Deleting a project from Chrome Rest client

Background

Making REST API calls from some client to test out stuff is normal for any programmer. Programmers use various REST clients. I have seen most of the programmer use either

1.  Postman or
2.  Advanced REST client.

I prefer the  later one. No particular reasons but I just started with Advanced REST client so continuing with the same.

So in Advanced REST client you can save your REST calls in a PROJECT. A project can have multiple REST calls associated with it. This is where I was stuck. I created multiple projects with some dups and wanted to clean it up. Advanced REST client allows you to delete the API calls from the project but not the project itself. So naturally you looks for workarounds and today I will show you one.

Solution is quite simple. You delete it yourself from the application database. All you need to know is where and how. We will see that now.

 Deleting a project from Chrome Rest client

I have created a dummy project named GREETINGS and have added a dummy API call called ADD_CUST. Deleting API call is easy. Just click the delete icon just besides the API selector on the top.

As you must have notices we do not have option to delete project itself. Lets see how we can do that by deleting project from App database. Follow below steps

1. Right click 
2. Inspect Element 
3. Go to resources Tab
4. Select restClient DB
5. view projects table

You will see all the projects listed there.

Notice the ID of the project here. it's 10. We will use it later.  For now lets just see how can we run SQLs here. For this click on the database -restClient in this case. You should see a prompt. You can enter your SQL queries here.

Run simple query :

• select * from projects;

You should see list of all available projects.

 Finally lets see how we can delete a project. Execute following query.

• delete from projects where id=10;

Your project should get deleted from the DB. You can verify it by running select again.

Now simply refresh the page and you should see your project gone!

 

SOAP vs REST Web Services

Background

        It's been around couple of months since this topic has confused me a lot. Every time I have discussion about webservices I always try to figure out difference between SOAP based web service (JAX-WS Java API) and REST based web service (JAX-RS) web service. When the picture is finally clear let me put it down for others.

Biggest point to note about SOAP and REST is that they cannot be compared!  Yes you read it right you cannot compare both. You can only do comparison of things that are different aspects of same thing and these are not. 

SOAP (Simple Object Access Protocol) is a protocol which uses XML to communicate data. Like any other protocol it has it's own rules. REST (Representational State transfer) on the other hand is an architecture (repeating architecture) meaning a way to design APIs. 

Given that we will see each in a bit more detail so that it gives us a clear picture of REST and SOAP based web service.

W3C defines a web service as

 "A software system designed to support interoperable machine-to-machine interaction over a network."

SOAP based web services

• SOAP stands for Simple Object access protocol.
• As the name suggests this is a protocol. 
• SOAP uses service interfaces and named operations to expose the APIS.
• SOAP requires more bandwidth that REST requests.
• You can only transfer (send/receive) XML payload from SOAP based webservices.
• Java API for this is JAX-WS
• SOAP has Web Services Description Language (WSDL) which describes how webservice will work and lets you get a reference to it.
• SOAP based messages have strict specifications to follow for implementation.
• SOAP defines it's own security standards WS-security..

REST based web services

• REST stands for Representational state transfer.
• This is an architectural approach.
• REST uses URLs and method types (HTTP verbs like GET, POST, PUT etc) to expose APIs.
• REST requires less bandwidth that SOAP based requests.
• In REST you can send and receive any type of payload (Json, XML, HTML).
• Java API for this is JAX-RS
• REST does not have strict restrictions like SOAP.
• For security REST calls rely in underlying protocol like HTTPS.

Generic Notes

1. HTTP verbs used by REST
1. GET
2. OPTION
3. POST
4. PUT
5. PATCH
6. DELETE
2. Careful with HTTP POST verb. It is not safe and not idempotent.So You cannot just resend.
3. In REST based requests , response may be cached. Server may return ETag header when accessed. Client send this tag on subsequent request to access resource. If resource is not changed server will return 304 (not modified).

Major differences between GET and POST

 

Related Links 

• SOAP vs REST (differences) (SO)
• SOAP (Wiki)
• REST (Wiki)
• REST vs SOAP (Java Papers)
• SOAP vs REST Web Services (Java T Point)
• REST APIs must be hypertext-driven
• Using HTTP POST and GET using cURL(OSFG)

 

Find the row with maximum number of 1s

Background

This is a data structure and an interview question. It goes as follows - 

"Given a boolean 2D array, where each row is sorted. Find the row with the maximum number of 1s."

or to put it another way each row will have either all 0's or all 1's or k 0's and (n-l) 1's given a 2D array of m*n. Also each k 0's are contiguous and same goes with (n-1) 1's. In short they are sorted. This is just to confuse the interviewee and do not give straight hint that we are dealing with sorted things here :)

Example is in screenshot below-

Note : This question is also in GeeksForGeeks site which has solution in C. I am just going to provide solution in Java. So lets get started.

