-
Notifications
You must be signed in to change notification settings - Fork 0
/
PrimeSum.java
63 lines (56 loc) · 1.31 KB
/
PrimeSum.java
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
package math;
import java.util.ArrayList;
import java.util.Arrays;
/**
* Given an even number ( greater than 2 ), return two prime numbers whose sum will be equal to given number.
* <p>
* <strong>NOTE</strong> A solution will always exist.
* Read <a href="https://en.wikipedia.org/wiki/Goldbach%27s_conjecture">Goldbach�s conjecture</a>.
* <p>
* <strong>Example:</strong>
*
* <pre>
* Input : 4
* Output: 2 + 2 = 4
* </pre>
*
* If there are more than one solutions possible, return the lexicographically smaller solution.
*
* <pre>
* If [a, b] is one solution with a <= b,
* and [c,d] is another solution with c <= d, then
*
* [a, b] < [c, d]
*
* If a < c OR a==c AND b < d.
* </pre>
*
* @author Tzipora Ziegler
* @see https://www.interviewbit.com/problems/prime-sum/
*/
public class PrimeSum {
public ArrayList<Integer> primesum(int a) {
int num1 = 2;
int num2 = a - num1;
while (num1 < num2) {
if (isPrime(num1) && isPrime(num2)) {
break;
}
num1++;
num2 = a - num1;
}
return new ArrayList<Integer>(Arrays.asList(num1, num2));
}
public boolean isPrime(int num) {
// check if n is a multiple of 2
if (num % 2 == 0 && num != 2) {
return false;
}
for (int i = 3; i <= Math.sqrt(num); i += 2) {
if (num % i == 0) {
return false;
}
}
return true;
}
}