##### Short Problem Definition:

A non-empty zero-indexed array A consisting of N integers is given. A peak is an array element which is larger than its neighbors.

##### Link

##### Complexity:

expected worst-case time complexity is O(N*log(log(N)));

expected worst-case space complexity is O(N)

##### Execution:

I first compute all peaks. Because each block must contain a peak I start from the end and try to find a integral divisor sized block. If each block contains a peak I return the size.

##### Solution:

def solution(A): peaks = [] for idx in xrange(1, len(A)-1): if A[idx-1] < A[idx] > A[idx+1]: peaks.append(idx) if len(peaks) == 0: return 0 for size in xrange(len(peaks), 0, -1): if len(A) % size == 0: block_size = len(A) // size found = [False] * size found_cnt = 0 for peak in peaks: block_nr = peak//block_size if found[block_nr] == False: found[block_nr] = True found_cnt += 1 if found_cnt == size: return size return 0

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