Brute Force

Brute force is the worst complexity approach you can take. It gives solution... just not in efficient way. I believe you should always start with this approach. This helps  you get a better hang of the problem and interviewer will know you can get things done. Efficiency is something you can always work on or get help of colleges or code reviewers (it will comes more fluently with experience). So always start with brute force approach. 

For this brute force approach would be to iterate over all columns in each row and keep a count of maximum no of 1's in each row and index of that row. Finally return it when you are done with the looping. Time complexity for this would be O(m*n) for a 2D array of dimension m*n. So lets see how this would go.

/**
 * @author athakur
 * Brute force approach
 */
public class BruteForce {
    
    public int compute(int[][] array) 
    {
        int maxRowOnesIndex = 0;
        int maxRowOnesCount = 0;
        for(int i=0;i<array.length;i++)
        {
            int localMaxOnesCount = 0;
            for(int j=0; j<array[i].length;j++)
            {
                if(array[i][j] == 1) {
                    localMaxOnesCount++;
                }
            }
            if(localMaxOnesCount > maxRowOnesCount)
            {
                maxRowOnesCount = localMaxOnesCount;
                maxRowOnesIndex = i;
            }
        }
        return maxRowOnesIndex;
    }
}

To use this methods class and test the working execute following code.

 /**
 * @author athakur
 * Starter code for computing row with maximum 1's in a 2D array
 */
public class Compute {
    public static void main(String args[])
    {
        int[][] array = new int[][]{{0,1,1,1},{0,0,1,1},{1,1,1,1},{0,0,0,0}};
        System.out.println(new BruteForce().compute(array));
    }
}

Output : 2

That's the brute force method. As mentioned earlier it's time complexity is O(m*n) which is like the worst complexity. Can we do better than this? Sure we can :)

What is it that we can leverage here? We know that each row in the 2D array is sorted. We can easily to a binary search to find the leftmost 1. Count of 1st will be no of columns minus index of leftmost one. Now guess what the time complexity is for this approach?

It is O(m*logn). We iterate over each row (total of m rows) and then do a binary search on each row with log n complexity.

/**
 * @author athakur
 * Binary Search approach
 */
public class BinarySearch {

    private int first(int[] array, int low, int high)
    {
        if(high >= low)
        {
            int mid = low + (high-low)/2;
            if((mid == 0 || array[mid-1]==0) && (array[mid] == 1))
            {
                return mid;
            }
            else if(array[mid] == 0) {
                return first(array, mid + 1, high);
            }
            else {
                return first(array, low, mid-1);
            }
        }
        return -1;
    }
   
    public int compute(int [][] array)
    {
        int maxRowOnesIndex = 0;
        int maxRowOnesCount = 0;
        
        for(int i=0; i< array.length;i++)
        {
            int onesIndex = first(array[i], 0, array[i].length-1);
            if(onesIndex != -1)
            {
                int localMaxOnesCount = array.length - onesIndex;
                if(localMaxOnesCount > maxRowOnesCount)
                {
                    maxRowOnesCount = localMaxOnesCount;
                    maxRowOnesIndex = i;
                }
            }
        }
        return maxRowOnesIndex;
    }
    
}

And run it with

/**
 * @author athakur
 * Starter code for computing row with maximum 1's in a 2D array
 */
public class Compute {
    public static void main(String args[])
    {
        int[][] array = new int[][]{{0,1,1,1},{0,0,1,1},{1,1,1,1},{0,0,0,0}};
        System.out.println(new BinarySearch().compute(array));
    }
}

Output :2

Note : The above solution can be optimized further. Instead of doing binary search in every row, we first check whether the row has more 1s than max so far. If the row has more 1s, then only count 1s in the row. Also, to count 1s in a row, we don’t do binary search in complete row, we do search in before the index of last max.

So basically lets say 1st row has m 1's we will check for each for the subsequent row i whether array[i][n-m-1] is 1.

Related Links

• Find the row with maximum number of 1s (GFG)

Installing Oracle instant database client in Ubuntu

Background

In this post I am going to show how to install oracle instant client (database client) on your ubuntu machine. After this you should be able to connect to your oracle server and execute queries on it from your machine using sqlplus.

Prerequisites

 Firstly find out whether your system is 32 bit or 64 bit. Easiest way to find out is to execute

• uname - a

If your machine is 64 bit you should notice something like x64 in the output. If you don't see it 32 bit like in my case (refer screenshot below)

Next make sure you have alien command installed. We will need to to convert rpm packages to deb. To install simply run

