LeetCode 18. 4Sum

Description

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

Given an array nums of n integers and an integer target, are there elements abc, and d in nums 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.

Example:

Given array nums = [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]
]

Explanation

For 4 sum, we need to find 3 other numbers add with nums[i] to equal to the target.

Then it becomes the 3 sum problem.

For 3 sum, we need to find 2 other numbers add with nums[j] to equal to the target – nums[i].

Then it becomes the 2 sum problem.

We sorted the input array. Then we can use two pointers techniques to solve.

Video Tutorial

Java Solution

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;
            }
            
            for (int j = i + 1; j < nums.length - 2; j++) {
                if (j != i + 1 && nums[j] == nums[j - 1]) {
                    continue;
                }
                
                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) {
                        left++;
                    } else if (sum > target) {
                        right--;
                    } else {
                        List<Integer> quadruplet = new ArrayList<>();
                        quadruplet.add(nums[i]);
                        quadruplet.add(nums[j]);
                        quadruplet.add(nums[left]);
                        quadruplet.add(nums[right]);
                        
                        result.add(quadruplet);
                        
                        left++;
                        right--;
                        
                        while (left < right && nums[left] == nums[left - 1]) {
                            left++;
                        }
                        
                        while (left < right && nums[right] == nums[right + 1]) {
                            right--;
                        }
                    }                    
                }                
            }            
        }
        
        return result;
        
        // Time Complexity: O(n^3)
        // Space Complexity: O(n^3)
    }
}

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