Add solution #2179

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,310 LeetCode solutions in JavaScript
+# 1,311 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1199,6 +1199,7 @@
 2140|[Solving Questions With Brainpower](./solutions/2140-solving-questions-with-brainpower.js)|Medium|
 2154|[Keep Multiplying Found Values by Two](./solutions/2154-keep-multiplying-found-values-by-two.js)|Easy|
 2161|[Partition Array According to Given Pivot](./solutions/2161-partition-array-according-to-given-pivot.js)|Medium|
+2179|[Count Good Triplets in an Array](./solutions/2179-count-good-triplets-in-an-array.js)|Hard|
 2185|[Counting Words With a Given Prefix](./solutions/2185-counting-words-with-a-given-prefix.js)|Easy|
 2206|[Divide Array Into Equal Pairs](./solutions/2206-divide-array-into-equal-pairs.js)|Easy|
 2215|[Find the Difference of Two Arrays](./solutions/2215-find-the-difference-of-two-arrays.js)|Easy|

solutions/2179-count-good-triplets-in-an-array.js

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+/**
+ * 2179. Count Good Triplets in an Array
+ * https://leetcode.com/problems/count-good-triplets-in-an-array/
+ * Difficulty: Hard
+ *
+ * You are given two 0-indexed arrays nums1 and nums2 of length n, both of which are permutations
+ * of [0, 1, ..., n - 1].
+ *
+ * A good triplet is a set of 3 distinct values which are present in increasing order by position
+ * both in nums1 and nums2. In other words, if we consider pos1v as the index of the value v in
+ * nums1 and pos2v as the index of the value v in nums2, then a good triplet will be a set (x, y, z)
+ * where 0 <= x, y, z <= n - 1, such that pos1x < pos1y < pos1z and pos2x < pos2y < pos2z.
+ *
+ * Return the total number of good triplets.
+ */
+
+/**
+ * @param {number[]} nums1
+ * @param {number[]} nums2
+ * @return {number}
+ */
+var goodTriplets = function(nums1, nums2) {
+  const length = nums1.length;
+  const pos1 = new Array(length);
+  const pos2 = new Array(length);
+
+  for (let i = 0; i < length; i++) {
+    pos1[nums1[i]] = i;
+    pos2[nums2[i]] = i;
+  }
+
+  const indices = new Array(length);
+  for (let i = 0; i < length; i++) {
+    indices[pos1[i]] = pos2[i];
+  }
+
+  const leftTree = new Array(length + 1).fill(0);
+  const rightTree = new Array(length + 1).fill(0);
+
+  let result = 0;
+  for (let i = length - 1; i >= 0; i--) {
+    const position = indices[i];
+    update(rightTree, position, 1);
+  }
+
+  for (let i = 0; i < length; i++) {
+    const position = indices[i];
+    update(rightTree, position, -1);
+    const smaller = query(leftTree, position);
+    const larger = query(rightTree, length - 1) - query(rightTree, position);
+    result += smaller * larger;
+    update(leftTree, position, 1);
+  }
+
+  return result;
+
+  function update(tree, index, delta) {
+    for (let i = index + 1; i <= length; i += i & -i) {
+      tree[i] += delta;
+    }
+  }
+
+  function query(tree, index) {
+    let sum = 0;
+    for (let i = index + 1; i > 0; i -= i & -i) {
+      sum += tree[i];
+    }
+    return sum;
+  }
+};