*3Sum Smaller

Given an array of n integers nums and a target, find the number of index triplets i, j, k with 0 <= i < j < k < n that satisfy the condition nums[i] + nums[j] + nums[k] < target.

For example, given nums = [-2, 0, 1, 3], and target = 2.

Return 2. Because there are two triplets which sums are less than 2:

[-2, 0, 1]
[-2, 0, 3]

Follow up:
Could you solve it in O(n2) runtime?

public class Solution {
public int threeSumSmaller(int[] nums, int target) {
    Arrays.sort(nums);
    int sum = 0;
    for (int i = 0; i < nums.length - 2; i++) {
        sum += twoSumSmaller(nums, i + 1, target - nums[i]);
    }
    return sum;
}

private int twoSumSmaller(int[] nums, int startIndex, int target) {
    int sum = 0;
    int left = startIndex;
    int right = nums.length - 1;
    while (left < right) {
        if (nums[left] + nums[right] < target) {
            sum += right - left;
            left++;
        } else {
            right--;
        }
    }
    return sum;
}
}