<p>Given an array <code>nums</code> of <code>n</code> integers, return <em>an array of all the <strong>unique</strong> quadruplets</em> <code>[nums[a], nums[b], nums[c], nums[d]]</code> such that:</p>

<ul>
	<li><code>0 &lt;= a, b, c, d&nbsp;&lt; n</code></li>
	<li><code>a</code>, <code>b</code>, <code>c</code>, and <code>d</code> are <strong>distinct</strong>.</li>
	<li><code>nums[a] + nums[b] + nums[c] + nums[d] == target</code></li>
</ul>

<p>You may return the answer in <strong>any order</strong>.</p>

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

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

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

<pre>
<strong>Input:</strong> nums = [2,2,2,2,2], target = 8
<strong>Output:</strong> [[2,2,2,2]]
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 200</code></li>
	<li><code>-10<sup>9</sup> &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
	<li><code>-10<sup>9</sup> &lt;= target &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution:
    def fourSum(self, nums: List[int], target: int) -> List[List[int]]:
        nums.sort()  # Sort the list to make it easier to skip duplicates and use two pointers
        res = []
        n = len(nums)
        
        # Iterate over the first two numbers in the quadruplet
        for i in range(n - 3):
            if i > 0 and nums[i] == nums[i - 1]:  # Skip duplicates for the first number
                continue
            for j in range(i + 1, n - 2):
                if j > i + 1 and nums[j] == nums[j - 1]:  # Skip duplicates for the second number
                    continue

                # Now, use two pointers to find the remaining two numbers
                left, right = j + 1, n - 1
                while left < right:
                    sum_ = nums[i] + nums[j] + nums[left] + nums[right]
                    if sum_ == target:
                        res.append([nums[i], nums[j], nums[left], nums[right]])

                        # Move the left and right pointers to avoid duplicates
                        while left < right and nums[left] == nums[left + 1]:
                            left += 1
                        while left < right and nums[right] == nums[right - 1]:
                            right -= 1

                        left += 1
                        right -= 1
                    elif sum_ < target:
                        left += 1
                    else:
                        right -= 1
                        
        return res

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