> For the complete documentation index, see [llms.txt](https://docs-57.gitbook.io/data-structure-and-algorithms/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://docs-57.gitbook.io/data-structure-and-algorithms/problems/array/3sum.md).

# 3Sum

{% hint style="info" %}
Array, Two Pointers
{% endhint %}

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
```

Assumptions:

* Each input have multiple solutions.
* The same element cannot be used twice.
* The answer can be returned in any order.

Questions to Clarify:

* **Negative Numbers**: Can the input array contain negative numbers? Yes.
* **Duplicate Numbers**: Can the input array contain duplicate numbers? Yes.
* **Unique Solution**: Is it guaranteed that there will always be exactly one solution? No.
* **Multiple Solutions**: Should we return all possible solutions, or just one? All possible unique solutions.
* **Indices Required**: Do we need to return the actual indices of the solution numbers, or just the numbers themselves? Actual indices of the solution numbers.

Additional Considerations:

* **Array Size**: What are the expected size limits for the input array?
* **Time and Space Complexity**: Are there any specific time or space complexity requirements for the solution?
* **Algorithm Choice**: Would a specific algorithm (e.g., hash table-based or sorting-based) be preferred?

### Solutions

Check sum of three elements of an array.

#### **Approach – Brute Force Technique**

Run two nested loops to check sum of three elements is equal to `0`. Also avoid duplicate solution.

**Steps**

1. **Sort the array**: The algorithm starts by sorting the input array. This is done to allow us to skip over duplicate values in the array, which ensures that the solution set does not contain duplicate triplets.
2. **Iterate over the array**: The algorithm then iterates over the sorted array with three nested loops, each starting from a different index. The outer loop (`i`) goes from `0` to `nums.Length - 2`, the middle loop (`j`) goes from `i + 1` to `nums.Length - 1`, and the inner loop (`k`) goes from `j + 1` to `nums.Length`.
3. **Skip duplicates**: At the start of each loop, the algorithm checks if the current number is the same as the previous one. If it is, the algorithm skips to the next iteration. This is done to avoid adding duplicate triplets to the result.
4. **Check for zero sum**: For each triplet (`nums[i]`, `nums[j]`, `nums[k]`), the algorithm checks if their sum is zero. If it is, the triplet is added to the result list.
5. **Return the result**: After all triplets have been checked, the algorithm returns the list of triplets that sum to zero.

```csharp
public class Solution
{
    public IList<IList<int>> ThreeSum(int[] nums)
    {
        IList<IList<int>> result = new List<IList<int>>();
        Array.Sort(nums);
        for (int i = 0; i < nums.Length - 2; i++)
        {
            if (i == 0 || (i > 0 && nums[i] != nums[i - 1]))
            {
                for (int j = i + 1; j < nums.Length - 1; j++)
                {
                    if (j == i + 1 || nums[j] != nums[j - 1])
                    {
                        for (int k = j + 1; k < nums.Length; k++)
                        {
                            if (k == j + 1 || nums[k] != nums[k - 1])
                            {
                                if (nums[i] + nums[j] + nums[k] == 0)
                                {
                                    result.Add(new List<int> { nums[i], nums[j], nums[k] });
                                }
                            }
                        }
                    }
                }
            }
        }
        return result;
    }
}

```

> Complexity

* **Time Complexity:** O(n^3)
* **Auxiliary Space:** O(1)

This algorithm works by checking every possible triplet in the array. Because of the sorting and the check for duplicates, each triplet in the result is unique. However, because it uses three nested loops, its time complexity is O(n^3), which means it can be slow for large inputs.

#### **Approach – Two Poniters Technique**

As we know that *two-poiner* approach is good if if array in sorted order. here's a step-by-step explanation of the optimized algorithm:

**Steps**

1. **Sort the array**: The algorithm starts by sorting the input array. This is done to allow us to skip over duplicate values in the array, which ensures that the solution set does not contain duplicate triplets.
2. **Iterate over the array**: The algorithm then iterates over the sorted array with a single loop (`i`), which goes from `0` to `nums.Length - 2`.
3. **Skip duplicates**: At the start of the loop, the algorithm checks if the current number is the same as the previous one. If it is, the algorithm skips to the next iteration. This is done to avoid adding duplicate triplets to the result.
4. **Initialize two pointers**: For each `i`, the algorithm initializes two pointers, `low` and `high`. `low` is initialized to `i + 1` and `high` is initialized to `nums.Length - 1`. The variable `sum` is initialized to `0 - nums[i]`.
5. **Move the pointers**: While `low` is less than `high`, the algorithm checks the sum of `nums[low]` and `nums[high]`. If their sum is equal to `sum`, it means we have found a triplet that sums to zero, so it adds the triplet to the result list, increments `low`, and decrements `high`. If their sum is less than `sum`, it increments `low`, and if their sum is greater than `sum`, it decrements `high`.
6. **Skip duplicates**: After finding a triplet, the algorithm again checks for duplicates by comparing `nums[low]` and `nums[high]` with their adjacent elements. If `nums[low]` is the same as `nums[low + 1]`, it increments `low`. If `nums[high]` is the same as `nums[high - 1]`, it decrements `high`.
7. **Return the result**: After all triplets have been checked, the algorithm returns the list of triplets that sum to zero.

```csharp
public class Solution
{
   public IList<IList<int>> ThreeSum(int[] nums)
   {
       IList<IList<int>> result = new List<IList<int>>();
       Array.Sort(nums);
       for (int i = 0; i < nums.Length - 2; i++)
       {
           if (i == 0 || (i > 0 && nums[i] != nums[i - 1]))
           {
               int low = i + 1, high = nums.Length - 1, sum = 0 - nums[i];
               while (low < high)
               {
                   if (nums[low] + nums[high] == sum)
                   {
                       result.Add(new List<int> { nums[i], nums[low], nums[high] });
                       while (low < high && nums[low] == nums[low + 1]) low++;
                       while (low < high && nums[high] == nums[high - 1]) high--;
                       low++; high--;
                   }
                   else if (nums[low] + nums[high] < sum)
                   {
                       low++;
                   }
                   else
                   {
                       high--;
                   }
               }
           }
       }
       return result;
   }
}
```

> Complexity

* **Time Complexity:** O(n^2)
* **Auxiliary Space:** O(1)