lesson4_3_MaxCounters.rb

# You are given N counters, initially set to 0, and you have two possible operations on them:

# increase(X) − counter X is increased by 1,
# max counter − all counters are set to the maximum value of any counter.
# A non-empty array A of M integers is given. This array represents consecutive operations:

# if A[K] = X, such that 1 ≤ X ≤ N, then operation K is increase(X),
# if A[K] = N + 1 then operation K is max counter.
# For example, given integer N = 5 and array A such that:

#     A[0] = 3
#     A[1] = 4
#     A[2] = 4
#     A[3] = 6
#     A[4] = 1
#     A[5] = 4
#     A[6] = 4
# the values of the counters after each consecutive operation will be:

#     (0, 0, 1, 0, 0)
#     (0, 0, 1, 1, 0)
#     (0, 0, 1, 2, 0)
#     (2, 2, 2, 2, 2)
#     (3, 2, 2, 2, 2)
#     (3, 2, 2, 3, 2)
#     (3, 2, 2, 4, 2)
# The goal is to calculate the value of every counter after all operations.

# Write a function:

# def solution(n, a)

# that, given an integer N and a non-empty array A consisting of M integers, returns a sequence of integers representing the values of the counters.

# Result array should be returned as an array of integers.

# For example, given:

#     A[0] = 3
#     A[1] = 4
#     A[2] = 4
#     A[3] = 6
#     A[4] = 1
#     A[5] = 4
#     A[6] = 4
# the function should return [3, 2, 2, 4, 2], as explained above.

# Write an efficient algorithm for the following assumptions:

# N and M are integers within the range [1..100,000];
# each element of array A is an integer within the range [1..N + 1].

def solution(n, a)
  max = 0
  offsets = a.inject(Hash.new(max)) do |acc, el|
    next Hash.new(max) if el == n+1
    acc[el] +=1
    max = acc[el] if max < acc[el]
    acc
  end
  (1..n).map{|i| offsets[i]}
end