Minor changes

data_structures/arrays/get_products_of_all_other_elements.rb

@@ -5,10 +5,13 @@
 # each element at index `i` of the new array is the product of
 # all the numbers in the original array except the one at `i`.
 
-# Generates prefix products
-# prefix_products[i] contains the product of all the elements to the left
-# Note: for the element at index '0', there are no elements to the left,
-# so the prefix_products[0] would be 1
+#
+# This file solves the algorithm in 3 approaches:
+#
+# 1. Brute force
+# 2. Left and Right product lists
+# 3. O(1) space approach
+#
 
 #
 # 1. Brute force solution
@@ -16,9 +19,11 @@
 def calculate_products_of_all_other_elements(nums)
   product_of_other_elements = Array.new(nums.length, 1)
 
-  nums.each_with_index do |_num1, i|
-    nums.each_with_index do |num2, j|
-      product_of_other_elements[i] = product_of_other_elements[i] * num2 if i != j
+  nums.count.times do |i|
+    nums.count.times do |j|
+      next if i == j
+
+      product_of_other_elements[i] = product_of_other_elements[i] * nums[j]
     end
   end
 
@@ -48,7 +53,7 @@ def build_prefix_products(nums)
 
   nums.each do |num|
     prefix_products << if prefix_products.count > 0
-                         (prefix_products.last * num)
+                         prefix_products.last * num
                        else
                          num
                        end
@@ -66,7 +71,7 @@ def build_suffix_products(nums)
 
   nums.reverse.each do |num|
     suffix_products << if suffix_products.count > 0
-                         (suffix_products.last * num)
+                         suffix_products.last * num
                        else
                          num
                        end
@@ -107,19 +112,13 @@ end
 puts(products([1, 2, 3]))
 # => [6, 3, 2]
 
-
 #
 # Approach 3: O(1) space approach
 #
 
-# Although the above solution is good enough to solve the problem since
-# we are not using division anymore, there's a follow-up component as
-# well which asks us to solve this using constant space. Understandably so,
-# the output array does not count towards the space complexity.
-# This approach is essentially an extension of the approach above.
+# This approach is essentially an extension of the approach 2.
 # Basically, we will be using the output array as one of L or R and we will
-# be constructing the other one on the fly. Let's look at the algorithm based
-# on this idea.
+# be constructing the other one on the fly.
 
 # Complexity analysis
 #
@@ -128,36 +127,35 @@ puts(products([1, 2, 3]))
 # answer.
 
 # Space complexity: O(1) since don't use any additional array for our
-# computations. The problem statement mentions that using the answeranswer
+# computations. The problem statement mentions that using the answer
 # array doesn't add to the space complexity.
 
 def products(nums)
   return [] if nums.size < 2
 
-  # The answer array to be returned
   res = [1]
 
   # answer[i] contains the product of all the elements to the left
   # Note: for the element at index '0', there are no elements to the left,
   # so the answer[0] would be 1
-  (0..(nums.size - 2)).each do |idx|
+  (0..(nums.size - 2)).each do |i|
     # answer[i - 1] already contains the product of elements to the left of 'i - 1'
     # Simply multiplying it with nums[i - 1] would give the product of all
     # elements to the left of index 'i'
-    num = nums[idx]
-    res << num * res[idx]
+    num = nums[i]
+    res << num * res[i]
   end
 
   # R contains the product of all the elements to the right
   # Note: for the element at index 'length - 1', there are no elements to the right,
   # so the R would be 1
   product = 1
-  (nums.size - 1).downto(1).each do |idx|
-      num = nums[idx]
-      # For the index 'i', R would contain the
-      # product of all elements to the right. We update R accordingly
-      res[idx - 1] *= (product * num)
-      product *= num
+  (nums.size - 1).downto(1).each do |i|
+    num = nums[i]
+    # For the index 'i', R would contain the
+    # product of all elements to the right. We update R accordingly
+    res[i - 1] *= (product * num)
+    product *= num
   end
 
   res

data_structures/linked_lists/doubly_linked_list.rb

@@ -146,7 +146,7 @@ class DoublyLinkedList
     print "#{head.val} --> "
     if head.next.nil?
       puts("nil\n")
-      return
+      nil
     else
       print_values(head.next)
     end