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I think 1534236469 out of range! Leetcode: 7. Reverse Integer
I can't pass the test input 1534236469. why? the return range is[Integer.MAX_VALUE, Integer.MIN_VALUE],others should return zero
class Solution {
public int reverse(int x) {
if(x > Integer.MAX_VALUE || x < Integer.MIN_VALUE)
return 0;
int ans = 0;
while(x != 0) {
ans = ans * 10 + (x % 10);
x /= 10;
}
return ans;
}
}
Thank for your help
The reverse of 1534236469 is 9646324351, which is larger than Integer.MAX_VALUE, so your code will result in numeric overflow and an incorrect result.
You can use long instead of int to fix the problem.
EDIT:
Your added if(x > Integer.MAX_VALUE || x < Integer.MIN_VALUE) condition is pointless, since x is int, so it will never be outside the valid range of ints.
Even if x is within the valid range, the reverse of x may be outside the range. If you want to detect that the reversed x is too large and return 0, you should use long internally:
class Solution {
public int reverse(int x) {
long ans = 0;
while(x != 0) {
ans = ans * 10 + (x % 10);
x /= 10;
}
if(ans > Integer.MAX_VALUE || ans < Integer.MIN_VALUE) {
return 0;
} else {
return (int) ans;
}
}
}
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I found this code on an old youtube video and am having the hardest time understanding the logical sequence for the output. I was hoping any of you can help clarify.
Let's assume I pass a value of 21 to the getLargestPrime method.
In line 17 the conditional statement checks whether 21 % 2 != 0, that's True. So it carries another iteration of the code. But this time m = 3, which 21 % 3 != 0, that's actually False. Why is the code still executing the else statement? This is question #1.
Question #2
How can the if statement inside the else ever execute? When does number become == 1?
I know that this is probably super basic for you guys but for some reason I cannot follow the sequence in the output.
public class LargestPrime {
public static int getLargestPrime(int number){
if(number < 0){
return -1;
}
int m = 2;
int ans = 0;
int numbern;
if(number == 1){
System.out.println("This number is not a prime");
} else{
while (ans == 0){
if(number % m !=0){
m = m +1;
System.out.println(m + " value of m");
} else {
numbern = number;
number = number / m;
if(number == 1){
System.out.println(numbern + " is the largest prime factor of your number");
ans++;
}
}
}
} return number;
}
}
This is the output:
3 value of m
4 value of m
5 value of m
6 value of m
7 value of m
7 is the largest prime factor of your number
1
21%3!=0 returns false. Because of this the jvm executes the code after else. In e.g. 21%2!=0, that is true, it executes the code after if.
It is 1, if number is as m.
This is when your calculation gets to an end because there cannot be any more numbers after your number.
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How to implement an efficient code in java to verify whether a given number is a prime number or not for number's of size greater than or equal to 12 digits?
Example
Input:
100123456789
Output: prime
Input:
101111111111
Output: prime
Input:
101740496633
Output: prime
Input:
111111111111
Output:not prime
Input:
157639024808
Output: not prime
I tried implementing the following algorithm to find whether it is prime or not. But it is not working taking too long for numbers with greater than or equal to 12 digits.
My Code
public static Boolean checkPrime(long a){
if(a%2==0)
return false;
for(int i=3;i<(int)Math.sqrt(a);i=i+2){
if(a%i==0){
return false;
}
}
return true;
}
'a' is the number to be checked
The above function returns true if the given number is prime or false if the given number is not prime.
How to find whether a given number is prime or not for numbers of size greater than 12?
