RodneyShag Java solution - passes 100% of test cases
From my HackerRank solutions.
import static java.lang.System.in; class Prime { void checkPrime(int... numbers) { for (int num : numbers) { if (isPrime(num)) { System.out.print(num + " "); } } System.out.println(); } boolean isPrime(int n) { if (n < 2) { return false; } else if (n == 2) { return true; } else if (n % 2 == 0) { return false; } int sqrt = (int) Math.sqrt(n); for (int i = 3; i <= sqrt; i += 2) { if (n % i == 0) { return false; } } return true; } }
Let me know if you have any questions.
Subhrodeep Hi, can you please explain why are you giving the 3rd "return true" statement in the isPrime method?
RodneyShag Hi. The idea is to see if our number n is divisible by any numbers (not including 1 and n). If it's divisible by any such numbers, it is not prime, and we return false. If it's not divisible by any of these numbers, we come out of the for loop, and return true since our number is prime.
Subhrodeep Thanks :)
lin1000 Hi, do you think this version will be faster? wanna hear your thought.
class Prime { void checkPrime(int... numbers) { for (int num : numbers) { if (isPrime(num)) { System.out.print(num + " "); } } System.out.println(); } boolean isPrime(int n) { if (n < 2) { return false; } for (int i = 2; i <= (int) Math.sqrt(n) && isPrime(i); i++) { if (n % i == 0) { return false; } } return true; } }
RodneyShag That will be slower. The extra isPrime() check is much slower then the 3 lines of code in the for loop in isPrime(). You can actually test out both versions and see the execution times in your submissions.
keerthikumar_n Yes.. the basic thing is you have reduced the no. of loops :)
Desolution Hi there, maybe eliminating those who can be divided by 2 will be better?
after checking if n == 2, we can examine from 3, and 4, 6, 8, 10, ... are not needed.
(n could be divided by 2 if it can be divided by 4, 6, 8, 10, ...)
Since then, we could increase our count by 2.
boolean isPrime(int n){ if (n == 2) return true; if (n < 2 || n % 2 == 0) return false; for (int i = 3; i <= (int) Math.sqrt(n); i += 2){ if (n % i == 0) return false; } return true; }
RodneyShag Your solution is twice as fast as mine. Updated mine to match. Thanks.
aadhilr Can you comment on this code?
boolean isPrime(int n){ if (n < 2) return false; for (int i = 2; i <=n/2; i++){ if (n % i == 0) return false; } return true; }
RodneyShag Nice job solving the problem. It's runtime is O(n). If you want a faster solution, see my original comment above, since it's O(n^(1/2)) runtime.
aadhilr Thank you for the help!
divy23 [deleted]
hripsime_ghazar1 [deleted]
wzrds3 Since the locked code involves repeat checks of primailty, I factored in some caching so the code doesn't have to repeatedly check values.
class Prime{ private static Set<Integer> prime_list = new HashSet<>(); public void checkPrime(int ... nums) { for (int num : nums) { if (prime_list.contains(num)) { System.out.print(num + " "); } else if (isPrime(num)) { prime_list.add(num); System.out.print(num + " "); } } System.out.println(); } private static boolean isPrime(int n) { if (n < 2) return false; else if (n < 4) return true; else if (n % 2 == 0) return false; int sqrt = (int) Math.sqrt(n); for (int i = 3; i <= sqrt; i += 2) { if (n % i == 0) return false; } return true; } }
gidan24 Can you explain me why you check all odd numbers until square root of n? I don't understand why square root and not n/2..
Anyway this is my working solution:
private boolean isPrime(int n) { if (n < 2) { return false; } for (int i = 2; i <= n / 2; i++) { if (n % i == 0) { return false; } } return true; }
RodneyShag because sqrt(n) * sqrt(n) = n.
For example, if you want the factors of 100.
1 * 100 2 * 50 4 * 25 5 * 20 10 * 10 <--- after this point the numbers (a,b) switch to become (b,a) 20 * 5 25 * 4 50 * 2 100 * 1
samarthbais1 [deleted]
llanesjavier The test case are bad because are printing unnecesary blanks at the end of the lines. I lost many point on this issue.
npsrihari98 use system.out.print instead of system.out.println for printing numbers
jagacode Solution.java:25: error: cannot find symbol BufferedReader br=new BufferedReader(new InputStreamReader(in)); ^ symbol: variable in location: class Solution 1 error
MYcode:
import java.io.; import java.util.; import java.text.; import java.math.; import java.util.regex.; import java.lang.reflect.; class Prime{ void checkPrime(int... a) { int flag = 0; for(int i=0;i
public static void main(String[] args) { try{ BufferedReader br=new BufferedReader(new InputStreamReader(in)); int n1=Integer.parseInt(br.readLine()); int n2=Integer.parseInt(br.readLine()); int n3=Integer.parseInt(br.readLine()); int n4=Integer.parseInt(br.readLine()); int n5=Integer.parseInt(br.readLine()); Prime ob=new Prime(); ob.checkPrime(n1); ob.checkPrime(n1,n2); ob.checkPrime(n1,n2,n3); ob.checkPrime(n1,n2,n3,n4,n5); Method[] methods=Prime.class.getDeclaredMethods(); Set<String> set=new HashSet<>(); boolean overload=false; for(int i=0;i<methods.length;i++) { if(set.contains(methods[i].getName())) { overload=true; break; } set.add(methods[i].getName()); } if(overload) { throw new Exception("Overloading not allowed"); } } catch(Exception e) { System.out.println(e); } }}
alexandrustana2 import static java.lang.System.in;
This should fix the problem
jagacode Thanks bub
jagacode Why static?...can you explain?
