Problem

Given an array S of n integers, find all unique quadruplets (a, b, c, d) in S such that a + b + c + d = target. The quadruplets must be in non-descending order, and the solution set must not contain duplicate quadruplets.

Note:

  • Elements in a quadruplet (a,b,c,d) must be in non-descending order. (ie, a ≤ b ≤ c ≤ d)

The solution set must not contain duplicate quadruplets.

Examples

Example 1:

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Input: nums = [1, 0, -1, 0, -2, 2]
Output: [[-2, -1, 1, 2], [-2, 0, 0, 2], [-1, 0, 0, 1]]
Explanation: The pair (6, 7) gives the highest product of 42.

Also make sure that the solution set is lexicographically sorted.

Solution[i] < Solution[j] iff Solution[i][0] < Solution[j][0] OR (Solution[i][0] == Solution[j][0] AND ... Solution[i][k] < Solution[j][k])

Solution

Method 1 - Sorting and Two pointer technique

Here is the approach:

  1. Sort the Array: Start by sorting the array. This will help ensure the quadruplets are generated in non-descending order.
  2. Nested Loop for Quadruplets: Use four nested loops to find quadruplets that sum to the target:
    • The outer two loops fix the first two elements of the quadruplet.
    • The inner two loops use a two-pointer approach to find the remaining two elements that sum to target - (a + b).
  3. Check for Duplicates: Use sets or condition checks to avoid duplicate quadruplets.

Code

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public class Solution {
    public List<List<Integer>> fourSum(int[] nums, int target) {
        List<List<Integer>> result = new ArrayList<>();
        Arrays.sort(nums);

        for (int i = 0; i < nums.length - 3; i++) {
            if (i > 0 && nums[i] == nums[i - 1]) continue;  // Avoid duplicates

            for (int j = i + 1; j < nums.length - 2; j++) {
                if (j > i + 1 && nums[j] == nums[j - 1]) continue;  // Avoid duplicates

                int left = j + 1;
                int right = nums.length - 1;

                while (left < right) {
                    int sum = nums[i] + nums[j] + nums[left] + nums[right];
                    if (sum == target) {
                        result.add(Arrays.asList(nums[i], nums[j], nums[left], nums[right]));
                        left++;
                        right--;

                        while (left < right && nums[left] == nums[left - 1]) left++;  // Avoid duplicates
                        while (left < right && nums[right] == nums[right + 1]) right--;  // Avoid duplicates

                    } else if (sum < target) {
                        left++;
                    } else {
                        right--;
                    }
                }
            }
        }

        return result;
    }

    public static void main(String[] args) {
        Solution sol = new Solution();
        int[] nums = {1, 0, -1, 0, -2, 2};
        int target = 0;
        List<List<Integer>> quadruplets = sol.fourSum(nums, target);
        for (List<Integer> quadruplet : quadruplets) {
            System.out.println(quadruplet);
        }
        // Expected output:
        // (-2, -1, 1, 2)
        // (-2, 0, 0, 2)
        // (-1, 0, 0, 1)
    }
}
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class Solution:
    def fourSum(self, nums: List[int], target: int) -> List[List[int]]:
        nums.sort()
        result = []

        for i in range(len(nums) - 3):
            if i > 0 and nums[i] == nums[i - 1]:
                continue  # Avoid duplicates

            for j in range(i + 1, len(nums) - 2):
                if j > i + 1 and nums[j] == nums[j - 1]:
                    continue  # Avoid duplicates

                left, right = j + 1, len(nums) - 1

                while left < right:
                    four_sum = nums[i] + nums[j] + nums[left] + nums[right]
                    if four_sum == target:
                        result.append([nums[i], nums[j], nums[left], nums[right]])
                        left += 1
                        right -= 1

                        while left < right and nums[left] == nums[left - 1]:
                            left += 1  # Avoid duplicates
                        while left < right and nums[right] == nums[right + 1]:
                            right -= 1  # Avoid duplicates

                    elif four_sum < target:
                        left += 1
                    else:
                        right -= 1

        return result

# Example usage
sol = Solution()
nums = [1, 0, -1, 0, -2, 2]
target = 0
quadruplets = sol.fourSum(nums, target)
for quadruplet in quadruplets:
 print(quadruplet)
# Expected output:
# [-2, -1, 1, 2]
# [-2, 0, 0, 2]
# [-1, 0, 0, 1]

Complexity

  • ⏰ Time complexity: O(n^3), where n is the number of elements in the array. This is because of the three nested loops iterating through the sorted array.
  • 🧺 Space complexity: O(k), where k is the number of unique quadruplets.