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Problem solution · Python

The Number of Good Subsets

The Number of Good Subsets: a Python solution using dynamic programming. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Dynamic programming
Source
walkccc LeetCode Solutions
Length
27 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Dynamic programming

For The Number of Good Subsets, the implementation records answers for smaller states and reuses them to build the requested result without repeating work.

  1. Define precisely what one DP state represents.
  2. Establish the base cases before transitions are evaluated.
  3. Process states in dependency order and combine only already-known values.

Code notes

  • 27 lines of Python from the credited upstream file 1994.py.
  • The implementation visibly relies on sequence storage, hash lookup, cached states.
  • No explicit loop blocks detected.

Complexity

Multiply the number of reachable states by the work performed for each transition, then include the stored state table in memory usage.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeThe Number of Good Subsets · PythonPython
Use this to learn the idea, then write your own version.
class Solution:  def numberOfGoodSubsets(self, nums: list[int]) -> int:    MOD = 1_000_000_007    primes = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29]    n = 1 << len(primes)    # dp[i] := the number of good subsets with set of primes = i bit mask    dp = [1] + [0] * (n - 1)    count = collections.Counter(nums)     for num, freq in count.items():      if num == 1:        continue      if any(num % squared == 0 for squared in [4, 9, 25]):        continue      numPrimesMask = sum(1 << i                          for i, prime in enumerate(primes)                          if num % prime == 0)      for primesMask in range(n):        # Skip since there're commen set of primes (becomes invalid subset)        if primesMask & numPrimesMask > 0:          continue        nextPrimesMask = numPrimesMask | primesMask        dp[nextPrimesMask] += dp[primesMask] * freq        dp[nextPrimesMask] %= MOD     return (1 << count[1]) * sum(dp[1:]) % MOD

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