MaxSliceSum

A non-empty array A consisting of N integers is given. A pair of integers (P, Q), such that 0 ≤ P ≤ Q < N, is called a slice of array A. The sum of a slice (P, Q) is the total of A[P] + A[P+1] + ... + A[Q].

Write a function:

    int solution(vector<int> &A);

that, given an array A consisting of N integers, returns the maximum sum of any slice of A.

For example, given array A such that:
A[0] = 3  A[1] = 2  A[2] = -6
A[3] = 4  A[4] = 0

the function should return 5 because:

        (3, 4) is a slice of A that has sum 4,
        (2, 2) is a slice of A that has sum −6,
        (0, 1) is a slice of A that has sum 5,
        no other slice of A has sum greater than (0, 1).

Write an efficient algorithm for the following assumptions:

        N is an integer within the range [1..1,000,000];
        each element of array A is an integer within the range [−1,000,000..1,000,000];
        the result will be an integer within the range [−2,147,483,648..2,147,483,647].

#include <cassert>

const size_t MAX_ARRAY_SIZE = 1000000;
const size_t MIN_ARRAY_SIZE = 1;
const int MIN_VALUE = -1000000;
const int MAX_VALUE = 1000000;
const int MAX_RESULT = 2147483647;
const int MIN_RESULT = -2147483647;

int solution(vector<int> &A) {
    // write your code in C++11 (g++ 4.8.2)
    assert(A.size() <= MAX_ARRAY_SIZE);
    assert(A.size() >= MIN_ARRAY_SIZE);
    
    if (A.size() == 1) return A[0];
    
    long long sum = 0;
    long long result = A[0];
    
    for (size_t i = 0; i < A.size(); i ++) {
        assert((A[i] >= MIN_VALUE) && (A[i] <= MAX_VALUE));
        
        if (sum + A[i] > 0) {
            sum = max<long long>(A[i], sum + A[i]);   
            result = max<long long>(sum, result);
        } else {
            sum = 0;
            result = max<long long>(A[i], result);
        }
    }
    
    assert((result >= MIN_RESULT) && (result <= MAX_RESULT));
    
    return int(result);
}