Overall, the names should be descriptive and there are several types of naming styles such as:

 apple -- lowercase
 apple_number -- lowercase with underscores
 APPLE -- uppercase
 APPLE_NUMBER -- uppercase with underscore
 CamelCase -- capitalized the first character of each words.
 mixedCase -- initial lower case

Module name will be short and all-lowercase:

socket

Class name will be using CamelCase

MyClass

Exception name:

NullException

Global Variable Name:

lowercase

Function Name:

lowercase_to_uppercase()

Function and method arguments:

self and cls should be the first argument for the class and instance method.

lowercase_to_uppercase(self,arg) -- Instance method
lowercase_to_uppercase(cls,arg) -- Class method

Constants:

Cap with underscores.

MAX_CAPACITY

By far, variables naming and white space and operators are covered for better code styles. Next part will cover several recommendations for exceptions and return statements.

So Let’s begin!

Aligned with opening delimiter such as parenthesis or bracket:

hello = hello_function_name(var_name, var_number,
                            var_telephone)

When if conditional has multi-lines: one space after “if”

 if (this_condition and
     this_condition):

In the case of List :

list = [
    1, 3, 4,
    5, 6, 7,
    ]

When there is a expression that we cannot use implicit continuation, backslashes are coming in:

with open('/path/hiiamfile') as file_read, \
     open('/path/hi,'w') as file_write:

backslash will explicitly specify the continual expression between lines.

The next one will be binary operators:

cost = (icecream_price
        + storage_overnight
        * (number_icecream + number_cups)
        )

After getting to know the braces, We have somewhat basic structures and now the white spaces are remained when it comes to perfect structures of the codes:

1. No spaces right after parentheses

compute(icecream_price[0], {storage_overnight: 3})

2. No spaces before semicolon or colon

if a == 3: compute icecream, storage; 

3. Assignments have one space around the operator

x = 1

However, after all these rules, when there are priorities operators it can be changes to the space around only the last operator that will perform:

a = b + 1
d = d*d + c*f
a = (j+v) * (c-d)

Also, when we indicate a argument and parameter value, assignment operator will not have space around it:

def compute(a, b=0):

Naming conventions tutorials of Python 3 will be on the next article!! See you in the next article!

Source: http://legacy.python.org/dev/peps/pep-0008/#indentation

A zero-indexed array A consisting of N integers is given. Rotation of the array means that each element is shifted right by one index, and the last element of the array is moved to the first place. For example, the rotation of array A = [3, 8, 9, 7, 6] is [6, 3, 8, 9, 7] (elements are shifted right by one index and 6 is moved to the first place).

The goal is to rotate array A K times; that is, each element of A will be shifted to the right K times.

Write a function:

def solution(A, K)

that, given a zero-indexed array A consisting of N integers and an integer K, returns the array A rotated K times.

For example, given

A = [3, 8, 9, 7, 6] K = 3

the function should return [9, 7, 6, 3, 8]. Three rotations were made:

[3, 8, 9, 7, 6] -> [6, 3, 8, 9, 7] [6, 3, 8, 9, 7] -> [7, 6, 3, 8, 9] [7, 6, 3, 8, 9] -> [9, 7, 6, 3, 8]

For another example, given

A = [0, 0, 0] K = 1

the function should return [0, 0, 0]

Given

A = [1, 2, 3, 4] K = 4

the function should return [1, 2, 3, 4]

Assume that:

• N and K are integers within the range [0..100];
• each element of array A is an integer within the range [−1,000..1,000].

In your solution, focus on correctness. The performance of your solution will not be the focus of the assessment.

Solving Process

Assumption, Test Coverage: The array length will be divided by the length of A in case it is bigger than the array length and display the array from K index of array till the end and from the beginning to the end.

It is like a Caesar’s enigma. Moving the index numbers for the alphabets with K numbers further. For example a = 1, b =2 , and c = 3 and such. If we move by 1 (K). z = 1, a =2, b = 3 ,c =4 and such.

