Add solution #2131

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,764 LeetCode solutions in JavaScript
+# 1,765 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1632,6 +1632,7 @@
 2127|[Maximum Employees to Be Invited to a Meeting](./solutions/2127-maximum-employees-to-be-invited-to-a-meeting.js)|Hard|
 2129|[Capitalize the Title](./solutions/2129-capitalize-the-title.js)|Easy|
 2130|[Maximum Twin Sum of a Linked List](./solutions/2130-maximum-twin-sum-of-a-linked-list.js)|Medium|
+2131|[Longest Palindrome by Concatenating Two Letter Words](./solutions/2131-longest-palindrome-by-concatenating-two-letter-words.js)|Medium|
 2140|[Solving Questions With Brainpower](./solutions/2140-solving-questions-with-brainpower.js)|Medium|
 2145|[Count the Hidden Sequences](./solutions/2145-count-the-hidden-sequences.js)|Medium|
 2154|[Keep Multiplying Found Values by Two](./solutions/2154-keep-multiplying-found-values-by-two.js)|Easy|

solutions/2131-longest-palindrome-by-concatenating-two-letter-words.js

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+/**
+ * 2131. Longest Palindrome by Concatenating Two Letter Words
+ * https://leetcode.com/problems/longest-palindrome-by-concatenating-two-letter-words/
+ * Difficulty: Medium
+ *
+ * You are given an array of strings words. Each element of words consists of two lowercase
+ * English letters.
+ *
+ * Create the longest possible palindrome by selecting some elements from words and concatenating
+ * them in any order. Each element can be selected at most once.
+ *
+ * Return the length of the longest palindrome that you can create. If it is impossible to create
+ * any palindrome, return 0.
+ *
+ * A palindrome is a string that reads the same forward and backward.
+ */
+
+/**
+ * @param {string[]} words
+ * @return {number}
+ */
+var longestPalindrome = function(words) {
+  const map = new Map();
+  let length = 0;
+  let hasCenter = false;
+
+  for (const word of words) {
+    map.set(word, (map.get(word) || 0) + 1);
+  }
+
+  for (const word of map.keys()) {
+    const reverse = word[1] + word[0];
+
+    if (word === reverse) {
+      const count = map.get(word);
+      length += Math.floor(count / 2) * 4;
+      if (count % 2 === 1) hasCenter = true;
+    } else if (map.has(reverse)) {
+      const pairs = Math.min(map.get(word), map.get(reverse));
+      length += pairs * 4;
+      map.set(word, 0);
+      map.set(reverse, 0);
+    }
+  }
+
+  return hasCenter ? length + 2 : length;
+};