DSA using JavaScript

Time Complexity: O(n^3)
Space Complexity: O(1)

Two Pointers

function threeSumBruteForce(nums) {
    let result = [];
    let n = nums.length;
    for (let i = 0; i < n - 2; i++) {
        for (let j = i + 1; j < n - 1; j++) {
            for (let k = j + 1; k < n; k++) {
                if (nums[i] + nums[j] + nums[k] === 0) {
                    let triplet = [nums[i], nums[j], nums[k]].sort((a, b) => a - b);
                    if (!result.some((t) => t[0] === triplet[0] && t[1] === triplet[1] && t[2] === triplet[2])) {
                        result.push(triplet);
                    }
                }
            }
        }
    }
    return result;
}

const nums = [-1, 0, 1, 2, -1, -4];
console.log(threeSumBruteForce(nums));

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

    for (let i = 0; i < nums.length - 2; i++) {
        if (i > 0 && nums[i] === nums[i - 1]) continue;  // Skip duplicates
        let left = i + 1, right = nums.length - 1;
        while (left < right) {
            let sum = nums[i] + nums[left] + nums[right];
            if (sum === 0) {
                result.push([nums[i], nums[left], nums[right]]);
                while (left < right && nums[left] === nums[left + 1]) left++;  // Skip duplicates
                while (left < right && nums[right] === nums[right - 1]) right--;  // Skip duplicates
                left++;
                right--;
            } else if (sum < 0) {
                left++;
            } else {
                right--;
            }
        }
    }
    return result;
}

const nums = [-1, 0, 1, 2, -1, -4];
console.log(threeSum(nums));