You can speed this up slightly by only checking 1/3 of values instead 1/2.
public static boolean checkPrime(long a) {
if (a % 2 == 0)
return a == 2;
for (int i = 3; i <= (int) Math.sqrt(a); i = i + 2) {
if (a % i == 0) {
return false;
}
}
return true;
}
public static boolean checkPrime2(long a) {
if (a % 2 == 0 || a % 3 == 0 || a % 5 == 0) {
return a <= 3 || a == 5;
}
for (int i = 6, max = (int) Math.sqrt(a); i <= max; i = i + 6) {
if (a % (i + 1) == 0 | a % (i + 5) == 0) {
return false;
}
}
return true;
}
public static void time(String desc, BooleanSupplier run) {
long start = System.nanoTime();
boolean result = run.getAsBoolean();
if (!result)
throw new AssertionError();
long time = System.nanoTime() - start;
System.out.printf("%s took %.3f mill-seconds%n", desc, time / 1e6);
}
public static void main(String... args) {
for (int i = 2; i < 1000; i++) {
boolean a = checkPrime(i);
boolean b = checkPrime2(i);
if (a != b)
throw new AssertionError(i);
}
for (int i = 0; i < 3; i++) {
time("checkPrime", () -> checkPrime(9999999998987L));
time("checkPrime2", () -> checkPrime2(9999999998987L));
}
}
prints
checkPrime took 26.887 mill-seconds
checkPrime2 took 13.878 mill-seconds
checkPrime took 25.527 mill-seconds
checkPrime2 took 11.286 mill-seconds
checkPrime took 16.799 mill-seconds
checkPrime2 took 9.929 mill-seconds
This is starting to get small enough that it's not clear that multiple threads would help.
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Here is my program, inside for loop Math.sqrt(n) - I think here is error.
import java.util.Scanner;
public class PrimeNoCheck {
public static void main(String[] args) {
Scanner s = new Scanner(System.in);
System.out.print("Enter a number : ");
int n = s.nextInt();
if (isPrime(n)) {
System.out.println(n + " is a prime number");
} else {
System.out.println(n + " is not a prime number");
}
}
public static boolean isPrime(int n) {
if (n <= 1) {
return false;
}
for (int i = 2; i < Math.sqrt(n); i++) {
if (n % i == 0) {
return false;
}
}
return true;
}
}
the algorithm is:
To check if n is a prime number, you have to try to divide it by all the numbers, from 2 (1 is useless), to sqrt(n), included.
You don't have to check after sqrt(n), because, if n can be divided by X,
greater than sqrt(n), it also can be divided by Y (n/X), lesser than sqrt(n), or by Z, lesser than Y, if Y is not prime.
The error in you code is that you stop just before sqrt(n).
With your code 4, 9, 25, ... are seen as prime, and every prime ^2 also.
just Replace that: <= instead of <
for (int i = 2; i <= Math.sqrt(n); i++) {
I am working on a prime factorization program implemented in Java.
The goal is to find the largest prime factor of 600851475143 (Project Euler problem 3).
I think I have most of it done, but I am getting a few errors.
Also my logic seems to be off, in particular the method that I have set up for checking to see if a number is prime.
public class PrimeFactor {
public static void main(String[] args) {
int count = 0;
for (int i = 0; i < Math.sqrt(600851475143L); i++) {
if (Prime(i) && i % Math.sqrt(600851475143L) == 0) {
count = i;
System.out.println(count);
}
}
}
public static boolean Prime(int n) {
boolean isPrime = false;
// A number is prime iff it is divisible by 1 and itself only
if (n % n == 0 && n % 1 == 0) {
isPrime = true;
}
return isPrime;
}
}
Edit
public class PrimeFactor {
public static void main(String[] args) {
for (int i = 2; i <= 600851475143L; i++) {
if (isPrime(i) == true) {
System.out.println(i);
}
}
}
public static boolean isPrime(int number) {
if (number == 1) return false;
if (number == 2) return true;
if (number % 2 == 0) return false;
for (int i = 3; i <= number; i++) {
if (number % i == 0) return false;
}
return true;
}
}
Why make it so complicated? You don't need do anything like isPrime(). Divide it's least divisor(prime) and do the loop from this prime. Here is my simple code :
public class PrimeFactor {
public static int largestPrimeFactor(long number) {
int i;
for (i = 2; i <= number; i++) {
if (number % i == 0) {
number /= i;
i--;
}
}
return i;
}
/**
* #param args
*/
public static void main(String[] args) {
System.out.println(largestPrimeFactor(13195));
System.out.println(largestPrimeFactor(600851475143L));
}
}
edit: I hope this doesn't sound incredibly condescending as an answer. I just really wanted to illustrate that from the computer's point of view, you have to check all possible numbers that could be factors of X to make sure it's prime. Computers don't know that it's composite just by looking at it, so you have to iterate
Example: Is X a prime number?