• sudo apt-get install alien 

 Installing Oracle instant database client

1. Download instant client files from oracle site.Select correct files depending on your operating system and architecture. For me it is Ubuntu and 32 bit. So I have downloaded following files
1. oracle-instantclient12.1-basic-12.1.0.2.0-1.i386.rpm
2. oracle-instantclient12.1-sqlplus-12.1.0.2.0-1.i386.rpm
3. oracle-instantclient12.1-devel-12.1.0.2.0-1.i386.rpm
2. Since only rpm files are available and Ubuntu works with debian we need to convert rpm packages to deb packages.
1. alien -k  oracle-instantclient12.1-basic-12.1.0.2.0-1.i386.rpm
2. alien -k oracle-instantclient12.1-sqlplus-12.1.0.2.0-1.i386.rpm
3. alien -k oracle-instantclient12.1-devel-12.1.0.2.0-1.i386.rpm
3. Now to install .deb packages use dpkg -i packageName
1. dpkg -i oracle-instantclient12.1-basic-12.1.0.2.0-1.i386
2. dpkg -i oracle-instantclient12.1-sqlplus-12.1.0.2.0-1.i386.rpm
3. dpkg -i oracle-instantclient12.1-devel-12.1.0.2.0-1.i386.rpm  
4. Now you need to set some environment variables. Open ~/.bashrc file and add export following environment variables (add following line to file)
1. export ORACLE_HOME=/usr/lib/oracle/12.1/client
2. export LD_LIBRARY_PATH=/usr/lib/oracle/12.1/client/lib/
3. export PATH=$PATH:$ORACLE_HOME/bin
5. To make source your changes are reflected in same terminal execute
1. source ~/.bashrc

And  you should be good to go. Verify your env variables.

Resolving errors

sometimes you may get following error

"sqlplus: error while loading shared libraries: libsqlplus.so: cannot open shared object file: No such file or directory"

in this case double check you have added LD_LIBRARY_PATH environment variable and it is pointing to write directory. Quick way to check is 

• echo $LD_LIBRARY_PATH
• cd <output_of_above_command>

Sqlplus should now be recognized and work as expected

If you see errors related to libaio.so.1 missing then install it using

• sudo apt-get install libaio1

Related Links

• Oracle Instant Client
• Oracle Sqlplus client on Mac 
• sqlplus remote connection giving ORA-21561: OID generation failed 
• Install Oracle instant client and sqlplus using Homebrew(OSFG)

Difference between StringBuilder and StringBuffer in Java

Background

In Java String class is immutable. So each concatenation or substring operation yields a new String.

Yes we all have heard String concatenation is bad to create long Strings. Concatenating large number of String literals will fill up the permgen area (where String pool resides) faster. One of the reason why SLF4J is preferred over LOG4J  . Incase you need to concatenate String literals you can use StringBuilder or StringBuffer instance instead. You can just keep appending your String literals to this instance and finally call toString() to get a single String instance. In this post we will see the difference between StringBuilder and StringBuffer.

StringBuilder and StringBuffer in Java

Their usages are quite simple.

        StringBuffer sBuffer = new StringBuffer("abc");
        sBuffer.append("pqr");
        System.out.println(sBuffer.toString());
        
        StringBuilder sBuilder = new StringBuilder("abc");
        sBuilder.append("pqr");
        System.out.println(sBuilder.toString());

As you can see the usage is similar but their difference is very important which we will see now. Just a note no new strings are created when append is called. Internally an array is maintained and String characters are appended to it.

Difference between StringBuilder and StringBuffer in Java

The most important difference between them is 

• StringBuffer is synchronized, StringBuilder is not.

Hence unless you have a multithreaded scenario to deal with always go for  StringBuilder and not StringBuffer. Even if you have multithread situation you can use a synchronized block around StringBuilder.

 You can say - "StringBuilder is intended as a drop in replacement for StringBuffer where synchronisation is not required"

Quoting the API docs for StringBuilder -

"A mutable sequence of characters. This class provides an API compatible with StringBuffer, but with no guarantee of synchronization. This class is designed for use as a drop-in replacement for StringBuffer in places where the string buffer was being used by a single thread (as is generally the case). Where possible, it is recommended that this class be used in preference to StringBuffer as it will be faster under most implementations."

A simple test for performance would be -

public class HelloWorld {
    public static void main(String args[]) throws IOException
    {        
        int limit = 1000000;
        long currTime;
        long timeTaken;

        {
            StringBuffer sb = new StringBuffer();
            currTime = System.currentTimeMillis();
            for (int i = limit; i --> 0 ;) {
                sb.append("");
            }
            timeTaken = System.currentTimeMillis() - currTime;
            System.out.println("StringBuffer Time : " + timeTaken);
        }
        {
            StringBuilder sb = new StringBuilder();
            currTime = System.currentTimeMillis();
            for (int i = limit; i --> 0 ;) {
                sb.append("");
            }
            timeTaken = System.currentTimeMillis() - currTime;
            System.out.println("StringBuilder Time : " + timeTaken);
        }   
    }
}

and this prints (one execution) -

StringBuffer Time : 29
StringBuilder Time : 6

Summary

So to summarize always use StringBuilder and not StringBuffer.

Related Links

• StringBuilder and StringBuffer(SO)