Number of Digit One LeetCode solution in Python

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# Runtime: 65 ms (Top 7.05%) | Memory: 13.8 MB (Top 68.36%)
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class Solution:
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    def countDigitOne(self, n: int) -> int:
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        num = str(n)[::-1]
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        count = 0
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        for i in range(len(num) - 1, -1, -1):
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            pv = 10**i  # placevalue
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            # mulitplicity of current digit (how many times it will be repeated)
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            mul = n // (pv * 10)
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            rem = n % pv  # remainder of current place value
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            count += (
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                mul * pv
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            )  # count for number of times 1 occurs in this place when the current digit is considered to be less than 1
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            # if the current digit is greater than 1 then additional number of 1's are added to the count (equivalent to the place value)
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            if num[i] > "1":
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                count += pv
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            # if the current digit is equal to 1 then additional number of 1's are added to the count (equivalent to the number modded by the current place value)
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            if num[i] == "1":
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                count += rem + 1
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        return count