In this guide, we will explore an algorithm to determine if any two numbers in a list add up to a given number `K`

. This problem is a common interview question and can be solved using various approaches. We will discuss a step-by-step solution using a hash set, which allows for faster lookups and reduced time complexity.

## Prerequisites

Before diving into the solution, make sure you are familiar with the following concepts:

## Solution: Using a Hash Set

The hash set approach has a time complexity of O(n), making it an efficient solution to this problem.

### Step 1: Create a function

First, let's create a function called `find_sum`

that takes two arguments: a list of integers `numbers`

and an integer `K`

.

```
def find_sum(numbers, K):
pass
```

### Step 2: Create an empty hash set

Inside the function, create an empty hash set called `seen_numbers`

. This will be used to store the numbers we have encountered in the list.

```
def find_sum(numbers, K):
seen_numbers = set()
```

### Step 3: Iterate through the list

Now, iterate through the list of numbers using a for loop. For each number, calculate the required complement that would add up to `K`

(i.e., `K`

minus the current number). Then, check if the complement is present in the `seen_numbers`

set.

If the complement is found, it means we have a pair of numbers that add up to `K`

. In this case, return `True`

. If the complement is not found, add the current number to the `seen_numbers`

set and continue to the next iteration.

```
def find_sum(numbers, K):
seen_numbers = set()
for num in numbers:
complement = K - num
if complement in seen_numbers:
return True
seen_numbers.add(num)
return False
```

That's it! Our function is now complete and can be used to determine if any two numbers in a list add up to a given number `K`

.

## Examples

Here are some examples of how to use the `find_sum`

function:

```
print(find_sum([1, 2, 3, 4, 5], 9)) # Output: True (4+5=9)
print(find_sum([10, 15, 3, 7], 17)) # Output: True (10+7=17)
print(find_sum([1, 3, 5, 7, 9], 2)) # Output: False (No pair adds up to 2)
```

## FAQ

### 1. Why use a hash set instead of a list for storing seen numbers?

A hash set allows for faster lookups (O(1) on average), making it more efficient than a list (O(n) for lookups).

### 2. Can this problem be solved using a different data structure?

Yes, there are various approaches to solving this problem, such as using a sorted list and applying the two-pointer technique, or using a dictionary to store the frequency of each number.

### 3. What is the time complexity of this solution?

The time complexity of this solution is O(n), where n is the length of the input list.

### 4. Can this solution handle negative numbers?

Yes, this solution can handle negative numbers in the input list.

### 5. Can this solution handle duplicate numbers in the list?

Yes, this solution can handle duplicate numbers in the input list.