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Count how many integers in [l, r] are not "special", where a "special" number has exactly two proper divisors — equivalently, it is a prime square (has exactly three total divisors). So the answer is (r−l+1) minus the number of primes p with p^2 in [l, r] (i.e., primes in [ceil(sqrt(l)), floor(sqrt(r))]).
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Leetcode 3233. Find the Count of Numbers Which Are Not Special
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