3Sum Leetcode problem

Description: given an array of numbers, return all unique triplets that add up to 0

Pseudocode

Sort the array
Declare an empty array result
Loop over the sorted array and for each iteration:
within the loop . . .
Check if the index is greater than 0 and if the number at index is equal to the number at the previous index - if it is, continue
declare left equal to i + 1
declare right and set to end of array
declare a while loop, and while left is less than right:
within the while loop . . .
declare sum, which adds up nums at i, left, and right.
If the sum is 0, push the unique values into result array and deduct right / increase left, handling unique cases like we did in step 1 of the loop.
if the sum is greater than 0, we should lower our sum of i, left, and right by decreasing our right pointer. Check that our newly decreased right pointer is different than the previous right pointer
If the sum is less than 0, we should increase our sum of i, left, and right by increasing our left pointer. Check that our newly increased left pointer is dfferent than our previous right pointer
Exit the while loop
Exit the for loop
return result

Code:

var threeSum = function(nums) {
    nums.sort((a, b) => a - b); 
    let result = [];
    for (let i = 0; i < nums.length; i++) {
        if (i > 0 && nums[i] === nums[i - 1]) continue;
        let left = i + 1;
        let right = nums.length - 1;
        while (left < right) {
            const sum = nums[i] + nums[left] + nums[right];
            if (sum === 0) {
                result.push([nums[i], nums[left], nums[right]]);
                right--;
                while (nums[right] === nums[right + 1]) right--;
                left++;
                while (nums[left] === nums[left - 1]) left++;
            } 

            if (sum > 0) {
                right--;
                while (nums[right] === nums[right + 1]) right--;
            } 

            if (sum < 0) {
                left++;
                while (nums[left] === nums[left - 1]) left++;
            }
        }
    }
    return result;
};