MinAbsSumOfTwo

Let A be a non-empty array consisting of N integers.

The abs sum of two for a pair of indices (P, Q) is the absolute value |A[P] + A[Q]|, for 0 ≤ P ≤ Q < N.

For example, the following array A:
  A[0] =  1
  A[1] =  4
  A[2] = -3

has pairs of indices (0, 0), (0, 1), (0, 2), (1, 1), (1, 2), (2, 2).
The abs sum of two for the pair (0, 0) is A[0] + A[0] = |1 + 1| = 2.
The abs sum of two for the pair (0, 1) is A[0] + A[1] = |1 + 4| = 5.
The abs sum of two for the pair (0, 2) is A[0] + A[2] = |1 + (−3)| = 2.
The abs sum of two for the pair (1, 1) is A[1] + A[1] = |4 + 4| = 8.
The abs sum of two for the pair (1, 2) is A[1] + A[2] = |4 + (−3)| = 1.
The abs sum of two for the pair (2, 2) is A[2] + A[2] = |(−3) + (−3)| = 6.

Write a function:

    int solution(vector<int> &A);

that, given a non-empty array A consisting of N integers, returns the minimal abs sum of two for any pair of indices in this array.

For example, given the following array A:
  A[0] =  1
  A[1] =  4
  A[2] = -3

the function should return 1, as explained above.

Given array A:
  A[0] = -8
  A[1] =  4
  A[2] =  5
  A[3] =-10
  A[4] =  3

the function should return |(−8) + 5| = 3.

Write an efficient algorithm for the following assumptions:

        N is an integer within the range [1..100,000];
        each element of array A is an integer within the range [−1,000,000,000..1,000,000,000].

#include <algorithm>

int solution(vector<int> &A) {
    // write your code in C++11 (g++ 4.8.2)
    if (A.size() == 1) return abs(A[0]) * 2;
    
    sort(A.begin(), A.end(), [](int a, int b) {
        return abs(a) < abs(b); });
    
    int min_sum = abs(A[0]) * 2;
    
    for (int i = 0; i < A.size() - 1; i++) {
        min_sum = min(min_sum, abs(A[i] + A[i + 1]));    
    }
    
    return min_sum;
}