Source: http://linuxwave.blogspot.com/2007/07/ssh-without-password.html
I was asked this question and I found this here when I googled for the same:
The init() method is typically used to perform servlet initialization–creating or loading objects that are used by the servlet in the handling of its requests. Why not use a constructor instead? Well, in JDK 1.0 (for which servlets were originally written), constructors for dynamically loaded Java classes (such as servlets) couldn’t accept arguments. So, in order to provide a new servlet any information about itself and its environment, a server had to call a servlet’s init() method and pass along an object that implements the ServletConfig interface. Also, Java doesn’t allow interfaces to declare constructors. This means that the javax.servlet.Servlet interface cannot declare a constructor that accepts a ServletConfig parameter. It has to declare another method, like init(). It’s still possible, of course, for you to define constructors for your servlets, but in the constructor you don’t have access to the ServletConfig object or the ability to throw a ServletException.
I started digging deep into java code where Servlet classes are loaded and found the code in org.apache.catalina.core.StandardWrapper.loadServlet().
It goes like this:
So if you put in any servlet initialization code in servlet’s default constructor, it should not use ServletConfig and ServletContext objects. This will throw NullpointerException, since ServerConfig information is not initialized the Servlet till this point. However, if you want to perform any initialization which does not refer to ServletContext / ServletConfig, you can safely do so in the default constructor.
Note: Any constructor other than the default constructor is useless as it is never called by the servlet-container.
Write a function to remove the duplicate items in-place (return the new size of the array)
Solution:
Complexity:
Complexity = complexity(sort()) + Complexity(remove_duplicates())
= O(N) + Complexity(remove_duplicates())
Hence depending on whether you use Bubble Sort or Merge Sort (?) the
overall complexity is going to be O(N^2) or O(NlogN)
Python Code:
#!/usr/bin/python
# Simple bubble sort for demo, O(N^2)
# You can use a better sorting algo
def sort( arr ):
L=len(arr)-1
for i in range(L,0,-1):
for j in range (0,i):
if arr[j] > arr[j+1]:
t = arr[j]
arr[j] = arr[j+1]
arr[j+1] = t
return arr
# Complexity O(N)
def remove_duplicates( arr ):
length = len(arr)
i=0;
while i < (length-1):
# check if two consequtive
# elements are equal ...
if arr[i] == arr[i+1] :
#delete next element
arr.pop(i+1)
#get new size of array
length = len(arr)
else:
# else move on to next
# element
i = i + 1
return arr
# Execute
a = [1,3,6,1,5,5,0,9,6,8,9,7]
arr = remove_duplicates(sort(a))
print arr
This is an easy one.
For machines it is always simpler to do bit-shifting than multiply.
What is bit-shifting one place to left, physically ?
Bit shifting one place right means multiplication by 2.
Example:
0010 << 1 = 0100, ( 2 * 2 = 4 ) 0100 << 1 = 1000, ( 4 * 2 = 8 ) 0011 << 1 = 0110, ( 3 * 2 = 6 )
Similarly, bit-shifting one place right means diuvision by 2.
Now what if you shit each bit 2 places ? It is simply bitshiting each bit by 1, 2 times – right ? Which is nothing but multiplying by 2 twice!
Hence, bit-shifting a number 2 places left multiplies it by 4.
That’s ok, but how is it related to multiplication by 7 ?
Elementary Watson!
For any given number N,
N*7 = N*( 1 + 2 + 4)
= N + N*2 + N*4
= N + N<<1 + N<<2
Where << is the bitwise left shift operator.
Here is example python code:
>>> def multiply_by_seven( n ): ... return (n<<1) + (n<<2) +n; ... >>> multiply_by_seven(2) 14 >>> multiply_by_seven(3) 21 >>> multiply_by_seven(7) 49 >>>
Similarly you can break multiplication by any number into powers of 2 and perform an efficient multiplication.
NOTE: Akshay just walked past me and said “Well, just multiply by 8 and subtract 1″ 
How do you efficiently right-shift an array with the least memory usage?
For example, the original array could be 2 3 4 5 6. after right-shift two positions, it should be 5 6 2 3 4. The function prototype is shift_right (int[] arret, int shiftCount).
OR
Algorithm to right-shift an array M places, where M < length(array)
Solution 1:
Here is the exact solution as given in “Programming Pearls, 2nd Edition” :
“Swap all the numbers M times”
For example: Say you have an array[1,2,3,4] and you have to shift
it right 2 places. The
--- Start ---
[1,2,3,4]
--- First Pass ---
[1,2,3,4]
^ ^
[2,1,3,4]
^ ^
[2,3,1,4]
^ ^
[2,3,4,1]
--- Second Pass ---
[2,3,4,1]
^ ^
[3,2,4,1]
^ ^
[3,4,2,1]
^ ^
[3,4,1,2]
--- Result ---
[3,4,1,2]
Complexity:
Not good! Take the worstcase. Say you have an array of length `n` and you have to shift all elements `n-1` places.
The complexity becomes `O(N*(N-1)) ≈ O(N^2)` as `n -> ∞`.