For the case where X = 67:
How do you check this?
I divide it by 2... it has a remainder of 1 (this also tells us that 67 is an odd number)
I divide it by 3... it has a remainder of 1
I divide it by 4... it has a remainder of 3
I divide it by 5... it has a remainder of 2
I divide it by 6... it has a remainder of 1
In fact, you will only get a remainder of 0 if the number is not prime.
Do you have to check every single number less than X to make sure it's prime? Nope. Not anymore, thanks to math (!)
Let's look at a smaller number, like 16.
16 is not prime.
why? because
2*8 = 16
4*4 = 16
So 16 is divisible evenly by more than just 1 and itself. (Although "1" is technically not a prime number, but that's technicalities, and I digress)
So we divide 16 by 1... of course this works, this works for every number
Divide 16 by 2... we get a remainder of 0 (8*2)
Divide 16 by 3... we get a remainder of 1
Divide 16 by 4... we get a remainder of 0 (4*4)
Divide 16 by 5... we get a remainder of 1
Divide 16 by 6... we get a remainder of 4
Divide 16 by 7... we get a remainder of 2
Divide 16 by 8... we get a remainder of 0 (8*2)
We really only need one remainder of 0 to tell us it's composite (the opposite of "prime" is "composite").
Checking if 16 is divisible by 2 is the same thing as checking if it's divisible by 8, because 2 and 8 multiply to give you 16.
We only need to check a portion of the spectrum (from 2 up to the square-root of X) because the largest number that we can multiply is sqrt(X), otherwise we are using the smaller numbers to get redundant answers.
Is 17 prime?
17 % 2 = 1
17 % 3 = 2
17 % 4 = 1 <--| approximately the square root of 17 [4.123...]
17 % 5 = 2 <--|
17 % 6 = 5
17 % 7 = 3
The results after sqrt(X), like 17 % 7 and so on, are redundant because they must necessarily multiply with something smaller than the sqrt(X) to yield X.
That is,
A * B = X
if A and B are both greater than sqrt(X) then
A*B will yield a number that is greater than X.
Thus, one of either A or B must be smaller than sqrt(X), and it is redundant to check both of these values since you only need to know if one of them divides X evenly (the even division gives you the other value as an answer)
I hope that helps.
edit: There are more sophisticated methods of checking primality and Java has a built-in "this number is probably prime" or "this number is definitely composite" method in the BigInteger class as I recently learned via another SO answer :]
You need to do some research on algorithms for factorizing large numbers; this wikipedia page looks like a good place to start. In the first paragraph, it states:
When the numbers are very large, no efficient integer factorization algorithm is publicly known ...
but it does list a number of special and general purpose algorithms. You need to pick one that will work well enough to deal with 12 decimal digit numbers. These numbers are too large for the most naive approach to work, but small enough that (for example) an approach based on enumerating the prime numbers starting from 2 would work. (Hint - start with the Sieve of Erasthones)
Here is very elegant answer - which uses brute force (not some fancy algorithm) but in a smart way - by lowering the limit as we find primes and devide composite by those primes...
It also prints only the primes - and just the primes, and if one prime is more then once in the product - it will print it as many times as that prime is in the product.
public class Factorization {
public static void main(String[] args) {
long composite = 600851475143L;
int limit = (int)Math.sqrt(composite)+1;
for (int i=3; i<limit; i+=2)
{
if (composite%i==0)
{
System.out.println(i);
composite = composite/i;
limit = (int)Math.sqrt(composite)+1;
i-=2; //this is so it could check same prime again
}
}
System.out.println(composite);
}
}
You want to iterate from 2 -> n-1 and make sure that n % i != 0. That's the most naive way to check for primality. As explained above, this is very very slow if the number is large.
