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Problem solution · Python

Find the Count of Good Integers

Find the Count of Good Integers: a Python solution using sorting and greedy selection. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Sorting and greedy selection
Source
walkccc LeetCode Solutions
Length
26 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

Sorting and greedy selection

For Find the Count of Good Integers, the implementation first exposes a useful order, then scans that order while making locally justified choices.

  1. Choose the key that reveals the greedy or grouping structure.
  2. Sort the relevant records by that key.
  3. Scan in order, maintaining the invariant that makes each local choice safe.

Code notes

  • 26 lines of Python from the credited upstream file 3272.py.
  • The implementation visibly relies on hash lookup, ordered lookup.
  • No explicit loop blocks detected.

Complexity

Sorting is typically the dominant term unless the subsequent scan uses a more expensive nested operation.

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 codeFind the Count of Good Integers · PythonPython
Use this to learn the idea, then write your own version.
class Solution:  def countGoodIntegers(self, n: int, k: int) -> int:    halfLength = (n + 1) // 2    minHalf = 10**(halfLength - 1)    maxHalf = 10**halfLength    ans = 0    seen = set()     for num in range(minHalf, maxHalf):      palindrome = str(num) + str(num)[::-1][n % 2:]      sortedDigits = ''.join(sorted(palindrome))      if int(palindrome) % k != 0 or sortedDigits in seen:        continue      seen.add(sortedDigits)      digitCount = collections.Counter(palindrome)      # Leading zeros are not allowed, so the first digit is special.      firstDigitChoices = n - digitCount['0']      permutations = firstDigitChoices * math.factorial(n - 1)      # For each repeated digit, divide by the factorial of the frequency since      # permutations that swap identical digits don't create a new number.      for freq in digitCount.values():        permutations //= math.factorial(freq)      ans += permutations     return ans

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