Subarray Sum Closest

Given an integer array, find a subarray with sum closest to zero.
Return the indexes of the first number and last number.

Example
Given [-3, 1, 1, -3, 5], return [0, 2], [1, 3], [1, 1], [2, 2] or [0, 4]

Challenge
O(nlogn) time

class Solution {
public:
    /**
     * @param nums: A list of integers
     * @return: A list of integers includes the index of the first number
     *          and the index of the last number
     */
    vector<int> subarraySumClosest(vector<int> nums){
        vector<int> result;
        if (nums.empty()) {
            return result;
        }

        const int num_size = nums.size();
        vector<pair<int, int> > sum_index(num_size + 1);

        for (int i = 0; i < num_size; ++i) {
            sum_index[i + 1].first = sum_index[i].first + nums[i];
            sum_index[i + 1].second = i + 1;
        }

        sort(sum_index.begin(), sum_index.end());

        int min_diff = INT_MAX;
        int closest_index = 1;
        for (int i = 1; i < num_size + 1; ++i) {
            int sum_diff = abs(sum_index[i].first - sum_index[i - 1].first);
            if (min_diff > sum_diff) {
                min_diff = sum_diff;
                closest_index = i;
            }
        }

        int left_index = min(sum_index[closest_index - 1].second,                             sum_index[closest_index].second);
        int right_index = -1 + max(sum_index[closest_index - 1].second,                                   sum_index[closest_index].second);
        result.push_back(left_index);
        result.push_back(right_index);
        return result;
    }
};