threeSum.js

/**
 * 3Sum -> Two pointers pattern
Medium

Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] where nums[i] + nums[j] + nums[k] == 0, and the indices i, j and k are all distinct.

The output should not contain any duplicate triplets. You may return the output and the triplets in any order.
 * 
 */

/**
 * Psuedo Code
 * Sort the array
 * Initialize an empty array for result
 * check for duplicates and skip
 * set left and right positions
 * while(left<right)
 * check for sum if it's equal to 0
 * check for duplicates
 * return
 *
 */

function threeSum(nums) {
  nums.sort((a, b) => a - b); //sort an array of numbers in ascending order(smallest to largest)

  let result = [];

  for (let i = 0; i < nums.length; i++) {
    //Skip duplicates i
    if (i > 0 && nums[i] === nums[i - 1]) continue;

    let left = i + 1;
    let 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]]);
        left++;
        right--;

        //Skip duplicates left & right
        while (left < right && nums[left] === nums[left - 1]) left++;
        while (left < right && nums[right] === nums[right + 1]) right--;
      } else if (sum < 0) {
        left++;
      } else {
        right--;
      }
    }
  }
  return result;
}

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