[LeetCode] 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?

Thought process:

Although the question requires that 0 <= i < j < k < n, we can still sort the array, because as long as we can find a triplet that satisfy nums[i] + nums[j] + nums[k] < target, we can always swap i, j, and k to make them in the right order.

So the idea is to sort the array and use two pointers.

Solution:

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

Time complexity: O(n^2).