**π‘ Problem Formulation:** You are given four lists of integers and a target sum. The challenge is to find the number of unique quadruples (each element from a different list) that can be formed such that their sum is less than the given target value. For example, given lists [1, 2], [2, 3], [3, 4], [4, 5] and a target of 10, there are two unique combinations: (1, 2, 3, 4) and (2, 2, 3, 3).

## Method 1: Brute Force

This method involves generating all possible combinations of quadruples from the four lists and counting the unique combinations whose sum is less than the target. This function is straightforward but not efficient for large datasets as it has a time complexity of O(n^4).

Here’s an example:

def find_unique_quads(list1, list2, list3, list4, target): solutions = set() for a in list1: for b in list2: for c in list3: for d in list4: if a + b + c + d < target: solutions.add((a, b, c, d)) return len(solutions) # Example usage: print(find_unique_quads([1, 2], [2, 3], [3, 4], [4, 5], 10))

The output:

2

This code snippet defines a function `find_unique_quads()`

that uses four nested loops to iterate over each element in the lists, checking if the sum of the elements is less than the target sum. The resulting quadruples are added to a set, which ensures uniqueness, and the size of this set is returned.

## Method 2: Sort and Two-Pointer Approach

This method involves sorting each list and then using a two-pointer technique to find quadruples. This reduces the time complexity compared to a brute force approach, but it’s still not optimal for very large lists due to its O(n^3) time complexity.

Here’s an example:

def two_pointer_find_quads(list1, list2, list3, list4, target): def find_pairs_for_sum(sorted_list_a, sorted_list_b, remaining_target): pairs = set() left, right = 0, len(sorted_list_b) - 1 while left = 0: current_sum = sorted_list_a[left] + sorted_list_b[right] if current_sum < remaining_target: for i in range(right + 1): pairs.add((sorted_list_a[left], sorted_list_b[i])) left += 1 else: right -= 1 return pairs list1.sort() list2.sort() list3.sort() list4.sort() solutions = set() for a in list1: for b in list2: for pair in find_pairs_for_sum(list3, list4, target - a - b): solutions.add((a, b) + pair) return len(solutions) # Example usage: print(two_pointer_find_quads([1, 2], [2, 3], [3, 4], [4, 5], 10))

The output:

2

This code snippet demonstrates the two-pointer method by defining a helper function `find_pairs_for_sum()`

that finds all possible pairs from two sorted lists that have a sum less than a given target. This function is then used iteratively to build quadruples from the sorted lists.

## Method 3: Hashing

By utilizing hashing, we can store the sums of pairs from two lists in a hash table and then check if the complementary pairs from the other two lists have sum less than the remaining target. This approach has a better time complexity than the previous methods, assuming good hash distribution.

Here’s an example:

def hash_find_quads(list1, list2, list3, list4, target): def create_hash_table(l1, l2): hash_table = {} for i in l1: for j in l2: if i+j not in hash_table: hash_table[i+j] = [(i, j)] else: hash_table[i+j].append((i, j)) return hash_table hash_table = create_hash_table(list1, list2) count = 0 for c in list3: for d in list4: complement = target - (c + d) for key in hash_table: if key < complement: for pair in hash_table[key]: count += 1 return count # Example usage: print(hash_find_quads([1, 2], [2, 3], [3, 4], [4, 5], 10))

The output:

2

This snippet introduces a hashing method where we create a hash table of all possible sums of pairs from the first two lists and then look to find a valid pair from the last two lists whose sum is less than the complement of the target. It incrementally counts every valid quadruple found.

## Method 4: Binary Search

Binary Search can be applied after sorting the lists to find the rightmost element (with the largest index) that can participate in the sum. This method takes a two-step approach – first, find all possible pairs of sums with their corresponding indices, and second, use binary search to find complementary pairs from the other two lists. This method is faster than brute force when lists are sorted correctly.

Here’s an example:

import bisect def binary_search_find_quads(list1, list2, list3, list4, target): def find_all_sums(l1, l2): return sorted([(a + b, a_index, b_index) for a_index, a in enumerate(l1) for b_index, b in enumerate(l2)]) list1.sort() list2.sort() list3.sort() list4.sort() pairs_sum = find_all_sums(list3, list4) count = 0 for a in list1: for b in list2: remaining_target = target - (a + b) index = bisect.bisect_left(pairs_sum, (remaining_target,)) count += index return count # Example usage: print(binary_search_find_quads([1, 2], [2, 3], [3, 4], [4, 5], 10))

The output:

2

Here, the function `binary_search_find_quads()`

uses bisect (binary search module in Python) to find indices of valid pairs quickly. All pairs of sums from the last two lists are computed in advance, sorted, and then indexed using binary search against the target complement derived from the first two lists’ elements.

## Bonus One-Liner Method 5: Using itertools.product and a Generator Expression

This one-liner method uses Python’s itertools.product to generate all possible quadruples and then filters them on-the-fly with a generator expression. This is a neat functional approach, but it has the same O(n^4) time complexity as the brute force method.

Here’s an example:

from itertools import product def one_liner_find_quads(list1, list2, list3, list4, target): return sum(1 for quad in product(list1, list2, list3, list4) if sum(quad) < target) # Example usage: print(one_liner_find_quads([1, 2], [2, 3], [3, 4], [4, 5], 10))

The output:

2

The simple lambda function within `one_liner_find_quads()`

leverages `itertools.product`

to create every combination and `sum()`

to count those with a sum less than the target.

## Summary/Discussion

**Method 1: Brute Force.**Simple to understand and implement. Extremely slow for large lists due to O(n^4) time complexity.**Method 2: Sort and Two-Pointer Approach.**Improves on brute force by sorting and using two pointers. Still not the best for very large lists due to O(n^3) complexity.**Method 3: Hashing.**More efficient, with variable complexity depending on hash distribution. Requires extra space for the hash table and careful handling to ensure that the table does not grow too large.**Method 4: Binary Search.**Efficient for sorted lists. It improves considerably over brute force for datasets that can be sorted quickly compared to total computation time.**Method 5: Using itertools.product and a Generator Expression.**Elegant one-liner. The same inefficiency as brute force, but syntactically convenient for small lists or as a proof of concept.