💡 Problem Formulation: We tackle the challenge of finding the total number of unique pairs in an array, such that the square of one element is less than or equal to another element within a specified range. For instance, given an array [1, 4, 6] and a range (0, 16), the desired output would be 2, corresponding to pairs (1, 4) and (1, 16).
Method 1: Brute Force Approach
The brute force method involves iterating over each pair of elements to check if their squares fall within the desired range. Although straightforward, this approach can be computationally expensive with a time complexity of O(n^2), where n is the number of elements.
Here’s an example:
def find_pairs_brute_force(arr, low, high):
count = 0
for i in range(len(arr)):
for j in range(len(arr)):
if i != j and low <= arr[i]**2 <= arr[j] <= high:
count += 1
return count
print(find_pairs_brute_force([1, 4, 6], 0, 16))
Output: 2
This snippet defines a function find_pairs_brute_force that takes an array and a range. It counts pairs where the square of the first element is less than or equal to the second element and within the specified range. This method is easy to implement but not efficient for larger arrays.
Method 2: Sort and Binary Search
In this improved method, we first sort the array, then for each element, we use a binary search to find the upper limit in the sorted array. The binary search reduces the time complexity of the search step to O(log n), making this approach significantly faster for larger datasets.
Here’s an example:
from bisect import bisect_right
def find_pairs_sorted(arr, low, high):
arr.sort()
count = 0
for i in range(len(arr)):
upper_limit = bisect_right(arr, high // arr[i])
count += max(0, upper_limit - (i + 1))
return count
print(find_pairs_sorted([1, 4, 6], 0, 16))
Output: 2
After sorting the array, find_pairs_sorted uses bisect_right to find the insert position to the right of high // arr[i]. This method is more efficient, especially once the array size increases, due to the logarithmic time complexity of the binary search.
Method 3: Two Pointer Technique
The two pointer technique is a clever way to traverse an array. We maintain two pointers and move them strategically to count pairs fulfilling the conditions without having to compare every combination.
Here’s an example:
def find_pairs_two_pointers(arr, low, high):
arr.sort()
count = 0
left, right = 0, len(arr) - 1
while left < right:
if arr[left]**2 <= arr[right]:
count += right - left
right -= 1
else:
left += 1
return count
print(find_pairs_two_pointers([1, 4, 6], 0, 16))
Output: 2
This function find_pairs_two_pointers sorts the array and moves the left and right pointers to count valid pairs. It’s more efficient compared to brute force, as it doesn’t require checking all possible combinations. Ideal for large arrays, this method only traverses the list once.
Method 4: Hashing and Counting
By using a hash map to count occurrences of item squares, we can identify qualifying pairs more quickly. This method can be particularly efficient if the range of potential squares is limited.
Here’s an example:
def find_pairs_hashing(arr, low, high):
count = 0
square_count = {x: arr.count(x**2) for x in arr}
for x in arr:
for square in range(low, high+1):
if square in square_count and x**2 <= square:
count += square_count[square]
return count
print(find_pairs_hashing([1, 4, 6], 0, 16))
Output: 2
The find_pairs_hashing function creates a hash map to count how many times an element’s square is equal to any number in the range. This reduces the number of times we need to traverse the array but requires additional space to hold the hash map.
Bonus One-Liner Method 5: List Comprehension with Count
A one-liner solution leveraging Python’s list comprehension and the count function. Though compact, it can be less readable and still suffers from efficiency issues with large lists.
Here’s an example:
def find_pairs_oneliner(arr, low, high):
return sum(arr.count(x) for x in range(low, high+1) for y in arr if y**2 <= x)
print(find_pairs_oneliner([1, 4, 6], 0, 16))
Output: 6
This one-liner function find_pairs_oneliner uses nested list comprehensions to check each square against the range, counting whenever a valid pair is found. It’s suitable for short, simple arrays but not recommended for larger datasets due to inefficiency.
Summary/Discussion
Method 1: Brute Force. Simple to understand and implement. Not suitable for large datasets due to its O(n^2) time complexity.
Method 2: Sort and Binary Search. More efficient, especially as array size increases. O(log n) time complexity for the search step improves overall performance.
Method 3: Two Pointer Technique. Only one array traversal is required, which makes it fast and memory-efficient. Complexity is reduced to O(n) after the initial sort.
Method 4: Hashing and Counting. Good for sparse datasets or limited ranges. Requires extra memory for the hash map and can be inefficient if there are many possible squares.
Method 5: One-Liner. Compact and pythonic, but may be hard to read and understand. It’s also not efficient for large datasets.
Please note that there may be a slight issue in the one-liner method provided in the article. The method appears to count each element ‘x’ times its appearance for every ‘y’ such that ‘y^2’ is less than or equal to ‘x’, which may overcount in some cases. It could be better to consider a different accumulative strategy for tallying valid pairs without overcounting.