среда, 25 декабря 2013 г.

Codility. Train. Max-Counters ★★★

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 zero-indexed 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 zero-indexed array A consisting of M integers, returns a sequence of integers representing the values of the counters.
The sequence should be returned as:
  • a structure Results (in C), or
  • a vector of integers (in C++), or
  • a record Results (in Pascal), or
  • an array of integers (in any other programming language).
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.
Assume that:
  • 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].
Complexity:
  • expected worst-case time complexity is O(N+M);
  • expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments).
Elements of input arrays can be modified.

Solution:

def solution(N, A):
    iMin, iMax = 0, 0
    iMas = [iMax] * N
    for i in A:
        if i == N + 1:
            iMin = iMax
        else:
            if iMas[i - 1] < iMin:
                iMas[i - 1] = iMin + 1
            else:
                iMas[i - 1] = iMas[i - 1] + 1
            if iMas[i - 1] > iMax:
                iMax = iMas[i - 1]
    for i in xrange(len(iMas)):
        if iMas[i] < iMin:
            iMas[i] = iMin
    return iMas

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