#15 - 3Sum

Medium • View on LeetCode • View on GitHub

Problem Description

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.

Solution

/**
 * @param {number[]} nums
 * @return {number[][]}
 */
var threeSum = function(nums) {
  const result = [];
  nums.sort((a, b) => a - b);

  for (let i = 0; i < nums.length - 2; i++) {
    if (i > 0 && nums[i] === nums[i - 1]) continue;
    let j = i + 1;
    let k = nums.length - 1;
    while (j < k) {
      const sum = nums[i] + nums[j] + nums[k];
      if (!sum) {
        result.push([nums[i], nums[j], nums[k]]);
        j++;
        k--;
        while (j < k && nums[j] === nums[j - 1]) {
          j++;
        }
        while (j < k && nums[k] === nums[k + 1]) {
          k--;
        }
      } else {
        sum < 0 ? j++ : k--;
      }
    }
  }

  return result;
};