# 18. 4Sum

[Link](https://leetcode.com/problems/4sum/)

Given an array `nums` of *n* integers and an integer `target`, are there elements *a*, *b*, *c*, and *d* in `nums` such that *a* + *b* + *c* + *d* = `target`? Find all unique quadruplets in the array which gives the sum of `target`.

**Note:**

The solution set must not contain duplicate quadruplets.

**Example:**

```
Given array nums = [1, 0, -1, 0, -2, 2], and target = 0.

A solution set is:
[
  [-1,  0, 0, 1],
  [-2, -1, 1, 2],
  [-2,  0, 0, 2]
]
```

**Solution**

Idea: reduce it to 2sum case, but select a base element first. Be aware the duplicate case for selection the base element, duplicate consecutive base element shall not be checked again

if k == 2 -> 2Sum

else if k > 2,  select a base element i first, and check the following elements which can make their (k-1) elements sum equal to target - nums\[i], reduce the problem to k-1 sum, and so on.

T: O(n^3)

```
class Solution {
    public List<List<Integer>> fourSum(int[] nums, int target) {
        List<List<Integer>> ret = new ArrayList<>();
        Arrays.sort(nums);
        for(int i = 0; i < nums.length-3; i++){
            if(i!=0 && nums[i] == nums[i-1]){
                continue;
            }
            
            for(int j = i+1; j < nums.length-2; j++){
                if(j != i+1 && nums[j] == nums[j-1]){
                    continue;
                }
                int k = j+1, l = nums.length-1;
                while(k < l){
                    int sum = nums[i] + nums[j] + nums[k] + nums[l];
                    if(sum > target){
                        l--;
                    }else if (sum < target){
                        k++;
                    }else {
                        ret.add(Arrays.asList(new Integer[]{nums[i], nums[j], nums[k], nums[l]}));
                        k++;
                        l--;
                        while(k < l && nums[k] == nums[k-1]) k++;
                        while(k < l && nums[l] == nums[l+1]) l--;
                    }
                }
            }
        }
        return ret;
    }
}
```

&#x20;
