3 Sum / 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.

const threeSum = (arr) => {
  let nums = [...arr].sort();
  const res = []; 

  for (let i = 0; i < nums.length - 2; i++) {
    // We can do - 2 cause our other 2 pointers will take care of the last 2 nums
    // Initialize 2 pointers for innerloop which will give sum 
    let l = i + 1,
      r = nums.length - 1;

    if (nums[i] === nums[i - 1]) continue; // check for dupes on the outter loop, skip if nums[i] is a num we already saw

    while (l < r) {
      let total = nums[i] + nums[l] + nums[r]; // Add up all our current pointer values

      if (total === 0) {
        res.push([nums[i], nums[l], nums[r]]); // Add if we found a solution

        while (nums[l] === nums[l + 1]) l++; // This skips dupes in the nums array
        while (nums[r] === nums[r - 1]) r--; // This skips dupes in the nums array

        // We need to do one extra move after skipping the dupes or else we will still be on a dupe
        l++;
        r--;
      } else if (total > 0) {
        // If its greater than 0, then we can move the r pointer down
        r--;
      } else {
        // its less than 0 and we need to move the left pointer up
        l++;
      }
    }
  }

  return res;
};