I'm trying to solve Three Sum (find all triplets that add to 0 within an array avoiding duplicate cases) but currently running into a bug I can't seem to find.

When the input is [-1,0,1,2,-1,-4], it works fine.

With this input however, [-1,0,1,2,-1,-4,-2,-3,3,0,4] I'm getting this array as output:

[[-1,-1,2],[-1,0,1],[-2,0,2],[-3,0,3],[-3,1,2],[-4,0,4],[-4,1,3]].

The correct output should be

[[-4,0,4],[-4,1,3],[-3,-1,4],[-3,0,3],[-3,1,2],[-2,-1,3],[-2,0,2],[-1,-1,2],[-1,0,1]]

So for some reasons my solution is omitting the triplets [-3,-1,4] and [-2,-1,3].

var threeSum = function (nums) {
    const sorted = nums.sort()
    const output = []
    for (let i = 0; i < sorted.length - 2; i++)
        if (i === 0 || (i > 0 && sorted[i] !== sorted[i - 1])) {
            let lower = i + 1
            let higher = sorted.length - 1

            while (lower < higher) {
                const currentSum = sorted[i] + sorted[lower] + sorted[higher];
                if (currentSum === 0) {
                    output.push([sorted[i], sorted[lower], sorted[higher]])
                    while (sorted[lower] === sorted[lower + 1]) lower++
                    while (sorted[higher] === sorted[higher - 1]) higher--
                    lower++
                    higher--
                }

                else if (currentSum < 0) {
                    lower++
                } else {
                    higher--
                }
            }
        }
    }
    return output
 };