Code Solution(mine)

def solution(A, K):     
   if K == 0 or len(A)/K ==0:
      return A
   if K > len(A):
      K = K/len(A)
   A = A[len(A)-K:len(A)] + (A[0:len(A)-K])
   return A

Big-O Calculation

def solution(A, K):     
   if K == 0 or len(A)/K ==0:
      return A
   if K > len(A):
      K = K/len(A)
   A = A[len(A)-K:len(A)] + (A[0:len(A)-K])
   return A
#O(1)

Other People’s code solution

def solution(A, K):

if len(A) == 0:

return A

K = K % len(A)

return A[-K:] + A[:-K]

O(1)

Learning Points

• Negative Index can be used to manipulate the array. Wow. This is what I need.

Codility Problem series

• Codility 1.1 BinaryGap
• Codility 2.1 OddOccurrencesinArray
• Codility 2.2 ArraysCyclirotation
• Codility 3.1 FrogJmp

A small frog wants to get to the other side of the road. The frog is currently located at position X and wants to get to a position greater than or equal to Y. The small frog always jumps a fixed distance, D.

Count the minimal number of jumps that the small frog must perform to reach its target.

Write a function:

def solution(X, Y, D)

that, given three integers X, Y and D, returns the minimal number of jumps from position X to a position equal to or greater than Y.

For example, given:

X = 10 Y = 85 D = 30

the function should return 3, because the frog will be positioned as follows:

• after the first jump, at position 10 + 30 = 40
• after the second jump, at position 10 + 30 + 30 = 70
• after the third jump, at position 10 + 30 + 30 + 30 = 100

Assume that:

• X, Y and D are integers within the range [1..1,000,000,000];
• X ≤ Y.

Complexity:

• expected worst-case time complexity is O(1);
• expected worst-case space complexity is O(1).

Solving Process

This is a simple division problem. It can be noted as N / K and if it has a remainder add one to quotient.

Code Solution(mine)

def solution(A, K):     
     return (Y - X)// D if (Y - X) % D == 0 else (Y - X) // D + 1

Big-O Calculation

def solution(A, K):     
     return (Y - X)// D if (Y - X) % D == 0 else (Y - X) // D + 1

#O(1)

Other People’s code solution

def solution(X, Y, D):

   distance = Y - X

   if distance % D == 0:

      return distance//D

   else:

      return distance//D + 1

#O(1)

Learning Points

• One line if statement can be used to simplify the answer.
• // operator

Codility Problem series

• Codility 1.1 BinaryGap
• Codility 2.1 OddOccurrencesinArray
• Codility 2.2 ArraysCyclirotation
• Codility 3.1 FrogJmp

A non-empty zero-indexed array A consisting of N integers is given. The array contains an odd number of elements, and each element of the array can be paired with another element that has the same value, except for one element that is left unpaired.

For example, in array A such that:

A[0] = 9 A[1] = 3 A[2] = 9 A[3] = 3 A[4] = 9 A[5] = 7 A[6] = 9

• the elements at indexes 0 and 2 have value 9,
• the elements at indexes 1 and 3 have value 3,
• the elements at indexes 4 and 6 have value 9,
• the element at index 5 has value 7 and is unpaired.

Write a function:

int solution(A);

that, given an array A consisting of N integers fulfilling the above conditions, returns the value of the unpaired element.

For example, given array A such that:

A[0] = 9 A[1] = 3 A[2] = 9 A[3] = 3 A[4] = 9 A[5] = 7 A[6] = 9

the function should return 7, as explained in the example above.

Assume that:

• N is an odd integer within the range [1..1,000,000];
• each element of array A is an integer within the range [1..1,000,000,000];
• all but one of the values in A occur an even number of times.

Complexity:

• expected worst-case time complexity is O(N);
• expected worst-case space complexity is O(1), beyond input storage (not counting the storage required for input arguments).

Solving Process

The array will be iterated by the unit of two elements. After sorting an array, Checking the paired element and if it is not paired, then the element’s value is the element the method returns.

It is like sorting rocks that have several height levels and spotting the only unpaired rock.

