Add LC 347. Top K Frequent Elements prompt and 2 solutions

Author: timothyshores

347. Top K Frequent Elements.js

@@ -0,0 +1,78 @@
+/*
+
+347. 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]
+
+
+Constraints:
+1 <= nums.length <= 105
+-104 <= nums[i] <= 104
+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[]}
+ */
+const topKFrequent = (nums, k) => {
+  const numsMap = {};
+
+  for (const num of nums) {
+    numsMap[num] = (numsMap[num] || 0) + 1;
+  }
+
+  return Object.keys(numsMap)
+    .sort((a, b) => numsMap[b] - numsMap[a])
+    .slice(0, k);
+};
+
+// Revised solution in O(n) time complexity
+const topKFrequentFollowUp = (nums, k) => {
+  const frequencyCounter = createfrequencyCounter(nums);
+  const frequencyBuckets = createFrequencyBuckets(frequencyCounter);
+  const maxFreq = Math.max(...Object.values(frequencyCounter));
+  return gatherTopKFrequent(frequencyBuckets, maxFreq, k);
+};
+
+const createfrequencyCounter = (nums) => {
+  const frequencyCounter = {};
+  for (const num of nums) frequencyCounter[num] = (frequencyCounter[num] || 0) + 1;
+  return frequencyCounter;
+};
+
+const createFrequencyBuckets = (frequencyCounter) => {
+  const freqBuckets = [];
+
+  for (const [uniqueNum, frequency] of Object.entries(frequencyCounter)) {
+    freqBuckets[frequency] = freqBuckets[frequency] || [];
+    freqBuckets[frequency].push(uniqueNum);
+  }
+
+  return freqBuckets;
+};
+
+const gatherTopKFrequent = (freqBuckets, maxFreq, k) => {
+  const mostFrequent = [];
+
+  for (let i = maxFreq; i >= 0; i--) {
+    if (freqBuckets[i]) mostFrequent.push(...freqBuckets[i]);
+    if (mostFrequent.length >= k) break;
+  }
+
+  return mostFrequent.slice(0, k);
+};

README.md

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