Solution Approach

The implementation uses bitmask dynamic programming to count all good subsets:

1. Initialize Constants and Data Structures:

• Define primes = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29] - all primes up to 30
• Use Counter to count frequency of each number in nums
• Create DP array f of size 2^10 (1024) where f[mask] represents the number of ways to form subsets using exactly the primes indicated by mask

2. Base Case:

• f[0] = pow(2, cnt[1]) - The empty prime set (no primes used) can be formed by taking any subset of 1s from the array. Since 1 doesn't contribute any prime factors, we have 2^cnt[1] ways.

3. Process Each Valid Number (2 to 30):

for x in range(2, 31):  if cnt[x] == 0 or x % 4 == 0 or x % 9 == 0 or x % 25 == 0:  continue

• Skip if the number doesn't appear in nums or contains repeated prime factors

4. Calculate Prime Mask for Current Number:

mask = 0 for i, p in enumerate(primes):  if x % p == 0:  mask |= 1 << i

• Create a bitmask representing which primes divide x
• If prime p at index i divides x, set bit i in the mask

5. Update DP States:

for state in range((1 << n) - 1, 0, -1):  if state & mask == mask:  f[state] = (f[state] + cnt[x] * f[state ^ mask]) % mod

• Iterate through states in reverse order (to avoid using updated values in the same iteration)
• state & mask == mask checks if state contains all primes in mask (i.e., we can remove mask from state)
• state ^ mask removes the primes of x from the current state, giving us the previous state
• We add cnt[x] * f[state ^ mask] because we can choose any of the cnt[x] occurrences of x to add to subsets represented by f[state ^ mask]

6. Calculate Final Answer:

return sum(f[i] for i in range(1, 1 << n)) % mod

• Sum all non-empty prime combinations (f[1] through f[1023])
• Exclude f[0] as it represents the empty subset (or subsets containing only 1s, which have no prime factors)

The algorithm efficiently counts all possible good subsets by tracking which primes have been used, ensuring no prime appears more than once in any subset's product.

Example Walkthrough

Let's walk through a small example with nums = [1, 2, 3, 4].

Step 1: Count frequencies

• cnt[1] = 1, cnt[2] = 1, cnt[3] = 1, cnt[4] = 1

Step 2: Initialize DP array

• f[0] = 2^1 = 2 (we have one 1 in the array, so 2 ways: include it or not)
• All other f[mask] = 0 initially

Step 3: Process valid numbers (2 to 30)

Processing x = 2:

• 2 is valid (not divisible by 4, 9, or 25)
• Prime mask: 2 = 2^1, so mask = 0b0000000001 (bit 0 set, representing prime 2)
• Update states:
  • For state = 0b0000000001 (has prime 2):
    • Check: state & mask == mask? Yes (0b01 & 0b01 = 0b01)
    • Previous state: state ^ mask = 0b01 ^ 0b01 = 0b00
    • Update: f[0b01] = (0 + 1 * f[0b00]) % mod = 1 * 2 = 2

Processing x = 3:

• 3 is valid
• Prime mask: 3 = 3^1, so mask = 0b0000000010 (bit 1 set, representing prime 3)
• Update states:
  • For state = 0b0000000011 (has primes 2 and 3):
    • Check: state & mask == mask? Yes (0b11 & 0b10 = 0b10)
    • Previous state: state ^ mask = 0b11 ^ 0b10 = 0b01
    • Update: f[0b11] = (0 + 1 * f[0b01]) % mod = 1 * 2 = 2
  • For state = 0b0000000010 (has prime 3):
    • Check: state & mask == mask? Yes (0b10 & 0b10 = 0b10)
    • Previous state: state ^ mask = 0b10 ^ 0b10 = 0b00
    • Update: f[0b10] = (0 + 1 * f[0b00]) % mod = 1 * 2 = 2

Processing x = 4:

• 4 = 2² is NOT valid (divisible by 4)
• Skip this number

Step 4: Calculate final answer

• Sum all non-empty states: f[0b01] + f[0b10] + f[0b11] = 2 + 2 + 2 = 6

Verification of the 6 good subsets:

1. [2] → product = 2 (prime factorization: 2¹) ✓
2. [1, 2] → product = 2 (prime factorization: 2¹) ✓
3. [3] → product = 3 (prime factorization: 3¹) ✓
4. [1, 3] → product = 3 (prime factorization: 3¹) ✓
5. [2, 3] → product = 6 (prime factorization: 2¹ × 3¹) ✓
6. [1, 2, 3] → product = 6 (prime factorization: 2¹ × 3¹) ✓

Note that subsets containing 4 like [4], [1, 4], [2, 4], etc., are NOT good because 4 = 2² has a repeated prime factor.

