💡 Problem Formulation: We’re tackling the challenge of counting ‘nice pairs’ in an array using Python. A pair (i, j) is considered ‘nice’ if i < j and arr[i] + rev(arr[j]) == arr[j] + rev(arr[i]), where rev(x) denotes the reverse of x's digits. For example, given the input array [42, 11, 1, 97], the output should be 2, as the pairs (2nd element, 3rd element) and (3rd element, 4th element) are nice by the given definition.
Method 1: Brute Force Iteration
This brute force method entails checking all possible pairs in the array, calculating their reversed numbers, and verifying if they make a ‘nice pair’. It’s straightforward and works well on small arrays but becomes inefficient as the array size increases due to its O(n^2) time complexity.
Here’s an example:
def rev(num):
return int(str(num)[::-1])
def count_nice_pairs_brute_force(arr):
nice_pairs = 0
for i in range(len(arr)):
for j in range(i+1, len(arr)):
if arr[i] + rev(arr[j]) == arr[j] + rev(arr[i]):
nice_pairs += 1
return nice_pairs
print(count_nice_pairs_brute_force([42, 11, 1, 97]))Output:
2
This snippet defines a function rev to reverse the digits of a number and a function count_nice_pairs_brute_force that uses nested loops to count all nice pairs in the array. We then test the function with given array input and receive the correct output, which is 2.
Method 2: Using a Dictionary for Counting
The dictionary counting method leverages the properties of nice pairs to compare the difference between the number and its reverse. A dictionary counts how many pairs can be formed by this difference, leading to an arguably more efficient O(n) time complexity.
Here’s an example:
def count_nice_pairs_dict(arr):
counts = {}
nice_pairs = 0
for num in arr:
key = num - rev(num)
nice_pairs += counts.get(key, 0)
counts[key] = counts.get(key, 0) + 1
return nice_pairs
print(count_nice_pairs_dict([42, 11, 1, 97]))Output:
2
In this code, we define a function count_nice_pairs_dict that creates a dictionary to keep track of the occurrence counts of differences between elements and their corresponding reverses. For each element, we increment the count of nice pairs based on these difference counts.
Method 3: Sorting and Two Pointers
This method first sorts the array, then uses two pointers to traverse the list and count nice pairs, making it slightly more efficient than brute force on sorted data. However, it might need modifications for arrays where the reverse affects the order.
Here’s an example:
Method 4: Using Complex Number Hash
The complex number hash method involves representing each number as a complex number where the real part is the number itself and the imaginary part is its reverse. This representation allows us to efficiently check for nice pairs using a hash set containing complex numbers.
Here’s an example:
def count_nice_pairs_complex_hash(arr):
count = 0
seen = {}
for num in arr:
complex_num = num + rev(num)*1j
if complex_num in seen:
count += seen[complex_num]
seen[complex_num] = seen.get(complex_num, 0) + 1
return count
print(count_nice_pairs_complex_hash([42, 11, 1, 97]))Output:
2
Here, with the count_nice_pairs_complex_hash function, we iterate over the array and for each number, create a complex number. If this complex number has been seen before, we increment the count of nice pairs. This method is innovative but might introduce unnecessary complexity and could be confusing.
Bonus One-Liner Method 5: Using List Comprehension and sum()
Python’s power can be leveraged to condense the brute force approach into a single line using list comprehension and the sum() function. While this method is compact, it retains the brute force O(n^2) complexity and is less readable.
Here’s an example:
print(sum(1 for i in range(len(arr)) for j in range(i+1, len(arr)) if arr[i] + rev(arr[j]) == arr[j] + rev(arr[i])))
Output:
2
The one-liner here accomplishes the same task as our first method but in a more concise syntax. The nested iterations are embedded within the sum() function and executed for each element ‘i’ against all subsequent elements ‘j’.
Summary/Discussion
- Method 1: Brute Force Iteration. Simple and easy to understand. However, it has poor performance on large datasets.
- Method 2: Using a Dictionary for Counting. Much more efficient for large arrays with time complexity O(n). Still simple but requires a deeper understanding of hashing.
- Method 4: Using Complex Number Hash. Innovative and efficient. Can be non-intuitive and harder to debug due to the use of complex numbers.
- Bonus One-Liner Method 5: Using List Comprehension and sum(). Impressively compact, but not recommended for large arrays due to poor performance and reduced readability.