_3Sum_Smaller.java

package medium;

import java.util.Arrays;

/**259. 3Sum Smaller

    Total Accepted: 13428
    Total Submissions: 33743
    Difficulty: Medium

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 _3Sum_Smaller {

    /**Basically, very similar to 3Sum, but the key is that you'll have to add result by (right-left), not just increment result by 1!*/
    public int threeSumSmaller(int[] nums, int target) {
        if(nums == null || nums.length == 0) return 0;
        int result = 0;
        Arrays.sort(nums);
        for(int i = 0; i < nums.length-2; i++){
            int left = i+1, right = nums.length-1;
            while(left < right){
                int sum = nums[i] + nums[left] + nums[right];
                if(sum < target) {
                    result += right-left;//this line is super cool!
                    left++;
                } else right--;
            }
        }
        return result;
    }

}