last operation. In the
+ * example above, answer is "ba".
+ */
+
+/**
+ * @param {string} s
+ * @return {string}
+ */
+var lastNonEmptyString = function(s) {
+  const charCount = new Array(26).fill(0);
+  let maxFrequency = 0;
+
+  for (const char of s) {
+    const index = char.charCodeAt(0) - 97;
+    charCount[index]++;
+    maxFrequency = Math.max(maxFrequency, charCount[index]);
+  }
+
+  let result = '';
+  for (let i = s.length - 1; i >= 0; i--) {
+    const index = s.charCodeAt(i) - 97;
+    if (charCount[index] === maxFrequency) {
+      result = s[i] + result;
+      charCount[index]--;
+    }
+  }
+
+  return result;
+};