Solve 1 - Two Sum (#6)

Author: terenceponce

README.md

@@ -5,3 +5,7 @@
 [![codecov](https://codecov.io/gh/terenceponce/leetcode/graph/badge.svg?token=i7gC0Uuurl)](https://codecov.io/gh/terenceponce/leetcode)
 
 This is a monorepo containing my solutions to LeetCode problems written in various languages.
+
+## List of Problems/Solutions
+
+- 1 - Two Sum - [Notes](notes/0001_two_sum.md) | [Elixir](elixir/lib/solutions/0001_two_sum/two_sum.ex) | [Ruby](ruby/lib/solutions/0001_two_sum/two_sum.rb)

elixir/lib/solutions/0001_two_sum/two_sum.ex

@@ -0,0 +1,19 @@
+defmodule LeetCode.Solutions.TwoSum do
+  @moduledoc false
+
+  @spec call(nums :: [integer], target :: integer) :: [integer]
+  def call(nums, target) do
+    nums
+    |> Enum.with_index()
+    |> Enum.reduce({%{}, []}, fn {num, index}, {map, result} ->
+      case Map.get(map, target - num) do
+        nil ->
+          {Map.put(map, num, index), result}
+
+        match ->
+          {map, [match, index]}
+      end
+    end)
+    |> elem(1)
+  end
+end

elixir/test/solutions/0001_two_sum/two_sum_test.exs

@@ -0,0 +1,17 @@
+defmodule LeetCode.Solutions.TwoSumTest do
+  use ExUnit.Case, async: true
+
+  alias LeetCode.Solutions.TwoSum
+
+  test "Case 1 works" do
+    assert TwoSum.call([2, 7, 11, 15], 9) == [0, 1]
+  end
+
+  test "Case 2 works" do
+    assert TwoSum.call([3, 2, 4], 6) == [1, 2]
+  end
+
+  test "Case 3 works" do
+    assert TwoSum.call([3, 3], 6) == [0, 1]
+  end
+end

notes/0001_two_sum.md

@@ -0,0 +1,122 @@
+# Two Sum
+
+**Link to Problem**: https://leetcode.com/problems/two-sum
+
+## Solutions
+
+- [Elixir](../elixir/lib/solutions/0001_two_sum/two_sum.ex)
+- [Ruby](../ruby/lib/solutions/0001_two_sum/two_sum.rb)
+
+## Description
+
+Given an array of integers `nums` and an integer `target`, return indices of the two numbers such that they add up to `target`.
+
+You may assume that each input would have **exactly one solution**, and you may not use the same element twice.
+
+You can return the answer in any order.
+
+## Examples
+
+### Example 1
+
+```
+Input: nums = [2,7,11,15], target = 9
+Output: [0,1]
+Explanation: Because nums[0] + nums[1] == 9, we return [0, 1].
+```
+
+### Example 2
+
+```
+Input: nums = [3,2,4], target = 6
+Output: [1,2]
+```
+
+### Example 3
+
+```
+Input: nums = [3,3], target = 6
+Output: [0,1]
+```
+
+## Solution
+
+This is my first ever LeetCode problem and I usually suck at algorithms, so this took me a long time to get.
+
+For people like me, the first solution I thought of was to iterate through each element in the list,
+compare it with all the other elements in the list, and try to add both elements to see if it matches
+the target.
+
+Unfortunately, while this does work, it is very slow because you have to go through the list multiple
+times in order to get the result.
+
+The optimal solution is to use a Hash Map which makes it possible to get the same answer while only
+going through the list once.
+
+Before we dive into the solution itself, I'd like to discuss something else first.
+
+One thing that I had to realize was that there was another way to figure out what elements it would
+take to reach the target. The obvious solution is to perform addition on all the elements until we
+get a sum that matches the target.
+
+Another way is to do it backwards and perform subtraction instead. We take the current element and
+subtract that from the target. From there, we try to figure out if there is an element that matches
+the difference.
+
+So how do we do that exactly? By creating a new variable and storing all the value and index of
+the current element if it doesn't satisfy the condition to match the target. In this case, we create
+a Hash Map (or Map in Elixir) since it allows us to store both value and index.
+
+In the case of Example 2, it would look like this:
+
+```
+[3, 2, 4]
+ ^
+ |
+we are here
+```
+
+Since the target is `6`, we will do `6 - 3` which is `3`. What we then do is look at our map
+to see if there is an element that has `3` as the value.
+
+At this point, our map is still empty, so there won't be any matches. The conclusion for this
+element is that we will skip it and go to the next one. However, we will also add this element
+into the map before we move to the next one:
+
+```
+{3: 0}
+```
+
+We put the value as the key, so we can use it for lookups and the index as the value, so we can get it later.
+
+After we stored the first element into the map, we will go to the next element:
+
+```
+[3, 2, 4]
+    ^
+    |
+we are here
+```
+
+At this point, we will try to do `6 - 2` which is `4`. We then look at the map to see if we
+have any matches.
+
+Since we only have `3` inside the map, there won't be any matches. So we repeat what we did
+in the previous element which is to add the current element to the map and move on to the next element:
+
+```
+{3: 0, 2: 1}
+```
+
+Afterwards, we go to the last element:
+
+```
+[3, 2, 4]
+       ^
+       |
+we are here
+```
+
+Obviously, we will try to do `6 - 4` which will result to `2`. If we do `Map.get(map, target - num)`,
+we will get `2` which has the value of `1`. From there, we should now be able to determine that the
+indices `[1, 2]` are the answer.

ruby/lib/solutions/0001_two_sum/two_sum.rb

@@ -0,0 +1,13 @@
+# frozen_string_literal: true
+
+def two_sum(nums, target)
+  hash = {}
+
+  nums.each_with_index do |num, index|
+    diff = target - num
+
+    return [hash[diff], index] if hash[diff]
+
+    hash[num] = index
+  end
+end

ruby/test/solutions/0001_two_sum/two_sum_test.rb

@@ -0,0 +1,18 @@
+# frozen_string_literal: true
+
+require 'test_helper'
+require_relative '../../../lib/solutions/0001_two_sum/two_sum'
+
+class TwoSumTest < Minitest::Test
+  def test_case_1_works
+    assert_equal [0, 1], two_sum([2, 7, 11, 15], 9)
+  end
+
+  def test_case_2_works
+    assert_equal [1, 2], two_sum([3, 2, 4], 6)
+  end
+
+  def test_case_3_works
+    assert_equal [0, 1], two_sum([3, 3], 6)
+  end
+end