ratan_singh98 because we are calling in public static void main(String[] args)
alexandrustana2 Because you used new BufferedReader(new InputStreamReader(in)); like this. the "in" is a static import from System.in
If you had not used static import the above statement would look like this new BufferedReader(new InputStreamReader(System.in));
nguyentuonglong1 thanks bro
skandaman This was v helpful. thanks
keerthikumar_n Thanks a lot :)
ozanapaydin You must import static java.lang.System.in, and you must have to import with "static" , It tells JVM that to create this from proram start until program ends. Also you can try if you wonder, you can not use any other variable name beside "in" on line written 'BufferedReader'
Singhalr31 can u just explain in detail how would static java.lang.System.in help??? i mean what it actually does??
Kotoamatsukami import static java.lang.System.*;
class Prime {
public void checkPrime(int... p){ boolean prime; for(int i=0;i<p.length;i++){ prime = true; int n = (int)java.lang.Math.sqrt(p[i]); if(p[i]>1) for(int j=2;j<n+1;j++){ if(p[i]%j==0){ prime = false; break; } } else prime = false; if(prime) { out.print(p[i]); out.print(" "); } } out.println(); }}
tao_zhang Mine works
import static java.lang.System.in; class Prime { void checkPrime(int... args){ String outPut = ""; for (int arg : args){ if (isPrime(arg)){ outPut += arg+" "; } } if (outPut != ""){ outPut = outPut.substring(0,outPut.length()-1); System.out.print(outPut); } System.out.println(); } boolean isPrime(int num){ if (num < 2){ return false; } for (int i = 2; i < (num/2)+1; i++){ if (num%i==0){ return false; } } return true; } }
rafa_fr Very nice the import static.
[deleted] Can improve
for (int i = 2; i < (num/2)+1; i++){ if (num%i==0){ return false; } }
as
for (int i = 2; i*i <= num; i++){ if (num%i==0){ return false; } }
will be faster
vlad_belon To avoid the i*i calculus in each loop:
long maxDivider = Math.round(Math.sqrt(number)); for (int divider = 2; divider <= maxDivider; divider++) { if (number % divider == 0) { return false; } }
RodneyShag Hi. For your isPrime() function, I see you iterated up to n/2. This gives you a runtime of O(n). Try iterating to square root of n like this for an O(n^(1/2)) runtime.
Hope this helps.
Chandra_Ganesh [deleted]
kalbatprog7 Sir how does the line checkprime(int... args), work? I am new to all this and I do not know what is int...? Please help!
rjvpillai364 It is basically called varargs or variable length arguments in java. The argument is specified with 3 periods(...) after the type. This syntax tells the compiler that the method can be called with 0 or more arguments. One thing to note here is that your method cannot have 2 varargs as arguments and varargs must always be last in method with more than 1 arguments. For example , void method(int... n1,int... n2) is invalid. void method(int... n1,int n2) is invalid, but void method(int n1,int... n2) is valid. It is logical since the compiler knows how many arguments to assign to varargs, thats why it should always be written last in the argument list
kalbatprog7 Thanks sir
psvishnu [deleted]
ericwilk7 This unit test is failing, anyone know why?
7
18
2
5
10
Output:
7
7
7 2
7 2 5
jarryd999 Mine is failing test cases, with results like yours. So I d/l cases and put the input into the "custom test input" and hit run code and I get different results that look successful. Any thoughts?
llanesjavier Test cases arebad. Included blank spaces at the end.
gustavoamaro Use String.split() method to remove end blank space ;)
kushalsharma0074 class Prime{ void checkPrime(int... a){
for(int i:a){ int check=0; for(int j=2;j<=i/2;j++){ if(i%j==0){ check=1; } } if(i==2 || i==3 ){ System.out.print(i+" "); } else if(check==0 && i!=1){ System.out.print(i+" "); } } lets try to ubderstand this System.out.println(); }}
abhijit_tambe import static java.lang.System.in; class Prime{ boolean p(int x){ if(x<=1) return false; for(int i=2;i<x;i++){ if(x%i==0) return false; } return true; } void checkPrime(int... a){ for(int x :a){ if(p(x)) System.out.print(x+" "); } System.out.println(); } }
manideepgarlapa1 class Prime{ public void checkPrime(int... intArgs){ int count=0; for(int i : intArgs){ if(i!=1){ count=0; for(int j=2;j<=i/2;j++){ if(i%j==0){ count++; }
} if(count==0){ System.out.print(i+" "); } } } System.out.println(); }}
jamestkirk84 class Prime { public void checkPrime(int...integers) { for (int number: integers) { if (checkInteger(number, 2)) System.out.print(number + " "); } System.out.println(); } private boolean checkInteger(int number, int divisor) { if (number == 1) return false; if (number/2 < divisor) return true; if (number % divisor == 0) return false; return checkOneInteger(number, divisor+1); } }
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