Count Primes LeetCode solution in Python

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class Solution:
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    def countPrimes(self, n: int) -> int:
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        # Prerequisite:
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        # What is prime number. What are they just the starting.
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        truth = [
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            True
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        ] * n  # making a list of lenght n. And keep all the values as True.
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        if n < 2:  # as 0 & 1 are not prime numbers.
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            return 0
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        truth[0], truth[1] = (
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            False,
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            False,
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        )  # as we added True in the truth list. So will make false for ) & 1 as they are not prime numbers.
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        i = 2  # As we know 0 & 1 are not prime.
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        while (
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            i * i < n
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        ):  # why we are doing it as i*i here is bcz lets say 5*2 = 10 is divisble by 2 as well as 5 so if 10 is already removed why to traverse a value which is already travered once. so in case of n=5 - 5<5. CONCLUSION : i<sqrt(n)
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            # why we are running the loop till n is bcz question says " prime numbers that are strictly less than n".
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            if truth[i] == True:
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                for j in range(
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                    i * i, n, i
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                ):  # if we have mutiple of a number in the range of n, we have to remove them as they can be prime. i.e 2 is prime, but its multiple in n = 10 are 4,6,8 they cant be prime. So we will make them false(means not a prime).
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                    truth[j] = False
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            i += 1  # increasing our iterator.
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        return truth.count(True)  # will count true value