[LeetCode] 4Sum

4Sum

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

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.

    For example, given array S = {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: 

Sorting helps us to deal with both constraints. Keep 4 pointers. We move the 1st pointer from left to right, and each time solve a 3SUM problem using the other 3 pointers in the subarray with index ranging from i + 1 to n.
Complexity: O(n^3). Since 3SUM can be solved in O(n^2) time. 
Further reading: [LeetCode] 3SUM

public class Solution {
    private ArrayList<Integer> quadruplet(
        int[] nums, 
        int i, 
        int j, 
        int k, 
        int l
    ) {
        ArrayList<Integer> li = new ArrayList<Integer>();
        li.add(nums[i]);
        li.add(nums[j]);
        li.add(nums[k]);
        li.add(nums[l]);
        return li;
    }
    
    public List<List<Integer>> fourSum(int[] nums, int target) {
        Arrays.sort(nums);
        ArrayList<List<Integer>> list = 
            new ArrayList<List<Integer>>();
        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;
                int l = nums.length - 1;
                while (k < l) {
                    int sum = nums[i] + nums[j] + nums[k] + nums[l];
                    if (sum == target) {
                        list.add(quadruplet(nums, i, j, k, l));
                        k++;
                        l--;
                        while (k < l && nums[k] == nums[k - 1]) {
                            k++;
                        }
                        while (k < l && nums[l] == nums[l + 1]) {
                            l--;
                        }
                    } else if (sum < target) {
                        k++;
                    } else {
                        l--;
                    }
                }
            }
        }
        return list;
    }
}