💡 Problem Formulation: We need to verify whether an array contains a specific number and its triple (three times the number). For instance, given an array [2, 3, 6, 9, 12], our script should confirm that the number 3 and its triple (9) are present.
Method 1: Brute Force Approach
Using a brute force approach involves two nested loops, checking for each element in the array if its triple is also present. It’s straightforward but not the most efficient method, with a time complexity of O(n^2).
Here’s an example:
def check_number_and_triple(arr):
for x in arr:
if x*3 in arr:
return True
return False
arr_example = [3, 5, 1, 9]
print(check_number_and_triple(arr_example))
Output: True
This code snippet defines a function check_number_and_triple() that takes an array as an argument. It iterates through the array and for each element, checks if its triple exists in the array using a nested loop. If it finds a matching triple, it returns True; otherwise, it returns False after completing the iteration.
Method 2: Using a Hash Set
This method improves on the brute force approach by using a hash set (Python’s set data structure), allowing for O(1) average time complexity for lookups, which results in an O(n) average time complexity for the whole function.
Here’s an example:
def check_number_and_triple(arr):
seen_numbers = set(arr)
for number in arr:
if number * 3 in seen_numbers:
return True
return False
arr_example = [7, 1, 5, 21]
print(check_number_and_triple(arr_example))
Output: True
Here, the function check_number_and_triple() first creates a set of all numbers in the array for faster lookups. It then iterates through the array and checks if the triple of each number is in the set, which is a much faster operation than checking in a list.
Method 3: Sorting and Binary Search
By first sorting the array, we can use binary search to look for the triple of each number, leveraging a time complexity of O(n log n) for sorting and O(log n) for each binary search, leading to an overall time complexity of O(n log n).
Here’s an example:
import bisect
def check_number_and_triple(arr):
arr.sort()
for number in arr:
index = bisect.bisect_left(arr, number * 3)
if index < len(arr) and arr[index] == number * 3:
return True
return False
arr_example = [2, 3, 14, 6]
print(check_number_and_triple(arr_example))
Output: True
We start by sorting the array, then for each element we use the bisect_left function from the bisect module to perform a binary search for the triple. If the triple is found within the array boundaries, the function returns True.
Method 4: Two-pointer Technique
The two-pointer technique is effective when dealing with a sorted array. We can use one pointer to iterate over the array and another one to find triples without re-checking already checked pairs. This method only works efficiently if the array is initially sorted, or if we sort it first.
Here’s an example:
def check_number_and_triple(arr):
arr.sort()
left = 0
while left < len(arr):
right = left + 1
while right < len(arr) and arr[right] / arr[left] <= 3:
if arr[right] / arr[left] == 3:
return True
right += 1
left += 1
return False
arr_example = [1, 2, 4, 8]
print(check_number_and_triple(arr_example))
Output: False
After sorting the array, we utilize two pointers. The left pointer goes through each element while the right pointer checks for the triple. If the triple is found, it returns True. If not, left is incremented and the search continues.
Bonus One-Liner Method 5: List Comprehension with Intersection
This is a more Pythonic method that uses list comprehension along with set intersection to check for the presence of both number and its triple in a single line. This approach is concise but may not be as clear or efficient for large lists.
Here’s an example:
arr_example = [8, 2, 4, 1, 3, 6]
print(bool(set(arr_example) & {x*3 for x in arr_example}))
Output: True
Here we convert the array into a set for O(1) lookup times. We then construct another set using list comprehension to get the triples of each element. Lastly, we check for the intersection of the two sets with the & operator which gives us True if there is a common element, indicating the number and its triple exist.
Summary/Discussion
- Method 1: Brute Force. Simple to understand and implement. Inefficient for large arrays due to high time complexity.
- Method 2: Hash Set. More efficient than brute force. Uses extra space for the set, but reduces overall time complexity.
- Method 3: Sorting and Binary Search. Efficient for sorted arrays. Initial sorting required, leading to increased time complexity if the input array is not initially sorted.
- Method 4: Two-pointer Technique. Works well with sorted lists and avoids redundant checks. Not as intuitive as other methods, and sorting is necessary if the input isn’t sorted already.
- Method 5: List Comprehension with Intersection. Pythonic and concise. Not necessarily performant for large lists and can be difficult to understand for beginners.