Short Problem Definition:
Find the minimal average of any slice containing at least two elements.
expected worst-case time complexity is
expected worst-case space complexity is
Every slice must be of size two or three. Slices of bigger sizes are created from such smaller slices. Therefore should any bigger slice have an optimal value, all sub-slices must be the same, for this case to hold true. Should this not be true, one of the sub-slices must be the optimal slice. The others being bigger. Therefore we check all possible slices of size 2/3 and return the smallest one. The first such slice is the correct one, do not use <=!
For other languages use BigInts or equal. This line (A[idx] + A[idx+1] + A[idx+2])/3.0 can easily overflow!
You can read the formal proof by Minh Tran Dao here.
def solution(A): min_idx = 0 min_value = 10001 for idx in xrange(0, len(A)-1): if (A[idx] + A[idx+1])/2.0 < min_value: min_idx = idx min_value = (A[idx] + A[idx+1])/2.0 if idx < len(A)-2 and (A[idx] + A[idx+1] + A[idx+2])/3.0 < min_value: min_idx = idx min_value = (A[idx] + A[idx+1] + A[idx+2])/3.0 return min_idx