In average case it will be O(mN), where `m` is the number of places to shift.
Solution 2:
Yet another solution.
Step 1: reverse the array by swapping all the elements (single pass, O(N))
Step 2: reverse all the elements from 0 to `M-1`
Step 3: reverse all the elements from `M` to `L-1` (where L is the length of the array)
Array : [1,2,3,4,5], M=3
First pass : [5,4,3,2,1] (all elements swapped reversing the array)
Second pass : [3,4,5,2,1] (reverse array from 0 to 3-1=2)
^ ^ ^
Third Pass : [3,4,5,1,2] (reverse elements 3 to 5)
^ ^
Result : [4,5,1,2,3]
As simple as that !!
Java Code: [ source: sun java fourums ]
int[] reverse(int[] a, int i, int j) { // reverse elements i, i+1 ... j-1
for (; --j > i; i++) {
int t= a[i]; a[i]= a[j]; a[j]= t;
}
return a; // convenience.
}
int[] shiftRight(int[] a, int i) {
return reverse(reverse(reverse(a, 0, a.length), 0, i), i, a.length);
}
Complexity: O(N)
Basically you traverse over the array twice, hence the complexity turns out to be O(2N) ≈ O(N)` as `n -> ∞`.
Problem : Given an unsorted array of integers, find all integer pairs in array which add up to 200.
Solution:
The fundamental is you have the greatest number in the lot in `head`. Now you start adding increments to it (starting from the smallest increment). If the result is smaller than the target number, you need to increase the increment and vice-versa.
Complexity:
Since assymptotically O(N^2) dominates O(N), the complexity of this solution is O(N^2) [ BAD! BAD!! BAD!!! ]
One good trick is to implement quikSort instead of bubbleSort. In that case the complexity becomes :
O(N) + O(NlogN) = O(NlogN)
I implemented bubble sort just for demonstration purposes. The optimized way is to sort it with quick sort.
My Tip: If you still don’t understand the fundamentals of this solution, use this text as a hint and try to do it on a piece of paper. It’s hard to explain but easier to realize 
package org.simplesoft.programmingproblems;
/**
* @author Amit Kumar Sharma
*/
public class Arrays {
/**
* Method to find all numbers in a sorted array
* of integers which add up to number `N`.
* For example "find all numbers in an array
* of integers which add up to 200"
* @param arr: Sorted array of integers
* @param n: number to which add up to
*/
public static void findNumbers(int[] arr, int n) {
// array is assumed to be sorted in descending order
int head = 0;
int tail = arr.length - 1;
int num1, num2;
while (tail != head - 1) {
num1 = arr[head];
num2 = arr[tail];
if (num1 + num2 > n) {
head++;
} else if (num1 + num2 < n) {
tail--;
} else if (num1 + num2 == n) {
System.out.println(num1 + " + " + num2 + " = " + n);
head++;
tail--;
}
}
}
/*
* Bubble sort of complexity O(n^2)
* If you push in quick sort here - the complexity of
* the overall problem becomes O(NlogN)
*/
public static void bubbleSort(int[] arr) {
for (int i = (arr.length - 1); i > 1; i--) {
for (int j = 0; j < i; j++) {
if (arr[j] < arr[j + 1]) {
// swap
int tmp = arr[j];
arr[j] = arr[j + 1];
arr[j + 1] = tmp;
}
}
}
}
/**
* Utility method to print integer arrays
*/
public static void echoArray(int[] arr) {
for (int i : arr) {
System.out.print(i + ", ");
}
System.out.println();
}
public static void main(String[] args) {
int[] arr = {
1, 8, 9,
23, 32, 16,
44, 83, 92,
62, 51, 12,
48, 50, 10,
22, 38, 15
};
echoArray(arr);
bubbleSort(arr);
findNumbers(arr, 60);
}
}
damn! damn!! damn!!!
I was developing a webservice using Apache-CXF (2.1.1) and Spring (2.5.5) using wsdl-first-approach.
