MaxDoubleSliceSum

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:

    int solution(vector<int> &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].

#include <string.h>

int solution(vector<int>& A)
{
 if (A.size() <= 3) return 0;

 int* firsts = new int[A.size()];
 int* seconds = new int[A.size()];

 memset(firsts, 0, sizeof(int) * A.size());
 memset(seconds, 0, sizeof(int) * A.size());

 for (unsigned int i = 1; i < A.size() - 1; ++i)
 {  
  firsts[i] = max(firsts[i - 1] + A[i], 0);
  int iInv = A.size() - i - 1;
  seconds[iInv] = max(seconds[iInv + 1] + A[iInv], 0);
 }

 int sumMax = 0;
 for (unsigned int i = 1; i < A.size() - 1; ++i)
 {
  sumMax = max(sumMax, firsts[i - 1] + seconds[i + 1]);
 }

 delete[] firsts;
 delete[] seconds;

 return sumMax;
}