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hello-algo/en/codes/ruby/chapter_computational_complexity/time_complexity.rb
Yudong Jin 14c314c08c Retranslate the complete English edition from Simplified Chinese (#2010)
* Retranslate the complete English edition from Simplified Chinese

* Polish three English explanations after reader-focused review
2026-10-31 03:35:51 +08:00

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3.6 KiB
Ruby

=begin
File: time_complexity.rb
Created Time: 2024-03-30
Author: Xuan Khoa Tu Nguyen (ngxktuzkai2000@gmail.com)
=end
### Constant complexity ###
def constant(n)
count = 0
size = 100000
(0...size).each { count += 1 }
count
end
### Linear complexity ###
def linear(n)
count = 0
(0...n).each { count += 1 }
count
end
### Linear complexity (array traversal)###
def array_traversal(nums)
count = 0
# The number of iterations is proportional to the array length
for num in nums
count += 1
end
count
end
### Quadratic complexity ###
def quadratic(n)
count = 0
# The number of iterations is proportional to the square of the input size n
for i in 0...n
for j in 0...n
count += 1
end
end
count
end
### Quadratic complexity (bubble sort)###
def bubble_sort(nums)
count = 0 # Counter
# Outer loop: the unsorted range is [0, i]
for i in (nums.length - 1).downto(0)
# Inner loop: move the largest element in the unsorted range [0, i] to its right end through swaps
for j in 0...i
if nums[j] > nums[j + 1]
# Swap nums[j] and nums[j + 1]
tmp = nums[j]
nums[j] = nums[j + 1]
nums[j + 1] = tmp
count += 3 # Swapping two elements takes 3 elementary operations
end
end
end
count
end
### Exponential complexity (iterative implementation)###
def exponential(n)
count, base = 0, 1
# Each cell divides into two per round, giving the sequence 1, 2, 4, 8, ..., 2^(n-1)
(0...n).each do
(0...base).each { count += 1 }
base *= 2
end
# count = 1 + 2 + 4 + 8 + .. + 2^(n-1) = 2^n - 1
count
end
### Exponential complexity (recursive implementation)###
def exp_recur(n)
return 1 if n == 1
exp_recur(n - 1) + exp_recur(n - 1) + 1
end
### Logarithmic complexity (iterative implementation)###
def logarithmic(n)
count = 0
while n > 1
n /= 2
count += 1
end
count
end
### Logarithmic complexity (recursive implementation)###
def log_recur(n)
return 0 unless n > 1
log_recur(n / 2) + 1
end
### Linearithmic complexity ###
def linear_log_recur(n)
return 1 unless n > 1
count = linear_log_recur(n / 2) + linear_log_recur(n / 2)
(0...n).each { count += 1 }
count
end
### Factorial complexity (recursive implementation)###
def factorial_recur(n)
return 1 if n == 0
count = 0
# Split one problem into n subproblems
(0...n).each { count += factorial_recur(n - 1) }
count
end
### Driver code ###
if __FILE__ == $0
# Try different values of n to see how the operation count grows for each complexity
n = 8
puts "Input size n = #{n}"
count = constant(n)
puts "Operation count for constant time = #{count}"
count = linear(n)
puts "Operation count for linear time = #{count}"
count = array_traversal(Array.new(n, 0))
puts "Operation count for linear time (array traversal) = #{count}"
count = quadratic(n)
puts "Operation count for quadratic time = #{count}"
nums = Array.new(n) { |i| n - i } # [n, n-1, ..., 2, 1]
count = bubble_sort(nums)
puts "Operation count for quadratic time (bubble sort) = #{count}"
count = exponential(n)
puts "Operation count for exponential time (iteration) = #{count}"
count = exp_recur(n)
puts "Operation count for exponential time (recursion) = #{count}"
count = logarithmic(n)
puts "Operation count for logarithmic time (iteration) = #{count}"
count = log_recur(n)
puts "Operation count for logarithmic time (recursion) = #{count}"
count = linear_log_recur(n)
puts "Operation count for linearithmic time (recursion) = #{count}"
count = factorial_recur(n)
puts "Operation count for factorial time (recursion) = #{count}"
end