All went well, wsdl validation passed, code compiled, war created, webservice hosted on apache-tomcat 6. And then as I tried to access the wsdl url, I got this error :
Apr 8, 2009 4:37:15 AM org.apache.catalina.core.StandardWrapperValve invoke ---- 8< ------------------------------------------------------ SEVERE: Servlet.service() for servlet CXFServlet threw exception javax.wsdl.WSDLException: WSDLException: faultCode=OTHER_ERROR: Can't find prefix for 'http://prediction.simplesoft.org/2009/xsd'. Namespace prefixes must be set on the Definition object using the addNamespace(...) method. at com.ibm.wsdl.util.xml.DOMUtils.getPrefix(Unknown Source) at com.ibm.wsdl.util.xml.DOMUtils.getQualifiedValue(Unknown Source) at com.ibm.wsdl.util.xml.DOMUtils.printQualifiedAttribute(Unknown Source) at com.ibm.wsdl.xml.WSDLWriterImpl.printParts(Unknown Source) at com.ibm.wsdl.xml.WSDLWriterImpl.printMessages(Unknown Source) at com.ibm.wsdl.xml.WSDLWriterImpl.printDefinition(Unknown Source) at com.ibm.wsdl.xml.WSDLWriterImpl.writeWSDL(Unknown Source) at com.ibm.wsdl.xml.WSDLWriterImpl.getDocument(Unknown Source) at org.apache.cxf.transport.http.WSDLQueryHandler.writeResponse(WSDLQueryHandler.java:168) at org.apache.cxf.transport.servlet.ServletController.invoke(ServletController.java:147) at org.apache.cxf.transport.servlet.AbstractCXFServlet.invoke(AbstractCXFServlet.java:174) at org.apache.cxf.transport.servlet.AbstractCXFServlet.doGet(AbstractCXFServlet.java:156) at javax.servlet.http.HttpServlet.service(HttpServlet.java:617) at javax.servlet.http.HttpServlet.service(HttpServlet.java:717) at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:290) at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:206) at org.springframework.orm.hibernate3.support.OpenSessionInViewFilter.doFilterInternal(OpenSessionInViewFilter.java:198) at org.springframework.web.filter.OncePerRequestFilter.doFilter(OncePerRequestFilter.java:76) at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:235) at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:206) at org.apache.catalina.core.StandardWrapperValve.invoke(StandardWrapperValve.java:233) at org.apache.catalina.core.StandardContextValve.invoke(StandardContextValve.java:191) at org.apache.catalina.core.StandardHostValve.invoke(StandardHostValve.java:128) at org.apache.catalina.valves.ErrorReportValve.invoke(ErrorReportValve.java:102) at org.apache.catalina.core.StandardEngineValve.invoke(StandardEngineValve.java:109) at org.apache.catalina.connector.CoyoteAdapter.service(CoyoteAdapter.java:286) at org.apache.coyote.http11.Http11Processor.process(Http11Processor.java:845) at org.apache.coyote.http11.Http11Protocol$Http11ConnectionHandler.process(Http11Protocol.java:583) at org.apache.tomcat.util.net.JIoEndpoint$Worker.run(JIoEndpoint.java:447) at java.lang.Thread.run(Thread.java:619) ---- 8< ------------------------------------------------------
I searched and searched through google and internet, found nothing that could fix this issue.
I finally found the silly bug : pretty improbable, but true:
The targetNamespace annotations for my endpoint (interface) and implementation class were different. This can not be, so I simply made them same and it all worked like cake-walk!
I am pasting the erred source below for all to see ( Don’t laugh 
)
The Endpoint >>
/**
* This class was generated by Apache CXF 2.1.1
* Fri Apr 03 17:22:14 IST 2009
* Generated source version: 2.1.1
*
*/
@WebService(targetNamespace = "http://prediction.simplesoft.org/ws", name = "PredictionPortType")
@XmlSeeAlso({org.simplesoft.prediction._2009.xsd.ObjectFactory.class})
@SOAPBinding(parameterStyle = SOAPBinding.ParameterStyle.BARE)
public interface PredictionPortType {
// And the rest of class definition ...
// ...
// ...
The Implementation >>
@WebService(serviceName = "PredictionService", portName = "PredictionPort",
targetNamespace = "http://prediction.simplesoft.org/2009/xsd", /*namespace is wrong, i.e, different from endpoint*/
endpointInterface = "org.simplesoft.prediction.ws.PredictionPortType")
public class PredictionServiceImpl implements PredictionPortType {
// And the rest of class definition ...
// ...
// ...
This little tutorial tells you how to send email using a server on which sendmail is installed.
STEP 1: telnet to mail server
telnet mail.example.net 25
STEP 2: Type the text starting with ‘#‘ [without # ofcourse ] when you see the following and press Enter
220 local ESMTP Sendmail 8.13.5/8.13.5; Wed, 15 Mar 2006 01:51:21 -0800 (PST) #HELO host.example.com 250 mail.example.net Hello host.example.com [192.0.2.1], pleased to meet you #MAIL FROM:<user@example.com> 250 2.1.0 <user@example.com>... Sender ok #RCPT TO:<postmaster@example.net> 250 2.1.5 <postmaster@example.net>... Recipient ok #DATA 354 Enter mail, end with "." on a line by itself #This is a test message #. 250 2.0.0 k2FApLlB020139 Message accepted for delivery #QUIT 221 2.0.0 mail.example.net closing connection
Step 3: Check status
Sendmail uses the mail facility of syslog for logging. The mail log is usually written to /var/log/maillog. You can test the status by tailing the log file.
Simple
public class Hello {
public static void main ( String[] args ) {
System.out.println("Hello World ");
}
}
Are static methods safe to be invoked by multiple threads simultaneously ?
Ok. So it goes like this. Picture this :
Now of in this process, the called method does not store the resulting value at a memory location which will not be cleared after the instruction sequence is completed then this method can be accessed by multiple threads at the same time without any problem. This can be so because each thread comes with it’s own data and these instructions are executed on the given data. This can be be true for both – instance methods as well as static methods.
Can Data Access Objects be Singleton ? If so, when ?
DAOs can be singleton provided none of the methods store state in any way. All the methods of the DAOs should just :
If so happens – even the following scenario will not result in a dicey result :
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