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Maximum Element-Sum of a Complete Subset of Indices

Difficulty: Hard

Problem Description

You are given a 1-indexed array nums. Your task is to select a complete subset from nums where every pair of selected indices multiplied is a perfect square, i.e. if you select a_i and a_j, i * j must be a perfect square. Return the sum of the complete subset with the maximum sum.

Key Insights

  • The product of two indices i and j is a perfect square if the prime factorization of both indices has even exponents for all prime factors.
  • We can categorize indices based on their prime factors and their exponents.
  • The maximum sum can be derived from indices that share the same prime factorization characteristics.

Space and Time Complexity

Time Complexity: O(n * sqrt(n))
Space Complexity: O(n)

Solution

To solve the problem, we can use the following approach:

  1. Prime Factorization: For each index, determine its prime factorization and group indices based on their prime factors. This allows us to find compatible indices whose products yield perfect squares.

  2. Grouping Indices: Use a hash map or similar structure to group indices that have the same prime factorization signature.

  3. Calculating Maximum Sum: For each group, calculate the sum of the corresponding values in nums and keep track of the maximum sum encountered.

This approach efficiently narrows down the potential subsets and calculates the maximum sum based on the properties of perfect squares.

Code Solutions

def max_element_sum(nums):
    from collections import defaultdict
    import math

    def prime_factors(n):
        factors = {}
        d = 2
        while d * d <= n:
            while (n % d) == 0:
                if d in factors:
                    factors[d] += 1
                else:
                    factors[d] = 1
                n //= d
            d += 1
        if n > 1:
            factors[n] = 1
        return factors

    groups = defaultdict(int)
    for i in range(len(nums)):
        index = i + 1
        factors = prime_factors(index)
        signature = frozenset((p, e % 2) for p, e in factors.items() if e % 2)  # Keep only odd exponents
        groups[signature] += nums[i]

    return max(groups.values())

# Example usage:
print(max_element_sum([8, 7, 3, 5, 7, 2, 4, 9]))  # Output: 16
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