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Problem solution · Python

Smallest Divisible Digit Product II

Smallest Divisible Digit Product II: a Python solution using hash-based lookup. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Hash-based lookup
Source
walkccc LeetCode Solutions
Length
88 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

Hash-based lookup

For Smallest Divisible Digit Product II, the implementation stores previously seen values or frequencies in a hash table for direct membership and lookup operations.

  1. Decide the key that represents the information needed later.
  2. Update its count or stored state while scanning the input.
  3. Use constant-time expected lookups to detect matches or assemble the result.

Code notes

  • 88 lines of Python from the credited upstream file 3348.py.
  • The implementation visibly relies on sequence storage, hash lookup.
  • No explicit loop blocks detected.

Complexity

Expected hash operations are constant time, but the surrounding scan and the number of stored keys determine total work and memory.

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 codeSmallest Divisible Digit Product II · PythonPython
Use this to learn the idea, then write your own version.
FACTOR_COUNTS = {    0: collections.Counter(),    1: collections.Counter(),    2: collections.Counter([2]),    3: collections.Counter([3]),    4: collections.Counter([2, 2]),    5: collections.Counter([5]),    6: collections.Counter([2, 3]),    7: collections.Counter([7]),    8: collections.Counter([2, 2, 2]),    9: collections.Counter([3, 3]),}  class Solution:  def smallestNumber(self, num: str, t: int) -> str:    primeCount, isDivisible = self._getPrimeCount(t)    if not isDivisible:      return '-1'     factorCount = self._getFactorCount(primeCount)    if sum(factorCount.values()) > len(num):      return ''.join(factor * freq for factor, freq in factorCount.items())     primeCountPrefix = sum((FACTOR_COUNTS[int(c)]                            for c in num), start=collections.Counter())    firstZeroIndex = next((i for i, d in enumerate(num) if d == '0'), len(num))    if firstZeroIndex == len(num) and primeCount <= primeCountPrefix:      return num     for i, c in reversed(list(enumerate(num))):      d = int(c)      # Remove the current digit's factors from primeCountPrefix.      primeCountPrefix -= FACTOR_COUNTS[d]      spaceAfterThisDigit = len(num) - 1 - i      if i <= firstZeroIndex:        for biggerDigit in range(d + 1, 10):          # Compute the required factors after replacing with a larger digit.          factorsAfterReplacement = self._getFactorCount(              primeCount - primeCountPrefix - FACTOR_COUNTS[biggerDigit]          )          # Check if the replacement is possible within the available space.          if sum(factorsAfterReplacement.values()) <= spaceAfterThisDigit:            # Fill extra space with '1', if any, and construct the result.            fillOnes = spaceAfterThisDigit - sum(                factorsAfterReplacement.values())            return (                num[:i]  # Keep the prefix unchanged.                + str(biggerDigit)  # Replace the current digit.                + '1' * fillOnes  # Fill remaining space with '1'.                + ''.join(factor * freq for factor,                          freq in factorsAfterReplacement.items())            )     # No solution of the same length exists, so we need to extend the number    # by prepending '1's and adding the required factors.    factorCount = self._getFactorCount(primeCount)    return (        '1' * (len(num) + 1 - sum(factorCount.values()))        + ''.join(factor * freq for factor, freq in factorCount.items())    )   def _getPrimeCount(self, t: int) -> tuple[dict[int, int], bool]:    """    Returns the count of prime factors of t and if t is divisible by 2, 3, 5, 7.    """    count = collections.Counter()    for prime in [2, 3, 5, 7]:      while t % prime == 0:        t //= prime        count[prime] += 1    return count, t == 1   def _getFactorCount(self, count: dict[int, int]) -> dict[str, int]:    """Returns the required factors to form the smallest number."""    count8, remaining2 = divmod(count[2], 3)  # 2^3 = 8    count9, count3 = divmod(count[3], 2)  # 3^2 = 9    count4, count2 = divmod(remaining2, 2)  # 2^2 = 4    # Combine 2 and 3 to 6 if both are present.    count2, count3, count6 = ((0, 0, 1) if count2 == 1 and count3 == 1                              else (count2, count3, 0))    # Combine 3 and 4 to 2 and 6 if both are present.    count2, count6, count3, count4 = ((1, 1, 0, 0)                                      if count3 == 1 and count4 == 1                                      else (count2, count6, count3, count4))    return {'2': count2, '3': count3, '4': count4, '5': count[5],            '6': count6, '7': count[7], '8': count8, '9': count9}

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