Code Solution(mine)

def solution(A):     
    if len(A) == 1:
         return A[0]
    A = sorted(A)
    for i in range(0 , len (A) , 2):
         if i+1 == len(A):
             return A[i]
         if A[i] != A[i+1]:
             return A[i]

Big-O Calculation

def solution(A):     
    if len(A) == 1:
         return A[0]
    A = sorted(A)       # O(n*log(N) or N)
    for i in range(0 , len (A) , 2): # O(N)
         if i+1 == len(A):
             return A[i]
         if A[i] != A[i+1]:
             return A[i]
# O(N*log(N) or O(N))

Other People’s code solution

def solution(A):

   result = 0

   for number in A:

     result ^= number

   return result
#O(N)

Learning Points

• XOR bitwise operator can detect overlapping values.
• https://codesays.com/2015/solution-to-odd-occurrences-in-array-by-codility/ Solution link

Codility Problem series

• Codility 1.1 BinaryGap
• Codility 2.1 OddOccurrencesinArray
• Codility 2.2 ArraysCyclirotation
• Codility 3.1 FrogJmp

A binary gap within a positive integer N is any maximal sequence of consecutive zeros that is surrounded by ones at both ends in the binary representation of N.

For example, number 9 has binary representation 1001 and contains a binary gap of length 2. The number 529 has binary representation 1000010001 and contains two binary gaps: one of length 4 and one of length 3. The number 20 has binary representation 10100 and contains one binary gap of length 1. The number 15 has binary representation 1111 and has no binary gaps.

Write a function:

int solution(int N);

that, given a positive integer N, returns the length of its longest binary gap. The function should return 0 if N doesn’t contain a binary gap.

For example, given N = 1041 the function should return 5, because N has binary representation 10000010001 and so its longest binary gap is of length 5.

Assume that:

• N is an integer within the range [1..2,147,483,647].

Complexity:

• expected worst-case time complexity is O(log(N));
• expected worst-case space complexity is O(1).

Solving Process

Every time the digit is 1, bin_counter will be updated and if it is bigger than the previous biggest one, then update it to bin_max. At last, return the bin_max.

Imagine a child recording a jump of a frog(0) without a stop(1). This is as easy as that as long as we know how to process a number to binary number, which is str(bin(N))[2:].

Code Solution(mine)

def solution(N):     
     bin_rep = str(bin(N))[2:]
     bin_gap = False
     bin_max = 0
     bin_counter = 0
     for symbol in bin_rep:
         if symbol == '1':
             if bin_max < bin_counter:
                 bin_max = bin_counter
             bin_gap = True
             bin_counter = 0
         elif bin_gap:
             bin_counter += 1
     return bin_max

Big-O Calculation

def solution(N):     
     bin_rep = str(bin(N))[2:]   # Constant 1
     bin_gap = False             
     bin_max = 0                 
     bin_counter = 0             
     for symbol in bin_rep:      # Constant N
         if symbol == '1':
             if bin_max < bin_counter:
                 bin_max = bin_counter
             bin_gap = True
             bin_counter = 0
         elif bin_gap:
             bin_counter += 1
     return bin_max
## O(N)

Other People’s code solution

def solution(N):
  return len(max(bin(N)[2:].strip('0').strip('1').split('1')))
# O(N)

Learning Points

• strip(str) and split(str) functions can be used for the string manipulation.

To be prepared for the upcoming interviews and also to further my learning according to the book “Pragmatic Thinking and Learning Refactor your WetWare” by Andy Hunt I would love to introduce what I play with to the public.

Codility Posts will be composed of all the lessons and my answers and alternative answers for the problem with the description of the problem using Python. Also, the readings I do for the interviews such as “Cracking the Coding Interview” and “Python Master the Art of Design Patterns” will be shared either.

Hmm, If I have some time , the daily process of the posts and the readings will be covered. The core principles of the learning process will be explained as an extension to my reading “Pragmatic Thinking and Learning Refactor your WetWare.”

Codility Problem series

• Codility 1.1 BinaryGap
• Codility 2.1 OddOccurrencesinArray
• Codility 2.2 ArraysCyclirotation
• Codility 3.1 FrogJmp