Description:
Count the number of prime numbers less than a non-negative number, n.
Credits:
Special thanks to @mithmatt for adding this problem and creating all test cases.
Hint:
Let’s start with a isPrime function. To determine if a number is prime, we need to check if it is not divisible by any number less than n. The runtime complexity of isPrime function would be O(n) and hence counting the total prime numbers up to n would be O(n2). Could we do better?
As we know the number must not be divisible by any number > n / 2, we can immediately cut the total iterations half by dividing only up to n / 2. Could we still do better?
Let’s write down all of 12’s factors:
2 × 6 = 12
3 × 4 = 12
4 × 3 = 12
6 × 2 = 12
As you can see, calculations of 4 × 3 and 6 × 2 are not necessary. Therefore, we only need to consider factors up to √n because, if n is divisible by some number p, then n = p × q and since p ≤ q, we could derive that p ≤ √n.
Our total runtime has now improved to O(n1.5), which is slightly better. Is there a faster approach?
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class Solution { public: int countPrimes(int n) { vector<int> book(n, 1); for(int i = 2; i * i < n; i++) { if(book[i] == 1) for(int j = i * i; j < n; j = j + i) book[j] = 0; } int cnt = 0; for(int i = 2; i < n; i++) { if(book[i] == 1) cnt++; } return cnt; } }; |
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