💡 Problem Formulation: You’ve been given a sorted list of integers and a target value k. The objective is to determine if any two numbers within the list can be added together to total k. For instance, given the list [1, 2, 3, 4] and k=7, the desired output would be True since 3 and 4 sum to 7.
Method 1: Two-Pointer Technique
This method utilizes the two-pointer technique. Given that the input list is already sorted, one pointer starts from the beginning while the other from the end. If their sum is less than k, the start pointer is moved up; if it’s more, the end pointer is moved down. This process repeats until the pointers meet or the correct sum is found.
Here’s an example:
def two_pointer_check(nums, k):
left, right = 0, len(nums) - 1
while left < right:
current_sum = nums[left] + nums[right]
if current_sum == k:
return True
elif current_sum < k:
left += 1
else:
right -= 1
return False
print(two_pointer_check([1, 2, 3, 4], 7))
Output:
True
This code snippet first establishes two pointers, one at the beginning and one at the end of the sorted list. It then enters a while loop that continually checks sums and adjusts the pointers until the pointers meet or the sum equal to k is found.
Method 2: Binary Search
In binary search, for each element x in the list, we look for k - x using a binary search. This method works effectively because the list is sorted, so binary search can be performed efficiently.
Here’s an example:
import bisect
def binary_search_check(nums, k):
for i in range(len(nums)):
complement_index = bisect.bisect_left(nums, k - nums[i], i+1)
if complement_index < len(nums) and nums[complement_index] == k - nums[i]:
return True
return False
print(binary_search_check([1, 2, 3, 4], 7))
Output:
True
The code uses the bisect module to perform binary search. It iterates over the sorted list, and for each element, it uses binary search to check if its complement to k exists in the remainder of the list.
Method 3: Hash Table
A hash table can be used to store previously seen elements. We iterate through the list, and for each element we check if k - element is in the hash table. If not, we add the element to the hash table. This is a particularly fast method with a time complexity of O(n).
Here’s an example:
def hash_table_check(nums, k):
seen = set()
for num in nums:
if k - num in seen:
return True
seen.add(num)
return False
print(hash_table_check([1, 2, 3, 4], 5))
Output:
True
In the provided code, a set is used to keep track of the elements we have seen so far. As we iterate over the list, we check if the complement of the current element with respect to k is in the set. If it is, we return True; if not, we continue and add the current element to the set.
Method 4: Brute Force
The brute force method involves checking every possible pair of numbers in the list until a pair sums to k or all pairs have been checked. This is the simplest approach but also the least efficient with a time complexity of O(n^2).
Here’s an example:
def brute_force_check(nums, k):
for i in range(len(nums)):
for j in range(i + 1, len(nums)):
if nums[i] + nums[j] == k:
return True
return False
print(brute_force_check([1, 2, 3, 4], 7))
Output:
True
This simple code illustrates the brute force approach. It uses nested loops to check every possible pair of numbers in the list, returning True as soon as a pair sums to k.
Bonus One-Liner Method 5: List Comprehension and ‘in’ Check
This one-liner uses list comprehension and Python’s in operator to check if the complement k - x exists in the slice of the list that follows the current element x. It’s a shorthand, less efficient form of the brute force method.
Here’s an example:
def one_liner_check(nums, k):
return any(k - num in nums[i+1:] for i, num in enumerate(nums))
print(one_liner_check([1, 2, 3, 4], 7))
Output:
True
In this succinct approach, we use any() to check if any element in the list has its complement in the remaining part of the list. This is an inherently less efficient method due to the slicing overhead.
Summary/Discussion
- Method 1: Two-Pointer Technique. Efficient for sorted lists. O(n) time complexity. Less code required and high readability.
- Method 2: Binary Search. Good alternative for sorted lists. O(n log n) time complexity because of binary search overhead per element.
- Method 3: Hash Table. Very efficient with an average case of O(n) time complexity. Additional memory usage for hash table.
- Method 4: Brute Force. Simplest to comprehend, but least efficient due to O(n^2) time complexity. Not recommended for large lists.
- Method 5: Bonus One-Liner. Readable and concise, but has slicing overhead which can lead to O(n^2) performance.