Problem Description
Given an array nums of n integers and an integer target, find all unique quadruplets in the array which gives the sum of target. Return the answer in any order.
Examples
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]]
Solutions
Python Solution
def fourSum(nums: List[int], target: int) -> List[List[int]]:
nums.sort()
n = len(nums)
result = []
for i in range(n - 3):
if i > 0 and nums[i] == nums[i - 1]:
continue
for j in range(i + 1, n - 2):
if j > i + 1 and nums[j] == nums[j - 1]:
continue
left = j + 1
right = n - 1
while left < right:
curr_sum = nums[i] + nums[j] + nums[left] + nums[right]
if curr_sum == target:
result.append([nums[i], nums[j], nums[left], nums[right]])
while left < right and nums[left] == nums[left + 1]:
left += 1
while left < right and nums[right] == nums[right - 1]:
right -= 1
left += 1
right -= 1
elif curr_sum < target:
left += 1
else:
right -= 1
return result
Java Solution
class Solution {
public List> fourSum(int[] nums, int target) {
Arrays.sort(nums);
List> result = new ArrayList<>();
int n = nums.length;
for (int i = 0; i < n - 3; i++) {
if (i > 0 && nums[i] == nums[i - 1]) continue;
for (int j = i + 1; j < n - 2; j++) {
if (j > i + 1 && nums[j] == nums[j - 1]) continue;
int left = j + 1;
int right = n - 1;
while (left < right) {
long sum = (long)nums[i] + nums[j] + nums[left] + nums[right];
if (sum == target) {
result.add(Arrays.asList(nums[i], nums[j], nums[left], nums[right]));
while (left < right && nums[left] == nums[left + 1]) left++;
while (left < right && nums[right] == nums[right - 1]) right--;
left++;
right--;
} else if (sum < target) {
left++;
} else {
right--;
}
}
}
}
return result;
}
}
C++ Solution
class Solution {
public:
vector> fourSum(vector& nums, int target) {
sort(nums.begin(), nums.end());
vector> result;
int n = nums.size();
for (int i = 0; i < n - 3; i++) {
if (i > 0 && nums[i] == nums[i - 1]) continue;
for (int j = i + 1; j < n - 2; j++) {
if (j > i + 1 && nums[j] == nums[j - 1]) continue;
int left = j + 1;
int right = n - 1;
while (left < right) {
long long sum = (long long)nums[i] + nums[j] + nums[left] + nums[right];
if (sum == target) {
result.push_back({nums[i], nums[j], nums[left], nums[right]});
while (left < right && nums[left] == nums[left + 1]) left++;
while (left < right && nums[right] == nums[right - 1]) right--;
left++;
right--;
} else if (sum < target) {
left++;
} else {
right--;
}
}
}
}
return result;
}
};
JavaScript Solution
/**
* @param {number[]} nums
* @param {number} target
* @return {number[][]}
*/
var fourSum = function(nums, target) {
nums.sort((a, b) => a - b);
const result = [];
const n = nums.length;
for (let i = 0; i < n - 3; i++) {
if (i > 0 && nums[i] === nums[i - 1]) continue;
for (let j = i + 1; j < n - 2; j++) {
if (j > i + 1 && nums[j] === nums[j - 1]) continue;
let left = j + 1;
let right = n - 1;
while (left < right) {
const sum = nums[i] + nums[j] + nums[left] + nums[right];
if (sum === target) {
result.push([nums[i], nums[j], nums[left], nums[right]]);
while (left < right && nums[left] === nums[left + 1]) left++;
while (left < right && nums[right] === nums[right - 1]) right--;
left++;
right--;
} else if (sum < target) {
left++;
} else {
right--;
}
}
}
}
return result;
};
C# Solution
public class Solution {
public IList> FourSum(int[] nums, int target) {
Array.Sort(nums);
var result = new List>();
int n = nums.Length;
for (int i = 0; i < n - 3; i++) {
if (i > 0 && nums[i] == nums[i - 1]) continue;
for (int j = i + 1; j < n - 2; j++) {
if (j > i + 1 && nums[j] == nums[j - 1]) continue;
int left = j + 1;
int right = n - 1;
while (left < right) {
long sum = (long)nums[i] + nums[j] + nums[left] + nums[right];
if (sum == target) {
result.Add(new List { nums[i], nums[j], nums[left], nums[right] });
while (left < right && nums[left] == nums[left + 1]) left++;
while (left < right && nums[right] == nums[right - 1]) right--;
left++;
right--;
} else if (sum < target) {
left++;
} else {
right--;
}
}
}
}
return result;
}
}
Complexity Analysis
- Time Complexity: O(n³) where n is the length of the input array
- Space Complexity: O(1) excluding the space required for output
Solution Explanation
This solution extends the 3Sum approach using sorting and two pointers. Here's how it works:
- Sort the array first to handle duplicates and use two-pointer technique
- Use two nested loops to fix first two numbers
- For the remaining two numbers:
- Use two pointers to find pairs that sum to target - (first + second)
- Skip duplicates to avoid duplicate quadruplets
- Handle integer overflow using long/long long types
Key points:
- Sorting helps avoid duplicates and makes the two-pointer approach efficient
- We skip duplicate numbers to avoid duplicate quadruplets
- Integer overflow is handled using appropriate data types
- The solution handles edge cases properly