5 Best Ways to Count Maximum Number of Distinct Pairs with Differences Larger Than a Target in Python

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πŸ’‘ Problem Formulation: Imagine you have an array of integers and you want to find the maximum number of unique pairs in which the absolute difference between each pair is greater than a specified target value. For example, given the list [1, 5, 3, 4, 2] and a target difference of 2, the desired output is 2 because there are two such pairs: (1, 4) and (1, 5).

Method 1: Using a Nested Loop

This approach involves iterating through the list with two nested loops, comparing each possible pair to determine if their difference exceeds the target. We can use a set to store the pairs to ensure their uniqueness.

Here’s an example:

def count_distinct_pairs(nums, target):
    pairs = set()
    for i in range(len(nums)):
        for j in range(i+1, len(nums)):
            if abs(nums[i] - nums[j]) > target:
                pairs.add((min(nums[i], nums[j]), max(nums[i], nums[j])))
    return len(pairs)

# Example usage
print(count_distinct_pairs([1, 5, 3, 4, 2], 2))

The output of this code snippet:

2

This nested loop method iterates over each element and its subsequent elements to check the difference between them. If the difference is greater than the target, the pair is added to a set which automatically handles the uniqueness. Finally, we return the number of unique pairs that have been found.

Method 2: Using Sorting

First, we sort the list of numbers. This will allow us to then iterate through the list just once, using two pointers to find pairs that meet the criteria, as once sorted, we can “slide” the pointers as needed to find qualifying pairs.

Here’s an example:

def count_distinct_pairs(nums, target):
    nums.sort()
    count = 0
    i, j = 0, 1
    while j < len(nums):
        if nums[j] - nums[i] > target:
            count +=1
            i += 1
        j += 1
    return count

# Example usage
print(count_distinct_pairs([1, 5, 3, 4, 2], 2))

The output of this code snippet:

2

By sorting the numbers first, we can efficiently find pairs by using two pointers. The first pointer starts at the beginning of the array, and the second starts just after the first. If the difference between the numbers at these pointers is greater than the target, we increment the count and move the first pointer up by one. Then, we continue to move the second pointer and repeat the process until we have compared every element.

Method 3: Using Hash Table

This method uses a hash table (implemented in Python as a dictionary) to store the numbers. As we iterate over the list, we check if there’s already a number in the hash table that, when paired with the current number, would have a difference larger than the target. We further employ a set to keep count of distinct pairs.

Here’s an example:

def count_distinct_pairs(nums, target):
    seen = {}
    pairs = set()
    for num in nums:
        if num - target in seen:
            pairs.add((num-target, num))
        if num + target in seen:
            pairs.add((num, num+target))
        seen[num] = True
    return len(pairs)

# Example usage
print(count_distinct_pairs([1, 5, 3, 4, 2], 2))

The output of this code snippet:

2

In this approach, we keep a hash table to record the existence of each number as we iterate through the list. For each number, we look for its potential partners that are greater or less than it by the target difference in the hash table. If a partner is found, we add the pair to a set to keep count of distinct pairs.

Method 4: Using Binary Search

After sorting the array of numbers, we perform a binary search for each number to find a pair with a difference greater than the target. This uses the fact that a sorted list can be efficiently searched to find potential partners.

Here’s an example:

import bisect

def count_distinct_pairs(nums, target):
    nums.sort()
    pairs = set()
    for i, num in enumerate(nums):
        partner_index = bisect.bisect_left(nums, num + target + 1)
        if partner_index < len(nums):
            pairs.add((num, nums[partner_index]))
    return len(pairs)

# Example usage
print(count_distinct_pairs([1, 5, 3, 4, 2], 2))

The output of this code snippet:

2

This method benefits from the sorted list property by finding potential partners through binary search, which significantly improves performance for larger lists. The `bisect_left` function finds the index where an element with a value greater than `num + target` should be inserted, effectively finding a partner for `num`.

Bonus One-Liner Method 5: List Comprehensions with a Set

This one-liner uses a list comprehension within a set to generate all pairs that satisfy the difference criterion in a concise and functional way.

Here’s an example:

def count_distinct_pairs(nums, target):
    return len({(min(x, y), max(x, y)) for x in nums for y in nums if x != y and abs(x - y) > target})

# Example usage
print(count_distinct_pairs([1, 5, 3, 4, 2], 2))

The output of this code snippet:

2

This one-liner works by generating all possible pairs using a set comprehension to ensure distinctness. We only include a pair if their absolute difference is greater than the target and they are not equal (to avoid pairs of the same element).

Summary/Discussion

  • Method 1: Using a Nested Loop. Simple implementation. Quadratic complexity may lead to performance issues with large lists.
  • Method 2: Using Sorting. More efficient than nested loop for large data sets, thanks to sorted array properties. However, it can still be somewhat inefficient since it does not use the full advantage of sorted arrays for reducing the search space.
  • Method 3: Using Hash Table. Efficient for data with a broad range of values. Might consume more memory because of hash table and set usage.
  • Method 4: Using Binary Search. Offers the best performance for large, sorted data sets. Requires understanding of binary search and sorted array manipulation.
  • Bonus Method 5: List Comprehensions with a Set. Elegant and concise one-liner. Has a clear disadvantage in terms of performance due to the full pair generation for large lists.