To find factors, you want something like:
long limit = sqrt(number);
for (long i=3; i<limit; i+=2)
if (number % i == 0)
print "factor = " , i;
In this case, the factors are all small enough (<7000) that finding them should take well under a second, even with naive code like this. Also note that this particular number has other, smaller, prime factors. For a brute force search like this, you can save a little work by dividing out the smaller factors as you find them, and then do a prime factorization of the smaller number that results. This has the advantage of only giving prime factors. Otherwise, you'll also get composite factors (e.g., this number has four prime factors, so the first method will print out not only the prime factors, but the products of various combinations of those prime factors).
If you want to optimize that a bit, you can use the sieve of Eratosthenes to find the prime numbers up to the square root, and then only attempt division by primes. In this case, the square root is ~775'000, and you only need one bit per number to signify whether it's prime. You also (normally) only want to store odd numbers (since you know immediately that all even numbers but two are composite), so you need ~775'000/2 bits = ~47 Kilobytes.
In this case, that has little real payoff though -- even a completely naive algorithm will appear to produce results instantly.
I think you're confused because there is no iff [if-and-only-if] operator.
Going to the square root of the integer in question is a good shortcut. All that remains is checking if the number within that loop divides evenly. That's simply [big number] % i == 0. There is no reason for your Prime function.
Since you are looking for the largest divisor, another trick would be to start from the highest integer less than the square root and go i--.
Like others have said, ultimately, this is brutally slow.
private static boolean isPrime(int k) throws IllegalArgumentException
{
int j;
if (k < 2) throw new IllegalArgumentException("All prime numbers are greater than 1.");
else {
for (j = 2; j < k; j++) {
if (k % j == 0) return false;
}
}
return true;
}
public static void primeFactorsOf(int n) {
boolean found = false;
if (isPrime(n) == true) System.out.print(n + " ");
else {
int i = 2;
while (found == false) {
if ((n % i == 0) && (isPrime(i))) {
System.out.print(i + ", ");
found = true;
} else i++;
}
primeFactorsOf(n / i);
}
}
For those answers which use a method isPrime(int) : boolean, there is a faster algorithm than the one previously implemented (which is something like)
private static boolean isPrime(long n) { //when n >= 2
for (int k = 2; k < n; k++)
if (n % k == 0) return false;
return true;
}
and it is this:
private static boolean isPrime(long n) { //when n >= 2
if (n == 2 || n == 3) return true;
if (n % 2 == 0 || n % 3 == 0) return false;
for (int k = 1; k <= (Math.floor(Math.sqrt(n)) + 1) / 6; k++)
if (n % (6 * k + 1) == 0 || n % (6 * k - 1) == 0) return false;
return true;
}
I made this algorithm using two facts:
We only need to check for n % k == 0 up to k <= Math.sqrt(n). This is true because for anything higher, factors merely "flip" ex. consider the case n = 15, where 3 * 5 = 5 * 3, and 5 > Math.sqrt(15). There is no need for this overlap of checking both 15 % 3 == 0 and 15 % 5 == 0, when we could just check one of these expressions.
All primes (excluding 2 and 3) can be expressed in the form (6 * k) + 1 or (6 * k) - 1, because any positive integer can be expressed in the form (6 * k) + n, where n = -1, 0, 1, 2, 3, or 4 and k is an integer <= 0, and the cases where n = 0, 2, 3, and 4 are all reducible.
Therefore, n is prime if it is not divisible by 2, 3, or some integer of the form 6k ± 1 <= Math.sqrt(n). Hence the above algorithm.
