three-sum.js

/*
Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] such that i != j, i != k, and j != k, and nums[i] + nums[j] + nums[k] == 0.

Notice that the solution set must not contain duplicate triplets.

 

Example 1:

Input: nums = [-1,0,1,2,-1,-4]
Output: [[-1,-1,2],[-1,0,1]]
Explanation: 
nums[0] + nums[1] + nums[2] = (-1) + 0 + 1 = 0.
nums[1] + nums[2] + nums[4] = 0 + 1 + (-1) = 0.
nums[0] + nums[3] + nums[4] = (-1) + 2 + (-1) = 0.
The distinct triplets are [-1,0,1] and [-1,-1,2].
Notice that the order of the output and the order of the triplets does not matter.

Example 2:

Input: nums = [0,1,1]
Output: []
Explanation: The only possible triplet does not sum up to 0.

Example 3:

Input: nums = [0,0,0]
Output: [[0,0,0]]
Explanation: The only possible triplet sums up to 0.


*/
var threeSum = function (nums) {
	let result = [];
	nums = nums.sort((a, b) => a - b);

	for (let i = 0; i < nums.length; i++) {
		if (i > 0 && nums[i] == nums[i - 1]) {
			// Check for the same value
			continue;
		}

		let l = i + 1;
		let r = nums.length - 1;

		while (l < r) {
			let threeSum = nums[i] + nums[l] + nums[r];
			if (threeSum > 0) {
				r -= 1;
			} else if (threeSum < 0) {
				l += 1;
			} else {
				result.push([nums[i], nums[l], nums[r]]);
				l += 1;
				while (nums[l] === nums[l - 1] && l < r) l += 1;
			}
		}
	}

	return result;
};