# 3Sum – Leetcode Solution

In this post, we are going to solve the 15. 3Sum problem of Leetcode. This problem 15. 3Sum is a Leetcode medium level problem. Let’s see code, 15. 3Sum.

Contents

## Problem

Given an integer array nums, return all the triplets `[nums[i], nums[j], nums[k]]` such that `i != j`, `i != k`, and `j != k`, and `nums[i] + nums[j] + nums[k] == 0`.

Notice that the solution set must not contain duplicate triplets.

### Example 1 :

``````
Input: nums = [-1,0,1,2,-1,-4]
Output: [[-1,-1,2],[-1,0,1]]
Explanation:
nums[0] + nums[1] + nums[2] = (-1) + 0 + 1 = 0.
nums[1] + nums[2] + nums[4] = 0 + 1 + (-1) = 0.
nums[0] + nums[3] + nums[4] = (-1) + 2 + (-1) = 0.
The distinct triplets are [-1,0,1] and [-1,-1,2].
Notice that the order of the output and the order of the triplets does not matter.
``````

### Example 2 :

``````
Input: nums = [0,1,1]
Output: []
Explanation: The only possible triplet does not sum up to 0.
``````

### Example 3 :

``````
Input: nums = [0,0,0]
Output: [[0,0,0]]
Explanation: The only possible triplet sums up to 0.
``````

### Constraints

• `3 <= nums.length <= 3000`
• `-105 <= nums[i] <= 105`

Now, let’s see the code of 15. 3Sum – Leetcode Solution.

# 3Sum – Leetcode Solution

### 15. 3Sum – Solution in Java

```class Solution {
public List<List<Integer>> threeSum(int[] arr) {
Arrays.sort(arr);
int n = arr.length;
List<List<Integer>> ans = new ArrayList<>();
for (int i = 0; i < n; i++) {
if (arr[i] > 0) break; // Since arr[i] <= arr[l] <= arr[r], if a[i] > 0 then sum=arr[i]+arr[l]+arr[r] > 0
int l = i + 1, r = n - 1;
while (l < r) {
int sum = arr[i] + arr[l] + arr[r];
if (sum < 0) l++;
else if (sum > 0) r--;
else {
while (l+1 <= r && arr[l] == arr[l+1]) l++; // Skip duplicate nums[l]
l++;
r--;
}
}
while (i+1 < n && arr[i+1] == arr[i]) i++; // Skip duplicate nums[i]
}
return ans;
}
}```

### 15. 3Sum – Solution in C++

```class Solution {
public:
vector<vector<int>> threeSum(vector<int>& nums) {
sort(nums.begin() , nums.end());
if(nums.size() < 3){
return {};
}
if(nums[0] > 0){
return {};
}
unordered_map<int , int> hashMap;
for(int i = 0 ; i < nums.size() ; ++i){
hashMap[nums[i]] = i;
}
for(int i = 0 ; i < nums.size() - 2 ; ++i){
if(nums[i] > 0){
break;
}
for(int j = i + 1 ; j < nums.size() - 1 ; ++j){
int required = -1*(nums[i] + nums[j]);
if(hashMap.count(required) && hashMap.find(required)->second > j){
}
j = hashMap.find(nums[j])->second;
}
i = hashMap.find(nums[i])->second;
}
}
};```

### 15. 3Sum– Solution in Python

```class Solution:
def threeSum(self, nums: List[int]) -> List[List[int]]:
nums.sort()
ans = []
n = len(nums)
i = 0
while i < n:
if nums[i] > 0: break  # Since arr[i] <= arr[l] <= arr[r], if arr[i] > 0 then sum=arr[i]+arr[l]+arr[r] > 0
l = i + 1
r = n - 1
while l < r:
sum3 = nums[i] + nums[l] + nums[r]
if sum3 == 0:
ans.append([nums[i], nums[l], nums[r]])
while l+1 < n and nums[l+1] == nums[l]: l += 1  # Skip duplicates nums[l]
l += 1
r -= 1
elif sum3 < 0: l += 1
else: r -= 1

while i+1 < n and nums[i+1] == nums[i]: i += 1  # Skip duplicates nums[i]
i += 1
return ans```

Note: This problem 15. 3Sum is generated by Leetcode but the solution is provided by CodingBroz. This tutorial is only for Educational and Learning purpose.