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In this post, you will find the solution for the 4Sum in C++, Java & Python-LeetCode problem. We are providing the correct and tested solutions to coding problems present on LeetCode. If you are not able to solve any problem, then you can take help from our Blog/website.
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Link for the Problem – 4Sum– LeetCode Problem
4Sum– LeetCode Problem
Problem:
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 < na,b,c, anddare 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
4Sum– LeetCode Solutions
class Solution {
public:
vector<vector<int>> fourSum(vector<int>& nums, int target) {
vector<vector<int>> ans;
vector<int> path;
sort(begin(nums), end(nums));
nSum(nums, 4, target, 0, nums.size() - 1, path, ans);
return ans;
}
private:
// in [l, r], find n numbers add up to the target
void nSum(const vector<int>& nums, int n, int target, int l, int r,
vector<int>& path, vector<vector<int>>& ans) {
if (r - l + 1 < n || target < nums[l] * n || target > nums[r] * n)
return;
if (n == 2) {
// very similar to the sub procedure in 15. 3Sum
while (l < r) {
const int sum = nums[l] + nums[r];
if (sum == target) {
path.push_back(nums[l]);
path.push_back(nums[r]);
ans.push_back(path);
path.pop_back();
path.pop_back();
++l;
--r;
while (l < r && nums[l] == nums[l - 1])
++l;
while (l < r && nums[r] == nums[r + 1])
--r;
} else if (sum < target) {
++l;
} else {
--r;
}
}
return;
}
for (int i = l; i <= r; ++i) {
if (i > l && nums[i] == nums[i - 1])
continue;
path.push_back(nums[i]);
nSum(nums, n - 1, target - nums[i], i + 1, r, path, ans);
path.pop_back();
}
}
};
class Solution {
public List<List<Integer>> fourSum(int[] nums, int target) {
List<List<Integer>> ans = new ArrayList<>();
Arrays.sort(nums);
nSum(nums, 4, target, 0, nums.length - 1, new ArrayList<>(), ans);
return ans;
}
// in [l, r], find n numbers add up to the target
private void nSum(int[] nums, int n, int target, int l, int r, List<Integer> path,
List<List<Integer>> ans) {
if (r - l + 1 < n || target < nums[l] * n || target > nums[r] * n)
return;
if (n == 2) {
// very similar to the sub procedure in 15. 3Sum
while (l < r) {
final int sum = nums[l] + nums[r];
if (sum == target) {
path.add(nums[l]);
path.add(nums[r]);
ans.add(new ArrayList<>(path));
path.remove(path.size() - 1);
path.remove(path.size() - 1);
++l;
--r;
while (l < r && nums[l] == nums[l - 1])
++l;
while (l < r && nums[r] == nums[r + 1])
--r;
} else if (sum < target) {
++l;
} else {
--r;
}
}
return;
}
for (int i = l; i <= r; ++i) {
if (i > l && nums[i] == nums[i - 1])
continue;
path.add(nums[i]);
nSum(nums, n - 1, target - nums[i], i + 1, r, path, ans);
path.remove(path.size() - 1);
}
}
}
class Solution:
def fourSum(self, nums: List[int], target: int):
def nSum(l: int, r: int, target: int, n: int, path: List[int], ans: List[List[int]]) -> None:
if r - l + 1 < n or n < 2 or target < nums[l] * n or target > nums[r] * n:
return
if n == 2:
while l < r:
sum = nums[l] + nums[r]
if sum == target:
ans.append(path + [nums[l], nums[r]])
l += 1
while nums[l] == nums[l - 1] and l < r:
l += 1
elif sum < target:
l += 1
else:
r -= 1
return
for i in range(l, r + 1):
if i > l and nums[i] == nums[i - 1]:
continue
nSum(i + 1, r, target - nums[i], n - 1, path + [nums[i]], ans)
ans = []
nums.sort()
nSum(0, len(nums) - 1, target, 4, [], ans)
return ans
