### 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:

Time Complexity : O(n)

Space Complexity: O(n)

```
function threeSum(nums) {
// If less than 3 elements then we can't have 3 numbers that add to 0
if(nums.length < 3) return []
const output = []
// Sort the array in descending order. Must add order function to sort. If not we will not get the right order. Sort converts everything to a string and sorts the array by charCode. Adding the order function to sort guarantees we will get the array in the proper descending order.
nums.sort((a,b) => a - b)
for(let i = 0; i < nums.length - 2;i++) {
// we don't want repeats, so skip numbers we've already seen
if (i > 0 && nums[i] === nums[i - 1]) continue
let left = i+1
let right = nums.length-1
// Current number at i will be added to the current sum. We will move a left and a right pointer in the subarray of elements to the right of i to try and get a sum that will equal 0
while (left < right) {
// Get the current sum with with number at i and numbers at the left and right pointers
const sum = nums[i] + nums[right] + nums[left]
// If we get 0 then we add all the numbers to output and move our left and right pointers to look for more numbers that will add to 0 with the current number at i
if(sum===0) {
output.push([nums[i], nums[left], nums[right]])
// We will move the pointers until we find a number that is not equal to each pointers current number
while(nums[left]===nums[left+1]) left++
while(nums[right]===nums[right+1]) right--
left++
right--
} else if (sum > 0) {
// If the sum is greater than 0 that means we need smaller numbers to get 0 so we move the right pointer to the left
right--
} else {
// If the sum is less than 0 that means we need higher numbers to get 0 so we move the left pointer to the right
left++
}
}
}
return output
};
```

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