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?

why sort array is ok?

because sort array won't change the fact that a good triplet still stands

Fix first number, then use two pointers, notice that when you find a i(fixed),j,k triplet, which means all combine between k-j is valid triplet.

public class Solution {
    public int threeSumSmaller(int[] nums, int target) {
        Arrays.sort(nums);// even though sorting scrambles the indice. however
                          // it doesn't change the number of good triplet.
        int count = 0;
        for(int i=0; i<nums.length-2; i++){
            int j = i+1, k = nums.length-1;
            while(j < k){
                if((nums[i] + nums[j] + nums[k]) >= target) k--;
                else{
                    count += k -j; // all the combine between k-j are good triplet.
                    j++;

                }
            }
        }

        return count;
    }
}