18. 4Sum

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.

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]]

Constraints:

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

Approach 01: Sorting and Two Pointers

class Solution {
public:
    vector> fourSum(vector& nums, int target) {
        sort(nums.begin(), nums.end());
        vector> res;
        int n = nums.size();
        for (int i = 0; i < n - 3; i++) {
            if (i > 0 && nums[i] == nums[i-1]) continue;
            for (int j = i + 1; j < n - 2; j++) {
                if (j > i + 1 && nums[j] == nums[j-1]) continue;
                int left = j + 1, right = n - 1;
                while (left < right) {
                    long long sum = (long long)nums[i] + nums[j] + nums[left] + nums[right];
                    if (sum == target) {
                        res.push_back({nums[i], nums[j], nums[left], nums[right]});
                        while (left < right && nums[left] == nums[left+1]) left++;
                        while (left < right && nums[right] == nums[right-1]) right--;
                        left++;
                        right--;
                    } else if (sum < target) left++;
                    else right--;
                }
            }
        }
        return res;
    }
};

class Solution:
    def fourSum(self, nums: List[int], target: int) -> List[List[int]]:
        nums.sort()
        res = []
        n = len(nums)
        for i in range(n - 3):
            if i > 0 and nums[i] == nums[i-1]:
                continue
            for j in range(i + 1, n - 2):
                if j > i + 1 and nums[j] == nums[j-1]:
                    continue
                left, right = j + 1, n - 1
                while left < right:
                    s = nums[i] + nums[j] + nums[left] + nums[right]
                    if s == target:
                        res.append([nums[i], nums[j], nums[left], nums[right]])
                        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 s < target:
                        left += 1
                    else:
                        right -= 1
        return res

class Solution {
    public List> fourSum(int[] nums, int target) {
        Arrays.sort(nums);
        List> res = new ArrayList<>();
        int n = nums.length;
        for (int i = 0; i < n - 3; i++) {
            if (i > 0 && nums[i] == nums[i - 1]) continue;
            for (int j = i + 1; j < n - 2; j++) {
                if (j > i + 1 && nums[j] == nums[j - 1]) continue;
                int left = j + 1, right = n - 1;
                while (left < right) {
                    long sum = (long)nums[i] + nums[j] + nums[left] + nums[right];
                    if (sum == target) {
                        res.add(Arrays.asList(nums[i], nums[j], nums[left], nums[right]));
                        while (left < right && nums[left] == nums[left + 1]) left++;
                        while (left < right && nums[right] == nums[right - 1]) right--;
                        left++;
                        right--;
                    } else if (sum < target) left++;
                    else right--;
                }
            }
        }
        return res;
    }
}