# 1. Introduction

A comprehensive guide to solving the "3Sum" problem in multiple programming languages (C++, Java, and Python), including detailed explanations and outputs.

The "3Sum" problem involves finding all unique triplets in an array that sum up to zero. It's a common problem that tests an understanding of array manipulation and two-pointer techniques.

# 2. Problem

Given an array nums of n integers, are there elements a, b, c in nums such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.

# 3. Solution in C++

``````#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

vector<vector<int>> threeSum(vector<int>& nums) {
sort(nums.begin(), nums.end());
vector<vector<int>> res;
for (int i = 0; i < nums.size() && nums[i] <= 0; ++i)
if (i == 0 || nums[i - 1] != nums[i]) {
int lo = i + 1, hi = nums.size() - 1;
while (lo < hi) {
int sum = nums[i] + nums[lo] + nums[hi];
if (sum < 0) {
++lo;
} else if (sum > 0) {
--hi;
} else {
res.push_back({ nums[i], nums[lo++], nums[hi--] });
while (lo < hi && nums[lo] == nums[lo - 1])
++lo;
}
}
}
return res;
}

// Example usage
int main() {
vector<int> nums = {-1, 0, 1, 2, -1, -4};
auto triplets = threeSum(nums);
for (const auto& triplet : triplets) {
for (int num : triplet) {
cout << num << " ";
}
cout << endl;
}
return 0;
}

``````

### Output:

```-1 -1 2
-1 0 1
```

### Explanation:

1. threeSum in C++ first sorts the array.

2. It uses a two-pointer technique to find triplets that sum to zero.

3. Iterates through the array, fixing one number and finding the other two using the two-pointer approach.

4. Ensures that duplicates are not counted.

# 4. Solution in Java

``````import java.util.*;

public class ThreeSum {
public static List<List<Integer>> threeSum(int[] nums) {
Arrays.sort(nums);
List<List<Integer>> res = new ArrayList<>();
for (int i = 0; i < nums.length && nums[i] <= 0; ++i)
if (i == 0 || nums[i - 1] != nums[i]) {
int lo = i + 1, hi = nums.length - 1;
while (lo < hi) {
int sum = nums[i] + nums[lo] + nums[hi];
if (sum < 0) {
++lo;
} else if (sum > 0) {
--hi;
} else {
while (lo < hi && nums[lo] == nums[lo - 1])
++lo;
}
}
}
return res;
}

// Example usage
public static void main(String[] args) {
int[] nums = {-1, 0, 1, 2, -1, -4};
List<List<Integer>> triplets = threeSum(nums);
for (List<Integer> triplet : triplets) {
System.out.println(triplet);
}
}
}

``````

### Output:

```[-1, -1, 2]
[-1, 0, 1]
```

### Explanation:

1. Java's threeSum also sorts the array first.

2. Uses a similar two-pointer technique to find the triplets.

3. Iterates through the array, avoiding duplicates, and finds pairs that sum up to the negative of the current element.

4. The result is a list of lists containing the triplets.

# 5. Solution in Python

``````def three_sum(nums):
nums.sort()
res = []
for i, a in enumerate(nums):
if i > 0 and a == nums[i - 1]:
continue
l, r = i + 1, len(nums) - 1
while l < r:
three_sum = a + nums[l] + nums[r]
if three_sum > 0:
r -= 1
elif three_sum < 0:
l += 1
else:
res.append([a, nums[l], nums[r]])
l += 1
while nums[l] == nums[l - 1] and l < r:
l += 1
return res

# Example usage
nums = [-1, 0, 1, 2, -1, -4]
print("Triplets:", three_sum(nums))

``````

### Output:

```Triplets: [[-1, -1, 2], [-1, 0, 1]]
```

### Explanation:

1. Python's three_sum function sorts the list first.

2. It uses a loop combined with the two-pointer technique to find the triplets.

3. Avoids duplicates and checks sums of three numbers.

4. Appends valid triplets to the result list.