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+# How to find prime Numbers efficiently
+
+**Translator: [shazi4399](https://github.com/shazi4399)**
+
+**Author: [labuladong](https://github.com/labuladong)**
+
+The definition of a prime number seems simple,which is said to be prime number if it can be divided by 1 and itself.
+
+However,don't think that the definition of prime numbers is simple. I am afraid that few people can write a prime-related algorithm that works really efficiently. Let's say you write a function like this:
+
+```java
+// Returns several primes in the interval [2, n) 
+int countPrimes(int n)
+
+// E.g. countPrimes (10) returns 4
+// Because 2,3,5,7 is prime numbers
+```
+
+How would you progrma this function? I think you maybe write like this:
+
+```java
+int countPrimes(int n) {
+    int count = 0;
+    for (int i = 2; i < n; i++)
+        if (isPrim(i)) count++;
+    return count;
+}
+
+// Determines whether integer n is prime
+boolean isPrime(int n) {
+    for (int i = 2; i < n; i++)
+        if (n % i == 0)
+            // There are other divisibility factors
+            return false;
+    return true;
+}
+```
+
+The time complexity is O (n ^ 2), which is a big problem.**First of all, the idea of using the isPrime function to assist is not efficient; and even if you want to use the isPrime function, there is computational redundancy in writing the algorithm**.
+
+Let's briefly talk about **how to write an algorithm if you want to determine whether a number is prime or not**. Just slightly modify the for loop condition in the isPrim code above:
+
+```java
+boolean isPrime(int n) {
+    for (int i = 2; i * i <= n; i++)
+        ...
+}
+```
+
+In other words, `i` does not need to traverse to` n`, but only to `sqrt (n)`. Why? let's take an example, suppose `n = 12`.
+
+```java
+12 = 2 × 6
+12 = 3 × 4
+12 = sqrt(12) × sqrt(12)
+12 = 4 × 3
+12 = 6 × 2
+```
+
+As you can see, the last two products are the reverse of the previous two, and the critical point of inversion is at `sqrt (n)`.
+
+In other words, if no divisible factor is found within the interval `[[2, sqrt (n)]`, you can directly conclude that `n` is a prime number, because in the interval `[[sqrt (n), n] ` Nor will you find a divisible factor.
+
+Now, the time complexity of the `isPrime` function is reduced to O (sqrt (N)), ** but we don't actually need this function to implement the` countPrimes` function. The above just hope that readers understand the meaning of `sqrt (n)`, because it will be used again later.
+
+
+### Efficient implementation `countPrimes`
+
+The core idea of efficiently solving this problem is to reverse the conventional idea above:
+
+First from 2, we know that 2 is a prime number, then 2 × 2 = 4, 3 × 2 = 6, 4 × 2 = 8 ... all are not prime numbers.
+
+Then we found that 3 is also a prime number, so 3 × 2 = 6, 3 × 3 = 9, 3 × 4 = 12 ... are also impossible to be prime numbers.
+
+Seeing this, do you understand the logic of this exclusion method a bit? First look at our first version of the code:
+
+```java
+int countPrimes(int n) {
+    boolean[] isPrim = new boolean[n];
+    // Initialize the arrays to true
+    Arrays.fill(isPrim, true);
+
+    for (int i = 2; i < n; i++) 
+        if (isPrim[i]) 
+            // Multiples of i cannot be prime
+            for (int j = 2 * i; j < n; j += i) 
+                    isPrim[j] = false;
+    
+    int count = 0;
+    for (int i = 2; i < n; i++)
+        if (isPrim[i]) count++;
+    
+    return count;
+}
+```
+
+If you can understand the above code, then you have mastered the overall idea, but there are two subtle areas that can be optimized.
+
+First of all, recall the `isPrime` function that just judges whether a number is prime. Due to the symmetry of the factors, the for loop only needs to traverse` [2, sqrt (n)] `. Here is similar, our outer for loop only needs to traverse to `sqrt (n)`:
+
+```java
+for (int i = 2; i * i < n; i++) 
+    if (isPrim[i]) 
+        ...
+```
+
+In addition, it is difficult to notice that the inner for loop can also be optimized. Our previous approach was:
+
+```java
+for (int j = 2 * i; j < n; j += i) 
+    isPrim[j] = false;
+```
+
+This can mark all integer multiples of `i` as` false`, but there is still computational redundancy.
+
+For example, when `n = 25` and` i = 4`, the algorithm will mark numbers such as 4 × 2 = 8, 4 × 3 = 12, and so on, but these two numbers have been marked by 2 × 4 and 3 × 4 that is `i = 2` and` i = 3`.
+
+We can optimize it slightly so that `j` traverses from the square of` i` instead of starting from `2 * i`:
+
+```java
+for (int j = i * i; j < n; j += i) 
+    isPrim[j] = false;
+```
+
+In this way, the algorithm for counting prime numbers is efficiently implemented. In fact, this algorithm has a name, which called Sieve of Eratosthenes. Take a look at the complete final code:
+
+```java
+int countPrimes(int n) {
+    boolean[] isPrim = new boolean[n];
+    Arrays.fill(isPrim, true);
+    for (int i = 2; i * i < n; i++) 
+        if (isPrim[i]) 
+            for (int j = i * i; j < n; j += i) 
+                isPrim[j] = false;
+    
+    int count = 0;
+    for (int i = 2; i < n; i++)
+        if (isPrim[i]) count++;
+    
+    return count;
+}
+```
+
+**The time complexity of this algorithm is difficult to calculate**.It is obvious that the time is related to these two nested for loops. The operands should be:
+
+  n/2 + n/3 + n/5 + n/7 + ...
+= n × (1/2 + 1/3 + 1/5 + 1/7...)
+
+In parentheses, ther is the inverse of the prime number .The final result is  O(N * loglogN),and readers interested in this can refer to the time complexity of the algorithm
+
+That is all about how to find prime Numbers.The seemingly simple problem does has a lot of details to polish
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