##### 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
```