Two sum solution is frequently queried by interviewers; this quandary is also recognised as two sum LeetCode. Let’s discover the two sum LeetCode solution and attain the two sum in C# archive.

In this post, learn how to tackle the Two Sum problem in C# with linear time complexity. By employing a hash map, we efficiently find the indices of two numbers in an array that add up to a specified target. Dive into the implementation details and master this essential algorithmic challenge.

In intermediate technical interviews, the Two Sum Solution often surfaces, gauging one’s grasp of time complexity in problem-solving. For beginners, I suggest exploring pattern and pyramid programs first. This question focuses on the efficiency of algorithms, essential for mastering coding interviews.

## Table of Contents

## Understand two sum problem

The initial step in finding two sum leetcode solution is breaking down the task into smaller components. Let’s start with this step first. If you’ve already completed it, kindly share it in the comments for other programmers to review.

### Question

Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.

You can return the answer in any order.

### Breakdown Question

Have you read the question? Do you have any thoughts on how to approach it? Let’s break it down together.

The interviewer presents an array of integers and a target number. Your task is to determine if any combination of array elements sums up to the given target.

For example, given an array [1, 2, 4, 5] and a target of 3, you need to identify which elements of the array need to be summed to reach the target. In this case, the answer would be [0, 1], representing the indices of the array elements. It’s crucial to return an array containing the index’s.

I trust you’ve understood the question and its breakdown. Additionally, you’re required to solve two sum problem with linear time complexity (O(n)).

## Implement logic to solve two sum problem

The Two Sum problem isn’t inherently difficult, but achieving it with linear time complexity poses a challenge. To tackle this, we’ll utilise a hash set. In this instance, using C#, we employ a dictionary hash set. You might wonder why we need a hash set. The answer is straightforward: hash sets offer built-in methods for efficiently checking element existence.

Let’s delve into the code to address this issue.

### Step 1

Let’s create an integer array named “array” and an integer target variable named “target”. We’ll then devise a method that returns a new integer array (the answer to the question), employing two parameters: the first being an integer array and the second being the target value.

```
static void Main(string[] args)
{
int[] array = { 2, 7, 11, 15 };
int target = 9;
Console.WriteLine(string.Join(", ", GetToSum(array, target)));
}
static int[] GetToSum(int[] array, int target)
{
return null;
}
```

### Step 2

Now, let’s create a Dictionary named “numIndices”. This hash set is responsible for checking duplicate elements. Then, initiate a for loop starting from 0 and iterating until the length of the “array”.

```
static int[] GetToSum(int[] array, int target)
{
Dictionary<int, int> numIndices = new Dictionary<int, int>();
for (int i = 0; i < array.Length; i++)
{
}
return null;
}
```

### Step 3

In each iteration of our loop, we obtain a number from the array. This number represents an element of A. Next, we need to find an element of B to determine if their sum equals our target value. To achieve this, let’s create a new integer named “secondElement” and assign it the value obtained by subtracting the target from element A (the value of the current iteration). This gives us element B.

Now that we have elements A and B, we check if element B exists in the Dictionary. If it does, we return an array containing element B and element A. This constitutes our answer.

```
static int[] GetToSum(int[] array, int target)
{
Dictionary<int, int> numIndexs = new Dictionary<int, int>();
for (int i = 0; i < array.Length; i++)
{
```**int secondElement = target - array[i];
if (numIndexs.ContainsKey(secondElement))
{
return new int[] { numIndexs[secondElement], i };
}**
}
return null;
}

### Step 4

Now, you might wonder what happens if element B is not present in our Dictionary. Indeed, you’re correct, and we must incorporate this logic as well.

Let’s add a conditional statement to check if our Dictionary does not contain element B. If this condition holds true, we’ll add element B to the Dictionary and proceed to the next iteration.

```
static int[] GetToSum(int[] array, int target)
{
Dictionary<int, int> numIndexs = new Dictionary<int, int>();
for (int i = 0; i < array.Length; i++)
{
int secondElement = target - array[i];
if (numIndexs.ContainsKey(secondElement))
{
return new int[] { numIndexs[secondElement], i };
}
```**if (!numIndexs.ContainsKey(array[i]))
{
numIndexs.Add(array[i], i);
}**
}
return null;
}

After completing all iterations, you’ll obtain your answer. If our array does not contain the targeted values A and B, then return null.

## Summarizing the two sum solution

Summarizing the Two Sum solution, we start by breaking down the problem into smaller steps. We utilize a hash set, specifically a Dictionary in C#, to efficiently check for the existence of elements.

In the code, we initialize an array and a target variable. Then, we iterate through the array, calculating the difference between the target and the current element to find the complementary value. If this complementary value exists in the Dictionary, we return the indices of the current element and its complement. If not, we add the current element to the Dictionary for future reference.

At the end of the iteration, if no valid pair is found, we return null. This approach ensures a linear time complexity, making it an efficient solution to the Two Sum problem.

i have found my solution from here. actually i did not find any in depth article on it yet.

Yes you are right it’s good

same here my friend this is actually good enough in depth logic explanation