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Problem solution · Python

Count Ways to Make Array With Product

Count Ways to Make Array With Product: 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
46 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 Count Ways to Make Array With Product, 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

  • 46 lines of Python from the credited upstream file 1735.py.
  • The implementation visibly relies on sequence storage, hash lookup.
  • 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 codeCount Ways to Make Array With Product · PythonPython
Use this to learn the idea, then write your own version.
class Solution:  def waysToFillArray(self, queries: list[list[int]]) -> list[int]:    MOD = 1_000_000_007    MAX = 10_000    minPrimeFactors = self._sieveEratosthenes(MAX + 1)     @functools.lru_cache(None)    def fact(i: int) -> int:      return 1 if i <= 1 else i * fact(i - 1) % MOD     @functools.lru_cache(None)    def inv(i: int) -> int:      return pow(i, MOD - 2, MOD)     @functools.lru_cache(None)    def nCk(n: int, k: int) -> int:      return fact(n) * inv(fact(k)) * inv(fact(n - k)) % MOD     ans = []     for n, k in queries:      res = 1      for freq in self._getPrimeFactorsCount(k, minPrimeFactors).values():        res = res * nCk(n - 1 + freq, freq) % MOD      ans.append(res)     return ans   def _sieveEratosthenes(self, n: int) -> list[int]:    """Gets the minimum prime factor of i, where 1 < i <= n."""    minPrimeFactors = [i for i in range(n + 1)]    for i in range(2, int(n**0.5) + 1):      if minPrimeFactors[i] == i:  # `i` is prime.        for j in range(i * i, n, i):          minPrimeFactors[j] = min(minPrimeFactors[j], i)    return minPrimeFactors   def _getPrimeFactorsCount(self, num: int, minPrimeFactors: list[int]) -> dict[int, int]:    count = collections.Counter()    while num > 1:      divisor = minPrimeFactors[num]      while num % divisor == 0:        num //= divisor        count[divisor] += 1    return count

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