AlgorithmsCodilityPatternsTutorials

The trick here is to first compute the maximum sums from beginning, and then from the end of the array. For example:

A =  [3, 2, 6, -1, 4, 5, -1, 2]

forward_computed_sums = [ 0, 2, 8, 7, 11, 16, 15, 17 ]

backwards_computed_sums = [ 19, 16, 14, 8, 9, 5, 0, 0 ]

You can rule out the first and the last item in array.

Now when we maximum sums for every position, we can easily calculate the maximum possible sum.

A non-empty zero-indexed 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:

function solution(A);
that, given a non-empty zero-indexed 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.

Assume that:

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].
Complexity:

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