add FindAllDuplicatesInAnArray - Day 6

Author: rajeshsantha

src/com/concept/scala/leetcode_30days_challenge_August2020/FindAllDuplicatesInAnArray.scala

@@ -0,0 +1,50 @@
+package com.concept.scala.leetcode_30days_challenge_August2020
+
+/** *
+ * Day 6
+ *
+ * @todo Given an array of integers, 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once.
+ * Find all the elements that appear twice in this array.
+ * Could you do it without extra space and in O(n) runtime?
+ * @example Example:
+ * Input:
+ * [4,3,2,7,8,2,3,1]
+ *
+ * Output:
+ * [2,3]
+ * @see https://leetcode.com/problems/find-all-duplicates-in-an-array/
+ * OR
+ * https://leetcode.com/explore/challenge/card/august-leetcoding-challenge/549/week-1-august-1st-august-7th/3414/
+ */
+object FindAllDuplicatesInAnArray {
+ def main(args: Array[String]): Unit = {
+ println(findDuplicates(Array(4, 3, 2, 7, 8, 2, 3, 1)).mkString(","))
+ }
+
+ def findDuplicates(nums: Array[Int]): List[Int] = nums diff nums.distinct toList
+
+ def findDuplicates2(nums: Array[Int]): List[Int] = nums.map(e => (e, nums.count(_ == e))).filter(_._2 == 2).map(_._1).distinct.toList
+
+ def findDuplicates3(nums: Array[Int]): List[Int] = {
+ nums.groupBy(identity).collect {
+ case (num, count) if count.lengthCompare(1) > 0 => num
+ }.toList
+ }
+
+ def findDuplicates4(nums: Array[Int]): List[Int] = {
+ nums.foldLeft(Map[Int, Int]()) { (acc, n) =>
+ val count = acc.get(n).fold(1)(_ + 1)
+ acc + (n -> count)
+ }.filter(_._2 == 2).keys.toList
+ }
+
+ def findDuplicates5(nums: Array[Int]): List[Int] = {
+ nums.indices.foreach(index => {
+ val indexToUpdate = nums(index) % (nums.length + 1) - 1
+ nums(indexToUpdate) = nums(indexToUpdate) + (nums.length + 1)
+ })
+ nums.indices.filter(index => nums(index) > 2 * (nums.length + 1)).map(_ + 1).toList
+ }
+
+}
+