How do you solve Count the Number of Square-Free Subsets on LeetCode?

Count the Number of Square-Free Subsets (LeetCode #2572) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * k), where k is the number of unique primes up to 30. time complexity and O(n) space complexity. Key insight: Understanding square-free integers is crucial.

What is the time complexity of Count the Number of Square-Free Subsets?

The optimal solution for Count the Number of Square-Free Subsets runs in O(n * k), where k is the number of unique primes up to 30. time and uses O(n) space. This complexity is efficient because we avoid recalculating products and only track unique prime factors, significantly reducing the number of checks needed.

What is the brute force solution for Count the Number of Square-Free Subsets?

The brute force approach for Count the Number of Square-Free Subsets is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves generating all possible non-empty subsets of the array and checking if the product of each subset is square-free. This is straightforward but inefficient for larger arrays due to the exponential growth of subsets. The optimal Optimal Solution approach improves this to O(n * k), where k is the number of unique primes up to 30..

What are the interview tips for Count the Number of Square-Free Subsets?

Explain your thought process clearly. Discuss edge cases, like duplicates or small arrays. Practice bit manipulation techniques.

What are common mistakes when solving Count the Number of Square-Free Subsets?

Not checking for square-free conditions correctly. Overcomplicating the subset generation process.

#2572

Count the Number of Square-Free Subsets

Medium

ArrayMathDynamic ProgrammingBit ManipulationNumber TheoryBitmaskBit ManipulationDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n * k), where k is the number of unique primes up to 30.
Space	O(1)	O(n)

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Intuition

Time O(n * k), where k is the number of unique primes up to 30.Space O(n)

The optimal solution uses bit manipulation and dynamic programming to efficiently count square-free subsets. By tracking prime factors and their occurrences, we can avoid recalculating products for each subset.

⚙️

Algorithm

3 steps

1Step 1: Precompute the prime numbers up to a certain limit (e.g., 30) and their corresponding masks.
2Step 2: For each number in nums, calculate its prime factorization and create a bitmask representing the primes involved.
3Step 3: Use a dynamic programming approach to count subsets while ensuring that no prime factor appears more than once.

solution.py31 lines

1# Full working Python code
2from collections import defaultdict
3
4MOD = 10**9 + 7
5
6def prime_factors_mask(num):
7    mask = 0
8    for i in range(2, 31):
9        if num % i == 0:
10            count = 0
11            while num % i == 0:
12                num //= i
13                count += 1
14            if count > 1:
15                return -1  # Not square-free
16            mask |= (1 << (i - 2))  # Set the bit for prime i
17    return mask
18
19def count_square_free_subsets(nums):
20    n = len(nums)
21    dp = defaultdict(int)
22    dp[0] = 1  # Empty subset
23    for num in nums:
24        mask = prime_factors_mask(num)
25        if mask == -1:
26            continue
27        for m in list(dp.keys()):
28            if m & mask == 0:  # No common prime factors
29                dp[m | mask] = (dp[m | mask] + dp[m]) % MOD
30    return (sum(dp.values()) - 1) % MOD  # Exclude the empty subset
31

ℹ

Complexity note: This complexity is efficient because we avoid recalculating products and only track unique prime factors, significantly reducing the number of checks needed.

1Understanding square-free integers is crucial.
2Efficiently managing prime factors can reduce complexity.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.