Add solution #1442

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,305 LeetCode solutions in JavaScript
+# 1,306 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1100,6 +1100,7 @@
 1438|[Longest Continuous Subarray With Absolute Diff Less Than or Equal to Limit](./solutions/1438-longest-continuous-subarray-with-absolute-diff-less-than-or-equal-to-limit.js)|Medium|
 1439|[Find the Kth Smallest Sum of a Matrix With Sorted Rows](./solutions/1439-find-the-kth-smallest-sum-of-a-matrix-with-sorted-rows.js)|Hard|
 1441|[Build an Array With Stack Operations](./solutions/1441-build-an-array-with-stack-operations.js)|Medium|
+1442|[Count Triplets That Can Form Two Arrays of Equal XOR](./solutions/1442-count-triplets-that-can-form-two-arrays-of-equal-xor.js)|Medium|
 1443|[Minimum Time to Collect All Apples in a Tree](./solutions/1443-minimum-time-to-collect-all-apples-in-a-tree.js)|Medium|
 1446|[Consecutive Characters](./solutions/1446-consecutive-characters.js)|Easy|
 1447|[Simplified Fractions](./solutions/1447-simplified-fractions.js)|Medium|

solutions/1442-count-triplets-that-can-form-two-arrays-of-equal-xor.js

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+/**
+ * 1442. Count Triplets That Can Form Two Arrays of Equal XOR
+ * https://leetcode.com/problems/count-triplets-that-can-form-two-arrays-of-equal-xor/
+ * Difficulty: Medium
+ *
+ * Given an array of integers arr.
+ *
+ * We want to select three indices i, j and k where (0 <= i < j <= k < arr.length).
+ *
+ * Let's define a and b as follows:
+ * - a = arr[i] ^ arr[i + 1] ^ ... ^ arr[j - 1]
+ * - b = arr[j] ^ arr[j + 1] ^ ... ^ arr[k]
+ *
+ * Note that ^ denotes the bitwise-xor operation.
+ *
+ * Return the number of triplets (i, j and k) Where a == b.
+ */
+
+/**
+ * @param {number[]} arr
+ * @return {number}
+ */
+var countTriplets = function(arr) {
+  const prefixXor = [0];
+  let result = 0;
+
+  for (const num of arr) {
+    prefixXor.push(prefixXor[prefixXor.length - 1] ^ num);
+  }
+
+  for (let i = 0; i < arr.length; i++) {
+    for (let k = i + 1; k < arr.length; k++) {
+      if ((prefixXor[k + 1] ^ prefixXor[i]) === 0) {
+        result += k - i;
+      }
+    }
+  }
+
+  return result;
+};