LeetCode 18. 4Sum

Question

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

Source: https://leetcode.com/problems/4sum/

Solution

Time Complexity O(n^3). In fact, this is a DFS problem. Remember to draw the DFS tree.

// two pointer search, take advantage of the orderliness of nums
// @pre: nums is sorted
private List<List<Integer>> twoSum(int[] nums, int target, int start) {
    List<List<Integer>> result = new ArrayList<>();
    int left = start;
    int right = nums.length - 1;
    while (left < right) {
        // a better way to skip duplicates
        // can perform skipping even if sum != 0
        if (left > start && nums[left] == nums[left - 1]) {
            left++;
            continue;
        }
        if (right < nums.length - 1 && nums[right] == nums[right + 1]) {
            right--;
            continue;
        }

        int sum = nums[left] + nums[right];
        if (sum == target) {
            result.add(new ArrayList<>(Arrays.asList(nums[left], nums[right])));
            left++;
            right--;
        } else if (sum > target) {
            right--;
        } else {
            left++;
        }
    }
    return result;
}

// dfs, depth is k
// @pre: nums is sorted
// return the k-sum combinations with given k
private List<List<Integer>> kSum(int[] nums, int k, int target, int start) {
    List<List<Integer>> result = new ArrayList<>();
    if (start >= nums.length) {
        return result;
    }
    // pruning
    double avg = ((double) target) / k;
    if (nums[start] > avg || nums[nums.length - 1] < avg) {
        return result;
    }
    if (k == 2) {
        return twoSum(nums, target, start);
    }
    for (int i = start; i < nums.length; i++) {
        // skip duplicates
        if (i != start && nums[i] == nums[i - 1]) {
            continue;
        }
        int n = nums[i];
        for (List<Integer> subset : kSum(nums, k - 1, target - n, i + 1)) {
            List<Integer> currLayerComb = new ArrayList<>();
            currLayerComb.add(n);
            currLayerComb.addAll(subset);
            result.add(currLayerComb);
        }
    }
    return result;
}

public List<List<Integer>> fourSum(int[] nums, int target) {
    Arrays.sort(nums);
    return kSum(nums, 4, target, 0);
}