C-Tests

CT-1: Get longest binary gap

class FindLongestBinaryGap
  def solution(n)
    binary_num = get_binary_number(n)

    new_arr = find_position_of_one_and_get_array(binary_num)

    res = no_gap(new_arr)
    return 0 if res == 0

    # get gap from newarr
    rev_Arr = new_arr.reverse
    get_max_binary_gap(rev_Arr)
  end

  def get_binary_number(n)
    num = n.to_s(2)
    arr =  num.to_s.split('');
  end

  def find_position_of_one_and_get_array(arr)
    newarr = []
    arr.each_with_index do |val, i|
      if val == 1.to_s
        # find index
        newarr.push(i);
      end
    end
    p newarr
  end

  def no_gap(binary)
    return 0 if binary.nil? || binary.size < 2
  end

  def get_max_binary_gap(rev_Arr)
    diff_array = Array.new
    for i in 0..(rev_Arr.size-1)  do
      first = rev_Arr[i].to_i
      second = rev_Arr[i+1].to_i
      if (first == second)
        next
      else
        diff = (first-second)
        diff_array.push(diff)
      end
    end
    diff_array.to_a;

    res = diff_array.to_a.max - 1
  end
end

res = FindLongestBinaryGap.new.solution(9);
puts res
res1 = FindLongestBinaryGap.new.solution(0);
puts res1
res2 = FindLongestBinaryGap.new.solution(529);
puts res2
res3 = FindLongestBinaryGap.new.solution(20);
puts res3
res4 = FindLongestBinaryGap.new.solution(15);
puts res4
res5 = FindLongestBinaryGap.new.solution(32);
puts res5

CT-2

A company has employed N developers (numbered from 0 to N−1) and wants to divide them into two teams. The first is a front-end team with F developers. The second is a back-end team with N−F developers. If the K-th developer is assigned to the front-end team then their contribution is A[K], and if they are assigned to the back-end team then their contribution is B[K]. What is the maximum sum of contributions the company can achieve?

Write a function:

def solution(a, b, f)

that, given two arrays A, B (consisting of N integers each) and the integer F, returns the maximum sum of contributions the company can achieve.

Examples:

1. Given A = [4, 2, 1], B = [2, 5, 3] and F = 2, the function should return 10. There should be two front-end developers and one back-end developer. The 0th and 2nd developers should be assigned to the front-end team (with contributions 4 and 1) and the 1st developer should be assigned to the back-end team (with contribution 5).

2. Given A = [7, 1, 4, 4], B = [5, 3, 4, 3] and F = 2, the function should return 18. The 0th and 3rd developers should be assigned to the front-end team and the 1st and 2nd developers should be assigned to the back-end team.

3. Given A = [5, 5, 5], B = [5, 5, 5] and F = 1, the function should return 15. The 0th developer can be assigned to the front-end team and the 1st and 2nd developers can be assigned to the back-end team.

Write an efficient algorithm for the following assumptions:

N is an integer within the range [1..200,000];
arrays A and B have equal lengths;
each element of array A is an integer within the range [0..1,000];
F is an integer within the range [0..N].


#### solution

# you can write to stdout for debugging purposes, e.g.
# puts "this is a debug message"

# you can write to stdout for debugging purposes, e.g.
# puts "this is a debug message"

def solution(a, b, f)
  # write your code in Ruby 2.2
  size_of_a = a.size
  size_of_b = b.size
  
  return 0 if size_of_a != size_of_b
  
  n = a.size
  contr_of_a = 0
  contr_of_b = 0
  total_contribution = 0
  case n
  when 1
    if f == 0
      contr_of_b = contr_of_b + b[0]
    elsif f==1
      contr_of_a = contr_of_a + a[0]
    end
    total_contribution = contr_of_a + contr_of_b
  else
    if f==1
      first = a[0]
      contr_of_a = contr_of_a + first
      for i in (1..n-1) do
        contr_of_b = contr_of_b + b[i]
      end
      total_contribution = contr_of_a + contr_of_b
    elsif f==2
      first = a[0] 
      last = a[n-1]
      contr_of_a = contr_of_a + first + last
      for i in (1..n-2) do
        contr_of_b = contr_of_b + b[i]
      end
      total_contribution = contr_of_a + contr_of_b
    else
      0  
    end
  end
    return total_contribution
end

CT3: CircularArray

Q:
An 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 an 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].

Solution:
# you can write to stdout for debugging purposes, e.g.
# puts "this is a debug message"

def solution(a, k)
  return a if a.size == 0
  for i in 1..k do
    p_arr = a.pop();
    a.unshift(p_arr);
    i = i+1;
  end
  return a
end

CT4: odd occurrence in an array

A non-empty 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:

def 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.

Write an efficient algorithm for the following assumptions:

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.

Solution:
# you can write to stdout for debugging purposes, e.g.
# puts "this is a debug message"

def solution(a)
  # write your code in Ruby 2.2
    new_arr = []
    @counts = nil
    a.each_with_object(Hash.new(0)) do |word, counts|
      counts[word] += 1
      @counts = counts
    end
    @counts.each do |count|
      unless count[1]%2 == 0
        new_arr << count[0]
      end
    end
    new_arr.first
end