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.
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
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 minus the current number). Then, check if the complement is present in the
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
Here are some examples of how to use the
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)
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.