Evaluating Efficiency

1. Read Big O Notation for Newbies with Ruby
2. Work through this quiz on Big O. Try out the code snippets and read the answers.
3. Do the assignment below and submit a PR with your answers.

Assignment - Determine the big O

1. Give the efficiency of each of the following code snippets and
2. Justify your answer

Examples

Examples

Problems for you

Snippet 1 - Big O:

def largest?(array, value)
  array.each do |item|
    return false if item > value
  end
  return true
end

Efficiency: O(n) or linear because the code runs upon each element in the array just one at a time to compare it to the given value.

Snippet 2 - Big O:

def info_dump(customers)
  puts "Customer Names: "
  customers.each do |customer|
    puts "#{customer[:name]}"
  end
  puts "Customer Locations: "
  customers.each do |customer|
    puts "#{customer[:country]}"
  end
end

Efficiency: O(n) or linear because the code runs upon each element in the array just one at a time to compare it to the given value.

Snippet 3 - Big O:

def first_element_is_red?(array)
  array[0] == 'red' ? true : false
end

Efficiency: O(1) because the code considers only the first element in the array regardless of the array size.

Snippet 4 - Big O:

def duplicates?(array)
  array.each_with_index do |item1, index1|
    array.each_with_index do |item2, index2|
      next if index1 == index2
      return true if item1 == item2
    end
  end
  false
end

Efficiency: O(n^2) because due to the nested loop in the array, we need to think every single element in the array so the element is counted as n*n times.

Snippet 5 - Big O:

words = [chocolate, coconut, rainbow]
endings = [cookie, pie, waffle]

words.each do |word|
  endings.each do |ending|
    puts word + ending
  end
end

Efficiency: O(n*m) or Quadratic due to nested loop. When the outer loop iterates, every single element in words array(n), inner loop will go through each element in array(m).

Snippet 6 - Big O:

numbers = # some array (you don't know contents)

def print_array(array)
    array.each {|num| puts num}
end

Efficiency: O(n) or Linear because the code runs upon each element in the array just one at a time to compare it to the given value.

Snippet 7 - Big O:

# this is insertion sort
(2...num.length).each do |j|
    key = num[j]
    i = j - 1
    while i > 0 and num[i] > key
        num[i+1] = num[i]
        i = i - 1
    end
    num[i+1] = key
end

Efficiency: O(n^2) because of nested loop and insertion sort have a complexity of it.

Snippet 8 - Big O:

# this is selection sort
n.times do |i|
  index_min = i
  (i + 1).upto(n-1) do |j|
    index_min = j if a[j] < a[index_min]
  end
  a[i], a[index_min] = a[index_min], a[i] if index_min != i
end

Efficiency: O(n^2) due to nested loop and selection sort has a complexity of O(n^2)