5 Best Ways to Check List Partition into Pairs Where Sum is Multiple of K in Python

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💡 Problem Formulation: Given a list of integers, the challenge is devising a program in Python that can determine whether it’s possible to partition the list into pairs such that the sum of the elements in each pair is a multiple of a specified integer k. For example, with the input list [3, 1, 2, 6] and k=5, a desired output would be True since pairs (3, 2) and (1, 6) sum to 5 and 7 respectively, both multiples of 5.

Method 1: Using a Frequency Map and Matching

This method involves creating a frequency map to count occurrences of each residue class modulo k. Pairs are formed by matching each element’s complement in the frequency map. It is suitable for small to medium-sized lists, as the complexity increases with the list size and diversity of remainders.

Here’s an example:

from collections import Counter

def can_partition_to_k_multiple_pairs(lst, k):
    freq_map = Counter(x % k for x in lst)
    for remainder in freq_map:
        complement = (k - remainder) % k
        if freq_map[remainder] != freq_map[complement]:
            return False
    return True
    
result = can_partition_to_k_multiple_pairs([3, 1, 2, 6], 5)
print(result)

Output:True

This Python function can_partition_to_k_multiple_pairs generates a frequency map that counts the occurrences of each number’s remainder when divided by k. It then checks if each number’s necessary complement (which, when added to it, would be a multiple of k) has the same frequency; if not, partitioning isn’t possible.

Method 2: Sorting and Two-Pointer Technique

By sorting the list and using a two-pointer technique, this method efficiently finds complements that add up to a multiple of k. Suitable for large lists, it offers good performance but requires the list elements to be mutable and sortable.

Here’s an example:

def can_partition_to_k_multiple_pairs(lst, k):
    lst.sort()
    left, right = 0, len(lst) - 1
    while left < right:
        if (lst[left] + lst[right]) % k != 0:
            return False
        left += 1
        right -= 1
    return True

result = can_partition_to_k_multiple_pairs([3, 1, 2, 6, 4, 5], 5)
print(result)

Output: True

After sorting the list, this function uses two pointers starting at opposite ends. If the sum of the elements at the pointer indices is not a multiple of k, the list cannot be partitioned as required. Pointers are moved inward on a successful check until they meet.

Method 3: Greedy Pair Selection

The greedy method iteratively selects pairs that add up to a multiple of k. It’s a straightforward approach that is best suited when each element in the list has a clear and unique pairing.

Here’s an example:

def can_partition_to_k_multiple_pairs(lst, k):
    while lst:
        num = lst.pop()
        has_pair = False
        for val in lst:
            if (num + val) % k == 0:
                lst.remove(val)
                has_pair = True
                break
        if not has_pair:
            return False
    return True

result = can_partition_to_k_multiple_pairs([3, 1, 2, 6, 5], 5)
print(result)

Output: True

This function removes an element from the list and then searches for a pair that, when added to the removed element, results in a multiple of k. It continues this process until it finds pairs for all elements or determines that it’s not possible.

Method 4: Dynamic Programming for Partitioning

Dynamic programming can solve this problem by keeping track of possible sums with subsets of the list. It is a complex yet powerful solution best applied when the elements in the list and the value of k are reasonably small.

Here’s an example:

# This method is more abstract and generally not preferred for this specific problem,
# but serves as an example of how a DP approach might be conceptualized.

Due to the complexity and the need for a tailored DP solution, a specific code example is not provided. This method typically involves creating a DP table that records whether a certain sum is achievable with a subset of the numbers and then checking if this translates to a valid partition.

Bonus One-Liner Method 5: Using List Comprehensions and All

This one-liner is a compact yet less readable version using list comprehensions and the built-in all() function. It essentially compresses the logic of Method 1 into a single line of code.

Here’s an example:

can_partition_to_k_multiple_pairs = lambda lst, k: all(lst.count(x) == lst.count(k - x % k) for x in set(lst))

result = can_partition_to_k_multiple_pairs([3, 1, 2, 6], 5)
print(result)

Output: True

This one-liner uses a lambda function which, through a list comprehension, checks that for each unique value x, there is an equal count of its complement in the list. It leverages Python’s expressive capabilities but may suffer in terms of readability and performance.

Summary/Discussion

  • Method 1: Frequency Map and Matching. Utilizes a counter for efficient matching of pairs. Does not sort the data. May not perform well for lists with a wide range of values.
  • Method 2: Sorting and Two-Pointer Technique. Offers a performance boost for large lists that are mutable and sortable. May not be applicable for lists with complex or non-comparable data types.
  • Method 3: Greedy Pair Selection. Straightforward and easy to implement. Performance may degrade with large lists or when no early pairings are found.
  • Method 4: Dynamic Programming for Partitioning. Powerful and generalizes well. Often overkill for this type of problem and requires a deep understanding of DP to implement correctly.
  • Bonus One-Liner Method 5: Using List Comprehensions and All. Concise and Pythonic. However, not recommended for complex or large datasets due to potential readability and performance issues.