Three Sum

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.

/**
 * @param {number[]} nums
 * @return {number[][]}
 */
var threeSum = function(nums) {
    nums.sort((a, b) => a - b)

    const res = [];
    const n = nums.length;

    for (let i = 0; i < n; i++) {
        if (i > 0 && nums[i] === nums[i - 1]) continue;

        let left = i + 1;
        let right = n - 1;

        while (left < right) {
            const sum = nums[i] + nums[left] + nums[right];

            if (sum === 0) {
                res.push([nums[i], nums[left], nums[right]]);

                while (left < right && nums[left] === nums[left + 1]) left++;
                while (left < right && nums[right] === nums[right - 1]) right--;

                left++;
                right--;
            } else if (sum < 0) {
                left++;
            } else {
                right--;
            }
        }

    }
    return res;
};