Add solution #1371

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,259 LeetCode solutions in JavaScript
+# 1,260 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1043,6 +1043,7 @@
 1367|[Linked List in Binary Tree](./solutions/1367-linked-list-in-binary-tree.js)|Medium|
 1368|[Minimum Cost to Make at Least One Valid Path in a Grid](./solutions/1368-minimum-cost-to-make-at-least-one-valid-path-in-a-grid.js)|Hard|
 1370|[Increasing Decreasing String](./solutions/1370-increasing-decreasing-string.js)|Easy|
+1371|[Find the Longest Substring Containing Vowels in Even Counts](./solutions/1371-find-the-longest-substring-containing-vowels-in-even-counts.js)|Medium|
 1372|[Longest ZigZag Path in a Binary Tree](./solutions/1372-longest-zigzag-path-in-a-binary-tree.js)|Medium|
 1374|[Generate a String With Characters That Have Odd Counts](./solutions/1374-generate-a-string-with-characters-that-have-odd-counts.js)|Easy|
 1380|[Lucky Numbers in a Matrix](./solutions/1380-lucky-numbers-in-a-matrix.js)|Easy|

solutions/1371-find-the-longest-substring-containing-vowels-in-even-counts.js

@@ -0,0 +1,32 @@
+/**
+ * 1371. Find the Longest Substring Containing Vowels in Even Counts
+ * https://leetcode.com/problems/find-the-longest-substring-containing-vowels-in-even-counts/
+ * Difficulty: Medium
+ *
+ * Given the string s, return the size of the longest substring containing each vowel an even number
+ * of times. That is, 'a', 'e', 'i', 'o', and 'u' must appear an even number of times.
+ */
+
+/**
+ * @param {string} s
+ * @return {number}
+ */
+var findTheLongestSubstring = function(s) {
+  const vowelMap = new Map([['a', 0], ['e', 1], ['i', 2], ['o', 3], ['u', 4]]);
+  const stateToIndex = new Map([[0, -1]]);
+  let state = 0;
+  let result = 0;
+
+  for (let i = 0; i < s.length; i++) {
+    if (vowelMap.has(s[i])) {
+      state ^= 1 << vowelMap.get(s[i]);
+    }
+    if (stateToIndex.has(state)) {
+      result = Math.max(result, i - stateToIndex.get(state));
+    } else {
+      stateToIndex.set(state, i);
+    }
+  }
+
+  return result;
+};