From bd92198fc8c729cb3deeb0b22c4345ff2b4959a0 Mon Sep 17 00:00:00 2001
From: Vitor Oliveira <vbrazo@gmail.com>
Date: Fri, 3 Sep 2021 12:33:56 -0700
Subject: [PATCH] add formatting and example outputs for quick debugging

---
 data_structures/arrays/3sum.rb | 144 +++++++++++++++++----------------
 1 file changed, 76 insertions(+), 68 deletions(-)

diff --git a/data_structures/arrays/3sum.rb b/data_structures/arrays/3sum.rb
index 1e4bf80..51801f2 100644
--- a/data_structures/arrays/3sum.rb
+++ b/data_structures/arrays/3sum.rb
@@ -1,80 +1,88 @@
-#Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] ..
-#.. such that i != j, i != k, and j != k, and nums[i] + nums[j] + nums[k] == 0.
-#Notice that the solution set must not contain duplicate triplets.
+# Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] ..
+# .. such that i != j, i != k, and j != k, and nums[i] + nums[j] + nums[k] == 0.
+# Notice that the solution set must not contain duplicate triplets.
 
-#Example 1:
-#Input: nums = [-1,0,1,2,-1,-4]
-#Output: [[-1,-1,2],[-1,0,1]]
+# Example 1:
+# Input: nums = [-1,0,1,2,-1,-4]
+# Output: [[-1,-1,2],[-1,0,1]]
 
-#Example 2:
-#Input: nums = []
-#Output: []
+# Example 2:
+# Input: nums = []
+# Output: []
 
-#Example 3:
-#Input: nums = [0]
-#Output: []
+# Example 3:
+# Input: nums = [0]
+# Output: []
 
-#Constraints:
-#0 <= nums.length <= 3000
+# Constraints:
+# 0 <= nums.length <= 3000
 #-105 <= nums[i] <= 105
 
-
-
-
-#Two Pointer Approach - O(n) Time / O(1) Space
-#Return edge cases.
-#Sort nums, and init ans array
-#For each |val, index| in nums:
-#if the current value is the same as last, then go to next iteration
-#init left and right pointers for two pointer search of the two sum in remaining elements of array
-#while left < right:
-#find current sum
-#if sum > 0, right -= 1
-#if sum < 0, left += 1
-#if it's 0, then add the values to the answer array, and set the left pointer to the next valid value ..
-#.. (left += 1 while nums[left] == nums[left - 1] && left < right)
-#Return ans[]      
-
+# Two Pointer Approach - O(n) Time / O(1) Space
+# Return edge cases.
+# Sort nums, and init ans array
+# For each |val, index| in nums:
+# if the current value is the same as last, then go to next iteration
+# init left and right pointers for two pointer search of the two sum in remaining elements of array
+# while left < right:
+# find current sum
+# if sum > 0, right -= 1
+# if sum < 0, left += 1
+# if it's 0, then add the values to the answer array, and set the left pointer to the next valid value ..
+# .. (left += 1 while nums[left] == nums[left - 1] && left < right)
+# Return ans[]
 
 # @param {Integer[]} nums
 # @return {Integer[][]}
 def three_sum(nums)
-    #return if length too short
-    return [] if nums.length < 3
-    
-    #sort nums, init ans array
-    nums, ans = nums.sort, []
-    
-    #loop through nums
-    nums.each_with_index do |val, ind|
-      #if the previous value is the same as current, then skip this iteration as it would create duplicates
-      next if ind > 0 && nums[ind] == nums[ind - 1]
-      
-      #init & run two pointer search
-      left, right = ind + 1, nums.length - 1
-      
-      while left < right
-        #find current sum
-        sum = val + nums[left] + nums[right]
-        
-        #decrease sum if it's too great, increase sum if it's too low
-        if sum > 0
-          right -= 1
-        elsif sum < 0
-          left += 1
-        #if it's zero, then add the answer to array and set left pointer to next valid value
-        else
-          ans << [val, nums[left], nums[right]]
-            
-          left += 1
-            
-          while nums[left] == nums[left - 1] && left < right
-            left += 1
-          end
-        end
+  # return if length too short
+  return [] if nums.length < 3
+
+  # sort nums, init ans array
+  nums = nums.sort
+  ans = []
+
+  # loop through nums
+  nums.each_with_index do |val, ind|
+    # if the previous value is the same as current, then skip this iteration as it would create duplicates
+    next if ind > 0 && nums[ind] == nums[ind - 1]
+
+    # init & run two pointer search
+    left = ind + 1
+    right = nums.length - 1
+
+    while left < right
+      # find current sum
+      sum = val + nums[left] + nums[right]
+
+      # decrease sum if it's too great, increase sum if it's too low
+      if sum > 0
+        right -= 1
+      elsif sum < 0
+        left += 1
+      # if it's zero, then add the answer to array and set left pointer to next valid value
+      else
+        ans << [val, nums[left], nums[right]]
+
+        left += 1
+
+        left += 1 while nums[left] == nums[left - 1] && left < right
       end
     end
-    
-    #return answer
-    ans
-end
\ No newline at end of file
+  end
+
+  # return answer
+  ans
+end
+
+nums = [-1, 0, 1, 2, -1, -4]
+print three_sum(nums)
+# Output: [[-1,-1,2],[-1,0,1]]
+
+nums = []
+print three_sum(nums)
+# Output: []
+
+nums = [0]
+print three_sum(nums)
+# Output: []