Last updated on July 19th, 2024 at 10:51 pm
Here, We see 4Sum LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.
List of all LeetCode Solution
Topics
Dynamic Programming, Recursion
Companies
Level of Question
Medium
4Sum LeetCode Solution
Table of Contents
Problem Statement
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]]
1. 4Sum Leetcode Solution C++
class Solution {
public:
vector<vector<int>> fourSum(vector<int>& nums, int target) {
vector<vector<int>> total;
int n = nums.size();
if(n<4) return total;
sort(nums.begin(),nums.end());
for(int i=0;i<n-3;i++)
{
if(i>0&&nums[i]==nums[i-1]) continue;
if(nums[i]+nums[i+1]+nums[i+2]+nums[i+3]>target) break;
if(nums[i]+nums[n-3]+nums[n-2]+nums[n-1]<target) continue;
for(int j=i+1;j<n-2;j++)
{
if(j>i+1&&nums[j]==nums[j-1]) continue;
if(nums[i]+nums[j]+nums[j+1]+nums[j+2]>target) break;
if(nums[i]+nums[j]+nums[n-2]+nums[n-1]<target) continue;
int left=j+1,right=n-1;
while(left<right){
int sum=nums[left]+nums[right]+nums[i]+nums[j];
if(sum<target) left++;
else if(sum>target) right--;
else{
total.push_back(vector<int>{nums[i],nums[j],nums[left],nums[right]});
do{left++;}while(nums[left]==nums[left-1]&&left<right);
do{right--;}while(nums[right]==nums[right+1]&&left<right);
}
}
}
}
return total;
}
};
2. 4Sum Leetcode Solution Java
class Solution {
public List<List<Integer>> fourSum(int[] nums, int target) {
ArrayList<List<Integer>> res = new ArrayList<List<Integer>>();
int len = nums.length;
if (nums == null || len < 4)
return res;
Arrays.sort(nums);
int max = nums[len - 1];
if (4 * nums[0] > target || 4 * max < target)
return res;
int i, z;
for (i = 0; i < len; i++) {
z = nums[i];
if (i > 0 && z == nums[i - 1])
continue;
if (z + 3 * max < target)
continue;
if (4 * z > target)
break;
if (4 * z == target) {
if (i + 3 < len && nums[i + 3] == z)
res.add(Arrays.asList(z, z, z, z));
break;
}
threeSumForFourSum(nums, target - z, i + 1, len - 1, res, z);
}
return res;
}
public void threeSumForFourSum(int[] nums, int target, int low, int high, ArrayList<List<Integer>> fourSumList,
int z1) {
if (low + 1 >= high)
return;
int max = nums[high];
if (3 * nums[low] > target || 3 * max < target)
return;
int i, z;
for (i = low; i < high - 1; i++) {
z = nums[i];
if (i > low && z == nums[i - 1])
continue;
if (z + 2 * max < target)
continue;
if (3 * z > target)
break;
if (3 * z == target) {
if (i + 1 < high && nums[i + 2] == z)
fourSumList.add(Arrays.asList(z1, z, z, z));
break;
}
twoSumForFourSum(nums, target - z, i + 1, high, fourSumList, z1, z);
}
}
public void twoSumForFourSum(int[] nums, int target, int low, int high, ArrayList<List<Integer>> fourSumList,
int z1, int z2) {
if (low >= high)
return;
if (2 * nums[low] > target || 2 * nums[high] < target)
return;
int i = low, j = high, sum, x;
while (i < j) {
sum = nums[i] + nums[j];
if (sum == target) {
fourSumList.add(Arrays.asList(z1, z2, nums[i], nums[j]));
x = nums[i];
while (++i < j && x == nums[i])
;
x = nums[j];
while (i < --j && x == nums[j])
;
}
if (sum < target)
i++;
if (sum > target)
j--;
}
return;
}
}
3. 4Sum Leetcode Solution JavaScript
var fourSum = function(nums, target) {
nums.sort((a, b) => a - b);
const result = []
for(let i = 0; i < nums.length - 3; i++) {
for(let j = i + 1; j < nums.length - 2; j++) {
let low = j + 1;
let high = nums.length - 1
while(low < high) {
const sum = nums[i] + nums[j] + nums[low] + nums[high];
if(sum === target) {
result.push([nums[i], nums[j], nums[low], nums[high]])
while(nums[low] === nums[low + 1]) low++;
while(nums[high] === nums[high - 1]) high--;
low++;
high--;
} else if(sum < target) {
low++
} else {
high--
}
}
while(nums[j] === nums[j + 1]) j++;
}
while(nums[i] === nums[i + 1]) i++;
}
return result
};
4. 4Sum Leetcode Solution Python
class Solution(object):
def fourSum(self, nums, target):
nums.sort()
results = []
self.findNsum(nums, target, 4, [], results)
return results
def findNsum(self, nums, target, N, result, results):
if len(nums) < N or N < 2: return
if N == 2:
l,r = 0,len(nums)-1
while l < r:
if nums[l] + nums[r] == target:
results.append(result + [nums[l], nums[r]])
l += 1
r -= 1
while l < r and nums[l] == nums[l - 1]:
l += 1
while r > l and nums[r] == nums[r + 1]:
r -= 1
elif nums[l] + nums[r] < target:
l += 1
else:
r -= 1
else:
for i in range(0, len(nums)-N+1):
if target < nums[i]*N or target > nums[-1]*N:
break
if i == 0 or i > 0 and nums[i-1] != nums[i]:
self.findNsum(nums[i+1:], target-nums[i], N-1, result+[nums[i]], results)
return