add TopKFrequentElements - Day 17

src/com/concept/scala/leetcode_30days_challenge_July2020/TopKFrequentElements.scala

@@ -0,0 +1,47 @@
+package com.concept.scala.leetcode_30days_challenge_July2020
+
+import scala.collection.mutable.ListBuffer
+
+/** *
+ * Day 17
+ *
+ * @todo Given a non-empty array of integers, return the k most frequent elements.
+ * @example Example 1:
+ *
+ * Input: nums = [1,1,1,2,2,3], k = 2
+ * Output: [1,2]
+ * Example 2:
+ *
+ * Input: nums = [1], k = 1
+ * Output: [1]
+ * @note You may assume k is always valid, 1 ≤ k ≤ number of unique elements.
+ * Your algorithm's time complexity must be better than O(n log n), where n is the array's size.
+ * It's guaranteed that the answer is unique, in other words the set of the top k frequent elements is unique.
+ * You can return the answer in any order.
+ *
+ */
+object TopKFrequentElements {
+ def main(args: Array[String]): Unit = {
+ println(topKFrequent(Array(1, 1, 1, 2, 2, 3), 2).mkString(","))
+ println(topKFrequent(Array(1), 1).mkString(","))
+
+ }
+
+
+ def topKFrequent(nums: Array[Int], k: Int): Array[Int] = {
+ def priorityLogic(topKList: ListBuffer[(Int, Int)], tuple: (Int, Int)): ListBuffer[(Int, Int)] = {
+ if (topKList.length < k)
+ topKList += tuple
+ else {
+ //val topKMin: (Int, Int) = topKList.sortBy(_._2).head
+ val topKMin: (Int, Int) = topKList.minBy(_._2)
+ if (tuple._2 > topKMin._2) topKList -= topKMin += tuple
+ }
+ topKList
+ }
+
+ nums.groupBy(identity).mapValues(_.length).foldLeft(ListBuffer[(Int, Int)]())(priorityLogic).map(_._1).toArray
+ }
+
+
+}