Added July 23 Challenge

Author: suriyakrishna

src/main/scala/com/leetcode/scala/monthly2020/july/SingleNumberIII.scala

@@ -0,0 +1,68 @@
+package com.leetcode.scala.monthly2020.july
+
+import scala.collection.mutable
+
+
+/*
+* Given an array of numbers nums, in which exactly two elements appear only once and all the other elements appear exactly
+* twice. Find the two elements that appear only once.
+*
+* */
+
+/*
+* Input:  [1,2,1,3,2,5]
+  Output: [3,5]
+*
+* Note:
+  1. The order of the result is not important. So in the above example, [5, 3] is also correct.
+  2. Your algorithm should run in linear runtime complexity. Could you implement it using only constant space complexity.
+*
+* */
+
+/*
+* Approach: Bit Manipulation
+*
+* */
+
+object SingleNumberIII {
+  def main(args: Array[String]): Unit = {
+    val nums = Array(1, 2, 1, 3, 2, 5)
+    val result = singleNumber(nums)
+    println(s"Result: ${result.mkString(" ")}, ${result.mkString(" ") == "5 3"}")
+  }
+
+  //Without Extra Space
+  def singleNumber(nums: Array[Int]): Array[Int] = {
+    var num1: Int = 0
+    for (num <- nums) {
+      num1 ^= num
+    }
+    num1 &= -num1
+    var result = Array.fill[Int](2)(0)
+    for (num <- nums) {
+      if ((num1 & num) == 0) {
+        result(0) ^= num
+      } else {
+        result(1) ^= num
+      }
+    }
+    result
+  }
+
+
+  // Using Extra Space
+  // Time Complexity - O(n)
+  // Space Complexity - O(n)
+
+  def singleNumberES(nums: Array[Int]): Array[Int] = {
+    val set: mutable.HashSet[Int] = mutable.HashSet()
+    for (num <- nums) {
+      if (set.contains(num)) {
+        set.remove(num)
+      } else {
+        set.add(num)
+      }
+    }
+    set.toArray
+  }
+}