add _3Sum - Day 8

Author: rajeshsantha

.idea/aws.xml

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+<?xml version="1.0" encoding="UTF-8"?>
+<module type="JAVA_MODULE" version="4">
+  <component name="NewModuleRootManager" inherit-compiler-output="true">
+    <exclude-output />
+    <content url="file://$MODULE_DIR$">
+      <sourceFolder url="file://$MODULE_DIR$/src" isTestSource="false" />
+    </content>
+    <orderEntry type="inheritedJdk" />
+    <orderEntry type="sourceFolder" forTests="false" />
+    <orderEntry type="library" name="scala-sdk-2.11.8" level="application" />
+  </component>
+</module>

.idea/codeStyles/Project.xml

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+package com.concept.scala.leetcode_30days_challenge_July2020
+
+import scala.collection.mutable
+import scala.collection.mutable.ListBuffer
+
+/** *
+  * Day 8
+  *
+  * @todo Given an array nums of n integers, are there elements a, b, c in nums such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.
+  * @note The solution set must not contain duplicate triplets.
+  * @example Given array nums = [-1, 0, 1, 2, -1, -4],
+  *
+  *          A solution set is:
+  *          [
+  *          [-1, 0, 1],
+  *          [-1, -1, 2]
+  *          ]
+  *
+  *
+  *
+  */
+object _3Sum {
+  def main(args: Array[String]): Unit = {
+    val nums = Array(-1, 0, 1, 2, -1, -4)
+    println(threeSumByHashMap(nums))
+  }
+
+  //MemoryLimitException
+  def threeSum(nums: Array[Int]): List[List[Int]] =
+    nums.sorted.combinations(3).map(x => if (x.sum == 0) x else 0).filter(x => x != 0).map(x => x.asInstanceOf[Array[Int]]).toArray.map(_.toList).toList
+
+  //RuntimeLimitException
+  def threeSumByHashMap(nums: Array[Int]): List[List[Int]] = {
+    import scala.collection.mutable
+    val pairs = mutable.HashMap[Int, List[Int]]()
+    nums.zipWithIndex.foreach({ case (x, idx) => pairs(x) = idx :: pairs.getOrElse(x, Nil) })
+
+    def twoSum(num1: Int, idx1: Int): List[List[Int]] = {
+      val res = for {
+        (num2, idx2) <- nums.iterator.zipWithIndex.collect({ case (x, idx) if idx != idx1 => x -> idx })
+        num3 <- pairs.getOrElse(-num1 - num2, Nil).filterNot(List(idx1, idx2).contains(_)).map(nums).take(1)
+      } yield {
+        (num1 :: num2 :: num3 :: Nil).sorted
+      }
+      res.toList
+    }
+
+    pairs.flatMap({ case (key, value :: tail) => twoSum(key, value) case _ => Nil })
+      .foldLeft(List[List[Int]]())({ case (acc, x) if !acc.contains(x) => x :: acc case (acc, _) => acc })
+
+
+  }
+
+
+}
+
+/*
+
+Java alternative
+class Solution {
+    public List<List<Integer>> threeSum(int[] nums) {
+            List<List<Integer>> result = new ArrayList<>();
+    Arrays.sort(nums);
+    for (int i = 0; i + 2 < nums.length; i++) {
+        if (nums[i] > 0) break;
+        if (i > 0 && nums[i] == nums[i - 1]) {
+            continue;
+        }
+        int j = i + 1, k = nums.length - 1;
+        int target = -nums[i];
+        while (j < k) {
+            if (nums[j] + nums[k] == target) {
+                result.add(Arrays.asList(nums[i], nums[j], nums[k]));
+                j++;
+                k--;
+                while (j < k && nums[j] == nums[j - 1]) j++;
+                while (j < k && nums[k] == nums[k + 1]) k--;
+            } else if (nums[j] + nums[k] > target) {
+                k--;
+            } else {
+                j++;
+            }
+        }
+    }
+    return result;
+
+    }
+}
+ */