Counting Potential Attack Vectors in Network Keys Based on Queries

Problem Understanding:

We are given an array keys and a 2D array queries. We need to find potential attack vectors (i, j, x) where keys[i] and keys[j] are divisible by x. Additionally, each query in queries specifies a range [a, b] within which x should fall.

Initial Breakdown:

  1. Loop through queries to handle each query individually.
  2. Loop through keys to identify pairs (i, j) where both elements are divisible by x.
  3. Filter the pairs based on the query range.

Subproblem Refinement:

  1. Parse a query [a, b] to identify the range for x.
  2. Generate all possible pairs (i, j) from keys.
  3. Check if keys[i] and keys[j] are divisible by x, where x lies within [a, b].

Task Identification:

  1. Parse a query to identify the range [a, b].
  2. Generate all possible pairs (i, j).
  3. Check divisibility of keys[i] and keys[j] by x.
  4. Check if x lies within [a, b].

Task Abstraction:

  1. Parsing the query: Extract a and b.
  2. Pair generation: Loop twice through keys array.
  3. Divisibility Check: Use the modulus operation.
  4. Range Check: Use conditional statements.

Method Naming:

  1. parseQuery
  2. generatePairs
  3. checkDivisibility
  4. validateRange

Subproblem Interactions:

  1. Parsing the query precedes all other operations.
  2. Pair generation can occur in parallel with range validation.
  3. Divisibility checks depend on the generated pairs and the query range.

Code:

Here is the code:

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def count_attack_vectors(n, keys, q, queries):
    results = []
    for query in queries:
        a, b = query
        count = 0
        for i in range(n):
            for j in range(n):
                for x in range(a, b+1):
                    if keys[i] % x == 0 and keys[j] % x == 0:
                        count += 1
        results.append(count)
    return results

n = 3
keys = [6, 9, 10]
q = 2
queries = [[2, 3], [9, 10]]

print(count_attack_vectors(n, keys, q, queries))

This code has a time complexity of O(n^3 * q) which is not optimal for larger arrays.

Optimization Steps:

  1. Pre-compute all pairs of divisible keys to avoid recomputation for each query.
  2. Use a more efficient data structure to store precomputed pairs for faster lookup.

By precomputing the divisible pairs, you could reduce the time complexity substantially, perhaps to O(n^2 + q).

Optimization Step 1: Reduce the Number of Iterations for x

In the original code, the innermost loop iterates through the range [a, b] for each query. This is inefficient. A first step in optimization is to precompute the factors for all keys, so that you can look them up instead of calculating them during each query.

Here’s how the code can be revised:

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from collections import defaultdict

def count_attack_vectors(n, keys, q, queries):
    # Precompute factors for all keys
    factors_dict = defaultdict(int)
    for key in keys:
        for i in range(1, int(key ** 0.5) + 1):
            if key % i == 0:
                factors_dict[i] += 1
                if i != key // i:
                    factors_dict[key // i] += 1

    results = []
    for query in queries:
        a, b = query
        count = 0
        for x in range(a, b + 1):
            count += factors_dict.get(x, 0) ** 2
        results.append(count)
    
    return results

n = 3
keys = [6, 9, 10]
q = 2
queries = [[2, 3], [9, 10]]

print(count_attack_vectors(n, keys, q, queries))

How the Optimization Works

  1. The factors_dict dictionary holds the count of keys that can be divided by each factor. This avoids having to compute the factors within each query, thus saving time.

  2. In each query, you only iterate through the range [a, b] once, checking the precomputed factors_dict.

This change significantly reduces the number of iterations, thereby improving the algorithm’s time complexity.

Optimization Step 2: Pre-calculate the Number of Pairs

After calculating the factors for each key, we can also pre-calculate the number of pairs (i, j) that can be formed by each factor, further reducing the need for nested loops in the query phase.

Here’s the updated code:

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from collections import defaultdict

def count_attack_vectors(n, keys, q, queries):
    # Precompute factors for all keys
    factors_count = defaultdict(int)
    for key in keys:
        for i in range(1, int(key ** 0.5) + 1):
            if key % i == 0:
                factors_count[i] += 1
                if i != key // i:
                    factors_count[key // i] += 1

    # Pre-calculate the number of pairs (i, j)
    pairs_count = {}
    for factor, count in factors_count.items():
        pairs_count[factor] = count * count
    
    results = []
    for query in queries:
        a, b = query
        count = 0
        for x in range(a, b + 1):
            count += pairs_count.get(x, 0)
        results.append(count)
    
    return results

n = 3
keys = [6, 9, 10]
q = 2
queries = [[2, 3], [9, 10]]

print(count_attack_vectors(n, keys, q, queries))

How the Second Optimization Works

  1. The pairs_count dictionary precalculates the number of pairs (i, j) that can be formed by each factor. This is done by simply squaring the count of keys that can be divided by that factor. This avoids having to iterate through all (i, j) pairs for each query.

  2. During each query, the code now only iterates through the range [a, b] once, using both factors_count and pairs_count.

With both optimizations, the algorithm should be considerably faster and more efficient.