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Problem solution · Python

Count the Number of Powerful Integers

Count the Number of Powerful Integers: a Python solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Direct simulation
Source
Kamyu LeetCode Solutions
Length
70 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

Direct simulation

For Count the Number of Powerful Integers, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 70 lines of Python from the credited upstream file count-the-number-of-powerful-integers.py.
  • The implementation keeps its working state in language-native values and containers.
  • No explicit loop blocks detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from Kamyu LeetCode Solutions by kamyu104 and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeCount the Number of Powerful Integers · PythonPython
Use this to learn the idea, then write your own version.
# Time:  O(logf)# Space: O(1) # math, combinatoricsclass Solution(object):    def numberOfPowerfulInt(self, start, finish, limit, s):        """        :type start: int        :type finish: int        :type limit: int        :type s: str        :rtype: int        """        def count(x):            def length(x):                result = 0                while x:                    x //= 10                    result += 1                return result             result = 0            n = length(x)            base = 10**n            l = n-len(s)            cnt = (limit+1)**l            for i in xrange(l):                base //= 10                curr = x//base%10                cnt //= limit+1                result += (min(curr-1, limit)-0+1)*cnt                if curr > limit:                    break            else:                if x%base >= int(s):                    result += 1            return result         return count(finish)-count(start-1)  # Time:  O(logf)# Space: O(logf)# math, combinatoricsclass Solution2(object):    def numberOfPowerfulInt(self, start, finish, limit, s):        """        :type start: int        :type finish: int        :type limit: int        :type s: str        :rtype: int        """        def count(x):            result = 0            str_x = str(x)            l = len(str_x)-len(s)            cnt = (limit+1)**l            for i in xrange(l):                cnt //= limit+1                result += (min(int(str_x[i])-1, limit)-0+1)*cnt                if int(str_x[i]) > limit:                    break            else:                if int(str_x[-len(s):]) >= int(s):                    result += 1            return result         return count(finish)-count(start-1)

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