Question

Given an array of integers, what is the maximum sum of a subset of the integers such that the integers in the subset were not originally next to each other?

Examples:

[3, 8, 4] => max sum is 8, since 8 > (3+4)
[12, 8, 9, 10] => max sum is 12 + 10 = 22, since this is greater than 12 + 9 and 8 + 10

I’m interested in figuring out the algorithm for doing this. Methodology / thinking process = greatly appreciated.

EDIT:
The integers range from 1 to 1000, inclusive. (Since this is entirely a learning exercise, I’d still be interested in learning about different methods if the integers ranged from, say, -1000 to 1000.)

Answer 1

Let you have array A {Ai, 1 <= i <= n }

F(i) – Maximum sum of subarray Aj { 1 <= j <= i }, then

F(0) = 0 – empty subarray

F(1) = A(1) – only first element

F(i) = max(F(i-2) + A(i), F(i-1)) , 2 <= i <= n

F(n) – answer

C++ implementation:

int GetMaximumSubarraySum(const vector<int>& a)
{
    // note that vector a have 1-based index
    vector<int> v(a.size());
    v[0] = 0;
    v[1] = a[1];
    for(int i =2; i < a.size(); i++)
        v[i] = max(v[i-2] + a[i], v[i-1]);

    return v.back();
}

Explanation:

First, main idea is to use dynamic programming.
We try to solve task for array with N element by using known answer for array with N-1 and N-2 first elements. If N = 0 the answer is 0 and if N = 1 the answer is A[1]. It’s clear. For N >= 2 we have 2 different ways:

1. Use element A[N], then the answer is A[N] + F[N-2] (because we can’t use A[N-1] element and F[N-2] is the best solution for subarray 1..N-2, we don’t care about if F[N-2] element is used or not, this is just the best solution for subarray 1..N-2.

2. Don’t use element A[N], then the answer is F[N-1] (because we can use A[N-1] element and F[N-1] is the best solution for subarray 1..N-1, also we don’t care about if F[N-1] element is used or not.

So we need to get max of this 2 situations.
To solve the task you need calculate F[N] in increasing order and memorize answers.

Let see at your example:

[12, 8, 9, 10]

F[0] = 0

F[1] = 12 – use 1-st element

F[2] = max(F[0]+A[2], F[1]) = max(8, 12) = 12 – use 1-st element

F[3] = max(F[1]+A[3], F[2]) = max(21, 12) = 21 – use 1-st, 3-rd element

F[4] = max(F[2]+A[4], F[3]) = max(22, 21) = 22 – use 1-st, 4-th element

The answer is F[4] = 22.