Add solution #1395

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,277 LeetCode solutions in JavaScript
+# 1,278 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1064,6 +1064,7 @@
 1391|[Check if There is a Valid Path in a Grid](./solutions/1391-check-if-there-is-a-valid-path-in-a-grid.js)|Medium|
 1392|[Longest Happy Prefix](./solutions/1392-longest-happy-prefix.js)|Hard|
 1394|[Find Lucky Integer in an Array](./solutions/1394-find-lucky-integer-in-an-array.js)|Easy|
+1395|[Count Number of Teams](./solutions/1395-count-number-of-teams.js)|Medium|
 1400|[Construct K Palindrome Strings](./solutions/1400-construct-k-palindrome-strings.js)|Medium|
 1402|[Reducing Dishes](./solutions/1402-reducing-dishes.js)|Hard|
 1408|[String Matching in an Array](./solutions/1408-string-matching-in-an-array.js)|Easy|

solutions/1395-count-number-of-teams.js

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+/**
+ * 1395. Count Number of Teams
+ * https://leetcode.com/problems/count-number-of-teams/
+ * Difficulty: Medium
+ *
+ * There are n soldiers standing in a line. Each soldier is assigned a unique rating value.
+ *
+ * You have to form a team of 3 soldiers amongst them under the following rules:
+ * - Choose 3 soldiers with index (i, j, k) with rating (rating[i], rating[j], rating[k]).
+ * - A team is valid if: (rating[i] < rating[j] < rating[k]) or (rating[i] > rating[j] > rating[k])
+ *   where (0 <= i < j < k < n).
+ *
+ * Return the number of teams you can form given the conditions. (soldiers can be part of
+ * multiple teams).
+ */
+
+/**
+ * @param {number[]} rating
+ * @return {number}
+ */
+var numTeams = function(rating) {
+  let result = 0;
+  const n = rating.length;
+
+  for (let j = 1; j < n - 1; j++) {
+    let leftLess = 0;
+    let leftGreater = 0;
+    let rightLess = 0;
+    let rightGreater = 0;
+
+    for (let i = 0; i < j; i++) {
+      if (rating[i] < rating[j]) leftLess++;
+      else leftGreater++;
+    }
+
+    for (let k = j + 1; k < n; k++) {
+      if (rating[k] < rating[j]) rightLess++;
+      else rightGreater++;
+    }
+
+    result += leftLess * rightGreater + leftGreater * rightLess;
+  }
+
+  return result;
+};