#2 Leetcode ( Q: (18) 4Sum )

Şerifhan Işıklı
2 min readMar 16, 2020

--

LeetCode 4Sum (Medium) Question.

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

Given an array nums of n integers and an integer target, are there elements a, b, c, 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]
]

Time Complexity: O(n³)
Space Complexity: O(n³)

Description

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.

Solution — Java:

class Solution {
public List<List<Integer>> fourSum(int[] nums, int target) {
List<List<Integer>> answer= 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> tetrad= new ArrayList<>();
tetrad.add(nums[i]);
tetrad.add(nums[j]);
tetrad.add(nums[left]);
tetrad.add(nums[right]);

answer.add(tetrad);

left++;
right--;

while (left < right && nums[left] == nums[left - 1]) {
left++;
}

while (left < right && nums[right] == nums[right + 1]) {
right--;
}
}
}
}
}

return answer;
// Serifhan Isıklı
}
}

Goal : 2 / 1000 ( LeetCode Question )

--

--

Şerifhan Işıklı
Şerifhan Işıklı

Written by Şerifhan Işıklı

Senior Software Engineer @Dogus Teknoloji.