Description

Given an array nums of n integers, return an array of all the unique quadruplets [nums[a], nums[b], nums[c], nums[d]] such that:

  • 0 <= a, b, c, d < n
  • a, b, c, and d are distinct.
  • nums[a] + nums[b] + nums[c] + nums[d] == target

You may return the answer in any order.

1
2
3
4
5
6
7
8
Example 1:

Input: nums = [1,0,-1,0,-2,2], target = 0
Output: [[-2,-1,1,2],[-2,0,0,2],[-1,0,0,1]]
Example 2:

Input: nums = [2,2,2,2,2], target = 8
Output: [[2,2,2,2]]
1
2
3
4
5
Constraints:

1 <= nums.length <= 200
-109 <= nums[i] <= 109
-109 <= target <= 109

Approach

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
class Solution:
    def fourSum(self, nums: List[int], target: int) -> List[List[int]]:
        nums.sort()
        n = len(nums)
        res = []

        for i in range(n - 3):
            if i and nums[i] == nums[i-1]:
                continue
            
            for j in range(i + 1, n - 2):
                if j > n + 1 and nums[j] == nums[j - 1]:
                    continue
                
                l = j + 1
                r = n - 1
                while l < r:
                    s = nums[i] + nums[j] + nums[l] + nums[r]
                    if s < target:
                        l += 1
                    elif s > target:
                        r -= 1
                    else:
                        if [nums[i], nums[j], nums[l], nums[r]] not in res:
                            res.append([nums[i], nums[j], nums[l], nums[r]])
                        l += 1
                        r -= 1
                        while l < r and nums[l] == nums[l - 1]: l += 1
                        while l < r and nums[r] == nums[r + 1]: r -= 1
        return res