The DP approach efficiently counts these by tracking which primes are used (via bitmask) and ensuring no prime appears twice.

Time and Space Complexity

Time Complexity: O(2^n * (n + 30)) where n = 10 (number of primes)

• Counting frequency of numbers in nums takes O(len(nums)) time
• Computing pow(2, cnt[1]) takes O(log(cnt[1])) time
• The outer loop iterates through numbers 2 to 30, which is O(30) iterations
• For each valid number x, we compute its prime mask in O(n) time by checking divisibility with each prime
• The inner loop iterates through all states from (1 << n) - 1 to 1, which is O(2^n) iterations
• Each state update operation takes O(1) time
• The final summation iterates through 2^n - 1 states
• Since n = 10 is a constant, the overall time complexity is O(2^10 * 40) = O(1024 * 40) which simplifies to O(1) for constant bounds, but more precisely O(2^n * n) for the algorithmic analysis

Space Complexity: O(2^n) where n = 10

• The primes array uses O(n) space
• The cnt Counter uses O(30) space at most (for numbers 1-30)
• The f array uses O(2^n) space to store DP states for all possible subsets of primes
• Since n = 10, the space complexity is O(2^10) = O(1024), which is effectively O(1) for constant bounds, but more precisely O(2^n) for the algorithmic analysis

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrect Handling of the Number 1

Pitfall: A common mistake is treating 1 like any other number when updating the DP states. Some might try to process 1 in the main loop along with numbers 2-30, which would incorrectly update the DP states since 1 has no prime factors.

Why it's wrong: The number 1 is special because:

• It has no prime factors (empty prime factorization)
• It can be freely added to any good subset without affecting whether the subset remains good
• Multiple 1s can be included without violating the square-free property

Correct approach:

# Handle 1s separately in the base case dp[0] = pow(2, count[1], MOD) # 2^count[1] ways to choose 1s  # Then process numbers 2-30 in the main loop for num in range(2, 31):  # ... process numbers with actual prime factors

2. Not Checking for Numbers with Repeated Prime Factors

Pitfall: Forgetting to skip numbers that inherently have repeated prime factors (like 4, 8, 9, 12, 16, 18, 20, 24, 25, 27, 28).

Why it's wrong: These numbers can never be part of a good subset because their prime factorization already contains repeated primes:

• 4 = 2² (prime 2 appears twice)
• 9 = 3² (prime 3 appears twice)
• 12 = 2² × 3 (prime 2 appears twice)

Correct approach:

# Must check for all perfect square factors up to 30 if num % 4 == 0 or num % 9 == 0 or num % 25 == 0:  continue # Skip numbers with repeated prime factors

3. Updating DP States in Wrong Order

Pitfall: Updating DP states in forward order (from 0 to 2^n - 1) instead of reverse order.

Why it's wrong: If we update in forward order, we might use already-updated values from the current iteration when calculating later states, leading to incorrect counting (elements being counted multiple times).

Correct approach:

# Update in reverse order to ensure we use values from previous iteration for state in range((1 << num_primes) - 1, 0, -1):  if state & prime_mask == prime_mask:  dp[state] = (dp[state] + count[num] * dp[state ^ prime_mask]) % MOD

4. Incorrectly Checking Prime Mask Compatibility

Pitfall: Using state & prime_mask == 0 to check if a number can be added to a state, or using state | prime_mask incorrectly.

Why it's wrong:

• state & prime_mask == 0 would check if there's NO overlap, but we're updating states that already contain the primes
• We need to check if the current state can have the prime_mask "removed" from it

Correct approach:

# Check if state contains all primes in prime_mask if state & prime_mask == prime_mask:  # state ^ prime_mask removes those primes from state  dp[state] = (dp[state] + count[num] * dp[state ^ prime_mask]) % MOD

5. Forgetting to Apply Modulo Operations

Pitfall: Not applying modulo at every arithmetic operation, especially when calculating powers or sums.

Why it's wrong: The numbers can grow very large, causing integer overflow. Even in Python which handles big integers, not using modulo can slow down calculations significantly.

Correct approach:

# Use modulo in pow function dp[0] = pow(2, count[1], MOD) # Third parameter applies modulo efficiently  # Apply modulo after each addition dp[state] = (dp[state] + count[num] * dp[state ^ prime_mask]) % MOD  # Apply modulo to final sum result = sum(dp[i] for i in range(1, 1 << num_primes)) % MOD