4Sum

Problem

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: 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

O(n^3)

public class Solution {
    public List<List<Integer>> fourSum(int[] nums, int target) {
        if (nums == null || nums.length < 4) return new ArrayList<>();
        List<List<Integer>> res = new ArrayList<>();
        int len = nums.length;
        Arrays.sort(nums);
        for (int i = 0; i < len - 3; i++) {
            if (nums[i] + nums[i + 1] + nums[i + 2] + nums[i + 3] > target) break; //check min
            if (nums[i] + nums[len - 1] + nums[len - 2] + nums[len - 3] < target) continue; //check current max
            if (i != 0 && nums[i] == nums[i - 1]) continue; //check duplicates of bound i

            for (int j = i + 1; j < len - 2; j++) {
                if (j != i + 1 && nums[j] == nums[j - 1]) continue; //check duplicates of bound j
                int low = j + 1, high = len - 1;

                while (low < high) {
                    if (nums[i] + nums[j] + nums[low] + nums[high] < target) low++;
                    else if (nums[i] + nums[j] + nums[low] + nums[high] > target) high--;
                    else {
                        res.add(Arrays.asList(nums[i], nums[j], nums[low], nums[high]));
                        while (low < high && nums[low] == nums[low + 1]) low++; //skip duplicates of binary search 
                        while (low < high && nums[high] == nums[high - 1]) high--; //skip duplicates of binary search 
                        low++;
                        high--;
                    }
                }
            }
        }
        return res;
    }
}

Aanalysis

Similar idea from 3Sum
The only difference is that we have one more loop because the problem requires 4 numbers
Notice that how we avoid duplicates in the second for loop, just checking whether j is the first iteration and duplication
We don't need the first two if statements because we already did them in the i loop and j starts from i + 1