# Problem Statement

Implement a method called my_count that takes a number and an array of numbers and returns the count representing the total number of elements in the array that is equal or greater than the given number.

# Problem Domain Analysis

Let's say our method takes two parameters, first parameter n for the number used for comparison and an array for the second parameter a that contains a list of numbers to be processed.

The result in this case = 3, since there are only three numbers 10, 15 and 25 that are equal or above 10. We can accomplish this in Ruby in just one line, in the irb:

``````\$irb
a = [2, 7, 10, 15, 25]
a.count{|x| x >= 10}
=> 3
``````

We need to implement our own version of the Ruby's count method called `my_count`. In this irb console session we only focus on specifying what needs to be done. The details of counting such as how many elements are in the list to be counted, when we need to terminate the loop etc has been hidden inside the count method. As you already know, what needs to be done is the focus of the problem domain analysis.

# Solution Domain Analysis

## Steps to Solve the Problem

``````while < n numbers have been examined do
1. Read the number
2. If current number satisfies the condition then add one to count.
end
``````

# Algorithm

1. Read the number of elements to be processed.
2. Initialize the count to 0
3. While there are still elements to be processed repeatedly do: a) Read the number b) If it satisfies the condition (>= 10) then add one to count.
4. Return the total number of elements that satisfies the condition.

# Logical Design

1. Read n and the array
2. Take the first element in the array
3. Check if element x >= n

Yes --> Increment counter

4. Go to the next element

# Skeleton Code

``````count = 0
i = 0

while i < n do

end
``````

## Terminating Condition

i == n is the terminating condition.

Consideration of the problem at its lowest limit (i.e., n=0) leads to a mechanism that can be extended to larger values of n by simple repetition. This is a very common technique in algorithm design.

index = 0
count = 0

# Pseudo Code

``````while index < (size of array - 1)
extract the element from the array
compare --> increment counter if yes
end

return count
``````

# Steps

## Step 1

Let's write our first test that is the minimum we have to do for making the test fail. Create `my_count_spec.rb` with the following contents:

``````describe MyCount do

it "should return 0 for n = 0 and an empty list" do
my_count = MyCount.new

result = my_count.calculate(0, [])

expect(result).to eql(0)
end

end
``````

## Step 2

To make the test fail for the right reason, add the MyCount class to the top of the `my_count_spec.rb` as follows:

``````class MyCount
def calculate(n, a)

end
end
``````

You will get expected 0 got nil error message.

## Step 3

Return 0 to make the test pass.

``````class MyCount
def calculate(n, a)
0
end
end
``````

## Step 4

Let's add the second test as follows:

``````describe MyCount do
# first test same as before

it "should return 1 for 0, [0]" do
my_count = MyCount.new

result = my_count.calculate(0, [0])

expect(result).to eql(1)
end
end
``````

This fails with the error 'expected 1, got 0' message.

## Step 5

The quick and dirty way to make both test pass is as follows:

``````class MyCount
def calculate(n, a)
if a.include? n
1
else
0
end
end
end
``````

## Step 6

Let's add the third spec that checks the opposite of what the second spec checked as follows:

``````describe MyCount do
# first two tests same as before

it "should return 0 for 1, [0]" do
my_count = MyCount.new

result = my_count.calculate(1, [0])

expect(result).to eql(0)
end
end
``````

This test passes because the include? method returns false.

## Step 7

Add the fourth spec as follows:

``````describe MyCount do
# first three tests same as before

it "should return 2 for 1, [1,1]" do
my_count = MyCount.new

result = my_count.calculate(1, [1,1])

expect(result).to eql(2)
end
end
``````

This fails with expected 2, got 1 error.

## Step 8

By reading the algorithm we developed earlier and the skeleton code, let's stop faking and implement the real solution as follows:

``````class MyCount
def calculate(n, a)
count = 0
index = 0

while index < a.size
if a[index] == n
count += 1
end
index += 1
end
count
end
end
``````

All the test passes.

## Warning

If you forget to increment the index or use the wrong terminating condition, your program will just hang.

## Step 9

Let's add the story test as follows:

``````describe MyCount do
# first four tests same as before

it "should return 5 for 3, [1,1,2,3,4,3,6,6,1]" do
my_count = MyCount.new

result = my_count.calculate(3, [1,1,2,3,4,3,6,6,1])

expect(result).to eql(5)
end
end
``````

This fails gloriously with the error message 'expected 5, got 2'.

## Step 10

This test has exposed a bug in our code. By inspecting our solution we find that the if condition is wrong. Correct it as follows:

``````if a[index] >= n
``````

All our tests passes.

Here is the final solution:

``````class MyCount
def calculate(n, a)
count = 0
index = 0

while index < a.size
if a[index] >= n
count += 1
end
index += 1
end
count
end
end

describe MyCount do

it "should return 0 for n = 0 and an empty list" do
my_count = MyCount.new

result = my_count.calculate(0, [])

expect(result).to eql(0)
end

it "should return 1 for 0, [0]" do
my_count = MyCount.new

result = my_count.calculate(0, [0])

expect(result).to eql(1)
end

it "should return 0 for 1, [0]" do
my_count = MyCount.new

result = my_count.calculate(1, [0])

expect(result).to eql(0)
end

it "should return 2 for 1, [1,1]" do
my_count = MyCount.new

result = my_count.calculate(1, [1,1])

expect(result).to eql(2)
end

it "should return 5 for 3, [1,1,2,3,4,3,6,6,1]" do
my_count = MyCount.new

result = my_count.calculate(3, [1,1,2,3,4,3,6,6,1])

expect(result).to eql(5)
end
end
``````

# Discussion

The code evolved from:

1. No code at all --> Returning nil
2. Returning nil --> Returning a constant
3. Returning a constant --> If-else condition
4. If-else condition --> While loop

While is a more generic construct than if-else. If-else is more generic than returning a constant and so on. We see that the code became more generic as our tests became more specific. For an in-depth discussion read about Transformation Priority Premise by Robert Martin on his blog.

Refactoring is changing the structure without changing the behavior. Transformation is changing the behavior with the least change in the structure. This forces us to write minimal code to pass the current test. Transformation and Refactoring are the two sides of the same coin. Red to Green is transformation. Green to Green is refactoring.

# Q & A

I am having difficulty writing a test. Do you have any suggestions?

Use the Arrange, Act, Assert structure. Also work backwards by writing the assert first, then write the code for 'Act' phase and finally for Arrange.

When do I use Transformation Priority Premise?

Transformation Priority Premise is not a silver bullet. It's a work in progress. Designing algorithms to solve a given problem and drawing flow charts will always help you write complex programs.

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