One of Maximum Slice Problems in Codility’s Lessons. [Link]

## Task Description

A non-empty array A consisting of N integers is given.

A triplet (X, Y, Z), such that 0 ≤ X < Y < Z < N, is called a *double slice*.

The *sum* of double slice (X, Y, Z) is the total of A[X + 1] + A[X + 2] + … + A[Y − 1] + A[Y + 1] + A[Y + 2] + … + A[Z − 1].

For example, array A such that:

A[0] = 3 A[1] = 2 A[2] = 6 A[3] = -1 A[4] = 4 A[5] = 5 A[6] = -1 A[7] = 2

contains the following example double slices:

- double slice (0, 3, 6), sum is 2 + 6 + 4 + 5 = 17,
- double slice (0, 3, 7), sum is 2 + 6 + 4 + 5 − 1 = 16,
- double slice (3, 4, 5), sum is 0.

The goal is to find the maximal sum of any double slice.

Write a function:

`def solution(A)`

that, given a non-empty array A consisting of N integers, returns the maximal sum of any double slice.

For example, given:

A[0] = 3 A[1] = 2 A[2] = 6 A[3] = -1 A[4] = 4 A[5] = 5 A[6] = -1 A[7] = 2

the function should return 17, because no double slice of array A has a sum of greater than 17.

Write an **efficient** algorithm for the following assumptions:

- N is an integer within the range [3..100,000];
- each element of array A is an integer within the range [−10,000..10,000].

## My Answer

- Detected time complexity:
**O(N)**

You can see the description in the comment.

```
def solution(A):
l_max_slice_sum = [0] * len(A)
r_max_slice_sum = [0] * len(A)
for i in range(1, len(A)-2): # A[X + 1] + A[X + 2] + ... + A[Y − 1]
# Let's assume that Y is equal to i+1.
# If l_max_slice_sum[i-1] + A[i] is negative, we assign X to i.
# It means that the slice sum is 0 because X and Y are consecutive indices.
l_max_slice_sum[i] = max(l_max_slice_sum[i-1] + A[i], 0)
for i in range(len(A)-2, 1, -1): # A[Y + 1] + A[Y + 2] + ... + A[Z − 1]
# We suppose that Y is equal to i-1.
# As aforementioned, Z will be assigned to i if r_max_slice_sum[i+1] + A[i]
# is negative, and it returns 0 because Y and Z becomes consecutive indices.
r_max_slice_sum[i] = max(r_max_slice_sum[i+1] + A[i], 0)
max_slice_sum = l_max_slice_sum[0] + r_max_slice_sum[2]
for i in range(1, len(A)-1):
# Let's say that i is the index of Y.
# l_max_slice_sum[i-1] is the max sum of the left slice, and
# r_max_slice_sum[i+1] is the max sum of the right slice.
max_slice_sum = max(max_slice_sum, l_max_slice_sum[i-1]+r_max_slice_sum[i+1])
return max_slice_sum
```