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Problem solution · Python

Count K-Reducible Numbers Less Than N

Count K-Reducible Numbers Less Than N: 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
34 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 K-Reducible Numbers Less Than N, 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

  • 34 lines of Python from the credited upstream file 3352.py.
  • The implementation visibly relies on 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 codeCount K-Reducible Numbers Less Than N · PythonPython
Use this to learn the idea, then write your own version.
class Solution:  def countKReducibleNumbers(self, s: str, k: int) -> int:    MOD = 1_000_000_007    ops = self._getOps(s)     @functools.lru_cache(None)    def dp(i: int, setBits: int, tight: bool) -> int:      """      Returns the number of positive integers less than n that are k-reducible,      considering the i-th digit, where `setBits` is the number of set bits in      the current number, and `tight` indicates if the current digit is      tightly bound.      """      if i == len(s):        return int(ops[setBits] < k and not tight)       res = 0      maxDigit = int(s[i]) if tight else 1       for d in range(maxDigit + 1):        nextTight = tight and (d == maxDigit)        res += dp(i + 1, setBits + d, nextTight)        res %= MOD      return res     return dp(0, 0, True) - 1  # - 0   def _getOps(self, s: str) -> int:    """Returns the number of operations to reduce a number to 0."""    ops = [0] * (len(s) + 1)    for num in range(2, len(s) + 1):      ops[num] = 1 + ops[num.bit_count()]    return ops

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