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Use .count as standard

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This commit is contained in:
Vitor Oliveira 2021-03-01 16:46:57 -08:00
parent 32706822fa
commit f072feacff
1 changed files with 5 additions and 5 deletions
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10
data_structures/arrays/get_products_of_all_other_elements.rb
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@ -17,7 +17,7 @@
# 1. Brute force solution
#
def calculate_products_of_all_other_elements(nums)
product_of_other_elements = Array.new(nums.length, 1)
product_of_other_elements = Array.new(nums.count, 1)
nums.count.times do |i|
nums.count.times do |j|
@ -90,7 +90,7 @@ def output(prefix_products, suffix_products, nums)
nums.count.times do |index|
result << if index == 0
suffix_products[index + 1]
elsif index == nums.length - 1
elsif index == nums.count - 1
prefix_products[index - 1]
else
(prefix_products[index - 1] * suffix_products[index + 1])
@ -131,14 +131,14 @@ puts(products([1, 2, 3]))
# array doesn't add to the space complexity.
def products(nums)
return [] if nums.size < 2
return [] if nums.count < 2
res = [1]
# res[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 res[0] would be 1
(0..(nums.size - 2)).each do |i|
(0..(nums.count - 2)).each do |i|
num = nums[i]
res << num * res[i]
end
@ -148,7 +148,7 @@ def products(nums)
# so the product would be 1
product = 1
(nums.size - 1).downto(1).each do |i|
(nums.count - 1).downto(1).each do |i|
num = nums[i]
# For the index 'i', product would contain the
# product of all elements to the right. We update product accordingly.