[LeetCode] 3Sum Smaller

Problem Description:

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(n^2) runtime?

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Sort nums first and then fix the left index (i) at each time while adjusting the middle and right indexes (j, k). The following code should be self-explanatory.

class Solution {
public:
    int threeSumSmaller(vector<int>& nums, int target) {
        sort(nums.begin(), nums.end());
        int n = nums.size(), ans = 0, i, j, k;
        for (int i = 0; i < n - 2; i++) {
            int j = i + 1, k = n - 1;
            while (j < k) {
                if (nums[i] + nums[j] + nums[k] >= target) k--;
                else {
                    ans += (k - j);
                    j++;
                }
            }
        }
        return ans;
    }
};