From ac016faadae83d58e1316a580a1e754b01029ba8 Mon Sep 17 00:00:00 2001
From: Prad Nukala <prad@sonr.io>
Date: Sat, 11 Jul 2026 23:08:07 -0400
Subject: [PATCH] feat(Medium/Array): add solution for top-k-frequent-elements
 problem

---
 Medium/Array/347.top-k-frequent-elements.js | 62 +++++++++++++++++++++
 1 file changed, 62 insertions(+)
 create mode 100644 Medium/Array/347.top-k-frequent-elements.js

diff --git a/Medium/Array/347.top-k-frequent-elements.js b/Medium/Array/347.top-k-frequent-elements.js
new file mode 100644
index 0000000..3936999
--- /dev/null
+++ b/Medium/Array/347.top-k-frequent-elements.js
@@ -0,0 +1,62 @@
+/*
+ * 347. Top K Frequent Elements
+ * Difficulty: Medium
+ * https://leetcode.com/problems/top-k-frequent-elements/
+ *
+ * ──────────────────────────────────────────────────
+ *
+ * Given an integer array nums and an integer k, return the k most
+ * frequent elements. You may return the answer in any order.
+ *
+ *
+ *
+ * Example 1:
+ *
+ * Input: nums = [1,1,1,2,2,3], k = 2
+ *
+ * Output: [1,2]
+ *
+ * Example 2:
+ *
+ * Input: nums = [1], k = 1
+ *
+ * Output: [1]
+ *
+ * Example 3:
+ *
+ * Input: nums = [1,2,1,2,1,2,3,1,3,2], k = 2
+ *
+ * Output: [1,2]
+ *
+ *
+ *
+ * Constraints:
+ *
+ * 	• 1 <= nums.length <= 10^5
+ *
+ * 	• -10^4 <= nums[i] <= 10^4
+ *
+ * 	• k is in the range [1, the number of unique elements in the array].
+ *
+ * 	• It is guaranteed that the answer is unique.
+ *
+ *
+ *
+ * Follow up: Your algorithm's time complexity must be better than O(n
+ * log n), where n is the array's size.
+ */
+
+/**
+ * @param {number[]} nums
+ * @param {number} k
+ * @return {number[]}
+ */
+var topKFrequent = function (nums, k) {
+  const freq = {};
+  for (let n of nums) freq[n] = (freq[n] || 0) + 1;
+
+  return Object.keys(freq)
+    .map(Number)
+    .sort((a, b) => freq[b] - freq[a])
+    .slice(0, k);
+};