Sort Squares of an Array

S
# Arrays - Sorted Squares

# Algorithm challenge description:
# Given an integer array nums sorted in non-decreasing order
# return an array of the squares of each number sorted in non-decreasing order.
# Input: [4, -1, -9, 2]
# Output: [1, 4, 16, 81]

#
# Approach 1: is using Ruby function (for sure)!
#

def sorted_squares(nums)
nums.map! { |num| num**2 }.sort
end
print(sorted_squares([4, -1, -9, 2]))

#
# Approach 2: is using bubble sort
#

def bubble_sort(array)
array_length = array.size
return array if array_length <= 1

loop do
swapped = false
(array_length - 1).times do |i|
if array[i] > array[i + 1]
array[i], array[i + 1] = array[i + 1], array[i]
swapped = true
end
end
break unless swapped
end
array
end

#
# Time complexity: O(n logn), where n is the length of the array.
# Space complexity: O(n) or O(logn)
#

def sorted_squares(nums)
# This takes O(n)
nums.map! { |num| num**2 }
# This can take Ο(n logn)
bubble_sort(nums)
end
print(sorted_squares([4, -1, -9, 2]))

#
# Approach 3: solving without ruby sort method. Using two-pointers
#
# Time complexity: O(n), where n is the length of the array.
# Space complexity: O(n), if you take output into account and O(1) otherwise.
#
def sorted_squares(nums)
p1 = 0
p2 = nums.length - 1
# since we're returing the result in ascending order,
# we'll fill in the array from the end
max_index = p2
output = []
while p1 < p2
nums1_square = nums[p1] * nums[p1]
nums2_square = nums[p2] * nums[p2]
if nums1_square < nums2_square
output[max_index] = nums2_square
p2 -= 1
elsif nums1_square > nums2_square
output[max_index] = nums1_square
p1 += 1
else
output[max_index] = nums1_square
max_index -= 1
output[max_index] = nums2_square
p1 += 1
p2 -= 1
end
max_index -= 1
end
# to account for any remaining value left in the input array
output[max_index] = nums[p1] * nums[p2] if p1 == p2
output
end

print(sorted_squares([4, -1, -9, 2]))