<p>Given an integer array nums, return all the triplets <code>[nums[i], nums[j], nums[k]]</code> such that <code>i != j</code>, <code>i != k</code>, and <code>j != k</code>, and <code>nums[i] + nums[j] + nums[k] == 0</code>.</p>

<p>Notice that the solution set must not contain duplicate triplets.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [-1,0,1,2,-1,-4]
<strong>Output:</strong> [[-1,-1,2],[-1,0,1]]
<strong>Explanation:</strong> 
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.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [0,1,1]
<strong>Output:</strong> []
<strong>Explanation:</strong> The only possible triplet does not sum up to 0.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [0,0,0]
<strong>Output:</strong> [[0,0,0]]
<strong>Explanation:</strong> The only possible triplet sums up to 0.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>3 &lt;= nums.length &lt;= 3000</code></li>
	<li><code>-10<sup>5</sup> &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
</ul>

class Solution:

    def threeSum2(self, nums: List[int]) -> List[List[int]]:
        n = len(nums)
        nums.sort()

        def twosum(i):
            low = i + 1
            high = len(nums) - 1
            while left < right:
                sum = nums[i] + nums[left] + nums[right]
                if sum == target:
                    result.append([nums[left], nums[right]])
                    left += 1
                    high -=1
                    # to avoid duplicates
                    while left < high and nums[left] == nums[left-1]:
                        left += 1 
                elif sum > target:
                    right -= 1
                else:
                    left += 1

        result = [] 
        for i in range(n - 2):
            if nums[i] > 0:
                break

            # To avoid duplicates
            if i != 0 and nums[i-1] == nums[i]:
                continue
            twosum(i)
        
        return result
            

    # Hashset
    def threeSum(self, nums: List[int]) -> List[List[int]]:
        n = len(nums)
        nums.sort()
        result = []
        print(nums)

        for i in range(n):
            # As the numbers are sorted, if current value is greater than 0,
            # the remaining numbers cannot sum to 0
            if nums[i] > 0:
                break
            
            # print(i, nums[i-1], nums[i])
            # Toi avoid duplicate triplets
            if i != 0 and nums[i-1] == nums[i]:
                continue

            tracker = set()
            # target = 0 - nums[i]
            j = i + 1
            while j < n:
                # num = nums[j]
                complement = 0 - nums[i] - nums[j]
                if complement in tracker:
                    result.append([nums[i], nums[j], complement])

                    # If there are duplicates in this subset
                    while j + 1 < n and nums[j] == nums[j + 1]:
                        j += 1
                tracker.add(nums[j])
                j += 1

        return result


    # Doesn't work
    def threeSum2(self, nums: List[int]) -> List[List[int]]:
        n = len(nums)
        nums.sort()
        result = set()

        for i in range(n):
            # As the numbers are sorted, if current value is greater than 0,
            # the remaining numbers cannot sum to 0
            if nums[i] > 0:
                break
            
            # print(i, nums[i-1], nums[i])
            if i != 0 and nums[i-1] == nums[i]:
                continue

            tracker = set()
            target = 0 - nums[i]
            for j in range(1, n):
                num = nums[j]
                if (target - num) in tracker:
                    new_res = sorted((nums[i], nums[j], target - num))
                    result.add(new_res)
                tracker.add(num)

        return result

        
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