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Problem solution · Python

Numbers With Repeated Digits

Numbers With Repeated Digits: 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
36 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 Numbers With Repeated Digits, 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

  • 36 lines of Python from the credited upstream file 1012.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 codeNumbers With Repeated Digits · PythonPython
Use this to learn the idea, then write your own version.
class Solution:  def numDupDigitsAtMostN(self, n: int) -> int:    return n - self._countSpecialNumbers(n)   # Same as 2376. Count Special Integers  def _countSpecialNumbers(self, n: int) -> int:    s = str(n)     @functools.lru_cache(None)    def dp(i: int, used: int, tight: bool) -> int:      """      Returns the number of special integers, considering the i-th digit, where      `used` is the bitmask of the used digits, and `tight` indicates if the      current digit is tightly bound.      """      if i == len(s):        return 1       res = 0      maxDigit = int(s[i]) if tight else 9       for d in range(maxDigit + 1):        # `d` is used.        if used >> d & 1:          continue        # Use `d` now.        nextTight = tight and (d == maxDigit)        if used == 0 and d == 0:  # Don't count leading 0s as used.          res += dp(i + 1, used, nextTight)        else:          res += dp(i + 1, used | 1 << d, nextTight)       return res     return dp(0, 0, True) - 1  # - 0

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