--
Wikipedia article on testing for primality
--
Edit: Thought I might as well post my full solution (*I did not use isPrime(), and my solution is nearly identical to the top answer, but I thought I should answer the actual question):
public class Euler3 {
public static void main(String[] args) {
long[] nums = {13195, 600851475143L};
for (num : nums)
System.out.println("Largest prime factor of " + num + ": " + lpf(num));
}
private static lpf(long n) {
long largestPrimeFactor = 1;
long maxPossibleFactor = n / 2;
for (long i = 2; i <= maxPossibleFactor; i++)
if (n % i == 0) {
n /= i;
largestPrimeFactor = i;
i--;
}
return largestPrimeFactor;
}
}
To find all prime factorization
import java.math.BigInteger;
import java.util.Scanner;
public class BigIntegerTest {
public static void main(String[] args) {
BigInteger myBigInteger = new BigInteger("65328734260653234260");//653234254
BigInteger originalBigInteger;
BigInteger oneAddedOriginalBigInteger;
originalBigInteger=myBigInteger;
oneAddedOriginalBigInteger=originalBigInteger.add(BigInteger.ONE);
BigInteger index;
BigInteger countBig;
for (index=new BigInteger("2"); index.compareTo(myBigInteger.add(BigInteger.ONE)) <0; index = index.add(BigInteger.ONE)){
countBig=BigInteger.ZERO;
while(myBigInteger.remainder(index) == BigInteger.ZERO ){
myBigInteger=myBigInteger.divide(index);
countBig=countBig.add(BigInteger.ONE);
}
if(countBig.equals(BigInteger.ZERO)) continue;
System.out.println(index+ "**" + countBig);
}
System.out.println("Program is ended!");
}
}
I got a very similar problem for my programming class. In my class it had to calculate for an inputted number. I used a solution very similar to Stijak. I edited my code to do the number from this problem instead of using an input.
Some differences from Stijak's code are these:
I considered even numbers in my code.
My code only prints the largest prime factor, not all factors.
I don't recalculate the factorLimit until I have divided all instances of the current factor off.
I had all the variables declared as long because I wanted the flexibility of using it for very large values of number. I found the worst case scenario was a very large prime number like 9223372036854775783, or a very large number with a prime number square root like 9223371994482243049. The more factors a number has the faster the algorithm runs. Therefore, the best case scenario would be numbers like 4611686018427387904 (2^62) or 6917529027641081856 (3*2^61) because both have 62 factors.
public class LargestPrimeFactor
{
public static void main (String[] args){
long number=600851475143L, factoredNumber=number, factor, factorLimit, maxPrimeFactor;
while(factoredNumber%2==0)
factoredNumber/=2;
factorLimit=(long)Math.sqrt(factoredNumber);
for(factor=3;factor<=factorLimit;factor+=2){
if(factoredNumber%factor==0){
do factoredNumber/=factor;
while(factoredNumber%factor==0);
factorLimit=(long)Math.sqrt(factoredNumber);
}
}
if(factoredNumber==1)
if(factor==3)
maxPrimeFactor=2;
else
maxPrimeFactor=factor-2;
else
maxPrimeFactor=factoredNumber;
if(maxPrimeFactor==number)
System.out.println("Number is prime.");
else
System.out.println("The largest prime factor is "+maxPrimeFactor);
}
}
public class Prime
{
int i;
public Prime( )
{
i = 2;
}
public boolean isPrime( int test )
{
int k;
if( test < 2 )
return false;
else if( test == 2 )
return true;
else if( ( test > 2 ) && ( test % 2 == 0 ) )
return false;
else
{
for( k = 3; k < ( test/2 ); k += 2 )
{
if( test % k == 0 )
return false;
}
}
return true;
}
public void primeFactors( int factorize )
{
if( isPrime( factorize ) )
{
System.out.println( factorize );
i = 2;
}
else
{
if( isPrime( i ) && ( factorize % i == 0 ) )
{
System.out.print( i+", " );
primeFactors( factorize / i );
}
else
{
i++;
primeFactors( factorize );
}
}
public static void main( String[ ] args )
{
Prime p = new Prime( );
p.primeFactors( 649 );
p.primeFactors( 144 );
p.primeFactors( 1001 );
}
}