Add solution #2799

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,435 LeetCode solutions in JavaScript
+# 1,436 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1408,6 +1408,7 @@
 2726|[Calculator with Method Chaining](./solutions/2726-calculator-with-method-chaining.js)|Easy|
 2727|[Is Object Empty](./solutions/2727-is-object-empty.js)|Easy|
 2780|[Minimum Index of a Valid Split](./solutions/2780-minimum-index-of-a-valid-split.js)|Medium|
+2799|[Count Complete Subarrays in an Array](./solutions/2799-count-complete-subarrays-in-an-array.js)|Medium|
 2818|[Apply Operations to Maximize Score](./solutions/2818-apply-operations-to-maximize-score.js)|Hard|
 2843|[Count Symmetric Integers](./solutions/2843-count-symmetric-integers.js)|Easy|
 2873|[Maximum Value of an Ordered Triplet I](./solutions/2873-maximum-value-of-an-ordered-triplet-i.js)|Easy|

solutions/2799-count-complete-subarrays-in-an-array.js

@@ -0,0 +1,42 @@
+/**
+ * 2799. Count Complete Subarrays in an Array
+ * https://leetcode.com/problems/count-complete-subarrays-in-an-array/
+ * Difficulty: Medium
+ *
+ * You are given an array nums consisting of positive integers.
+ *
+ * We call a subarray of an array complete if the following condition is satisfied:
+ * - The number of distinct elements in the subarray is equal to the number of distinct
+ *   elements in the whole array.
+ *
+ * Return the number of complete subarrays.
+ *
+ * A subarray is a contiguous non-empty part of an array.
+ */
+
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+var countCompleteSubarrays = function(nums) {
+  const totalDistinct = new Set(nums).size;
+  const frequency = new Map();
+  let completeCount = 0;
+  let left = 0;
+
+  for (let right = 0; right < nums.length; right++) {
+    frequency.set(nums[right], (frequency.get(nums[right]) || 0) + 1);
+
+    while (frequency.size === totalDistinct) {
+      completeCount += nums.length - right;
+      const leftNum = nums[left];
+      frequency.set(leftNum, frequency.get(leftNum) - 1);
+      if (frequency.get(leftNum) === 0) {
+        frequency.delete(leftNum);
+      }
+      left++;
+    }
+  }
+
+  return completeCount;
+};