Count Primes – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
Given an integer n, return the number of prime numbers that are strictly less than n.
Example 1:
Input: n = 10 Output: 4 Explanation: There are 4 prime numbers less than 10, they are 2, 3, 5, 7.
Example 2:
Input: n = 0 Output: 0
Example 3:
Input: n = 1 Output: 0
Constraints:
0 <= n <= 5 * 106
C++ Count Primes LeetCode Solution
int countPrimes(int n) {
if (n<=2) return 0;
vector<bool> passed(n, false);
int sum = 1;
int upper = sqrt(n);
for (int i=3; i<n; i+=2) {
if (!passed[i]) {
sum++;
//avoid overflow
if (i>upper) continue;
for (int j=i*i; j<n; j+=i) {
passed[j] = true;
}
}
}
return sum;
}
Java Count Primes LeetCode Solution
public class Solution {
public int countPrimes(int n) {
boolean[] notPrime = new boolean[n];
int count = 0;
for (int i = 2; i < n; i++) {
if (notPrime[i] == false) {
count++;
for (int j = 2; i*j < n; j++) {
notPrime[i*j] = true;
}
}
}
return count;
}
}
Python 3 Count Primes LeetCode Solution
class Solution:
# @param {integer} n
# @return {integer}
def countPrimes(self, n):
if n < 3:
return 0
primes = [True] * n
primes[0] = primes[1] = False
for i in range(2, int(n ** 0.5) + 1):
if primes[i]:
primes[i * i: n: i] = [False] * len(primes[i * i: n: i])
return sum(primes)
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