package com.fishercoder.solutions;

import java.util.Arrays;

/**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? */
public class _259 {

    /**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;
            int right = nums.length - 1;
            while (left < right) {
                int sum = nums[i] + nums[left] + nums[right];
                if (sum < target) {
                    result += right - left;//this line is key!
                    left++;
                } else {
                    right--;
                }
            }
        }
        return result;
    }
}

package com.fishercoder.solutions;

import java.util.Arrays;

public class _259 {

    public static class Solution1 {
        /**
         * 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;
                int right = nums.length - 1;
                while (left < right) {
                    int sum = nums[i] + nums[left] + nums[right];
                    if (sum < target) {
                        result += right - left;//this line is key!
                        left++;
                    } else {
                        right--;
                    }
                }
            }
            return result;
        }
    }
}

refactor 259