<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]]:
        # first, we sort like usual to allow us to eleminate duplicates
        nums.sort()
        # stores final result
        res = []
        # stores current candidate
        candidate = []

        # generic k-sum solution
        def kSum(k, start, target):
            # The purpose of this is to narrow it down to only
            # k=2 so we can use the 2sum approach. We build a
            # candidate using recursion until k == 2, then run
            # the base case.
            if k != 2:
                for i in range(start, len(nums)):
                    if i != start and nums[i] == nums[i-1]:
                        continue

                    candidate.append(nums[i])
                    kSum(k - 1, i + 1, target - nums[i])
                    candidate.pop()
                
                return

            # base case, here we just use the 2sum approach
            l = start
            r = len(nums)-1

            while l < r:
                v = nums[l] + nums[r]
                if v < target:
                    l += 1
                elif v > target:
                    r -= 1
                else:
                    res.append(candidate + [nums[l], nums[r]])

                    l += 1
                    while l < r and nums[l] == nums[l-1]:
                        l += 1

        # run algorithm to populate res
        kSum(4, 0, target)
        return res
16800