259. 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?

Solution:

Sort the list such that we can use two pointers merge from two sides to the central.

For each element num[i], we check if there are pairs from num[i + 1] to num[num.length - 1], such that the sum of three elements is smaller than target.

If we find num[i] + num[start] + num[end] < target, we know that

num[i] + num[start] + num[x] < target, where x > start and x < end.

Therefore, for num[i] and num[start], there are end - start solutions that ends up with sum < target.



Code:

public class Solution {
    public int threeSumSmaller(int[] nums, int target) {
        if (nums == null || nums.length == 0) {
            return 0;
        }
        Arrays.sort(nums);
        int count = 0;
        for (int i = 0; i < nums.length - 2; i++) {
            int start = i + 1;
            int end = nums.length - 1;
            while (start < end) {
                int sum = nums[i] + nums[start] + nums[end];
                if (sum < target) {
                    count += end - start;
                    start++;
                }
                else {
                    end--;
                }
            }
        }
        return count;
    }
}