Chrome vs Firefox : jQuery faceoff (how range selection behaves differently in these 2 browsers)

Hello Buddies.

Long time since last post.

I want to share an interesting scenario I encountered few days back related to UI.

Scenario –

There are 2 multiple select boxes available on the JSP, in such a way that you can filter the data from the first one, using one regex, select a range of filtered out options and push them towards the second one.

Seems easy in the first shot. Isn’t it?

I used jQuery for this and implemented it, for some POC, as follows -

https://github.com/namitsharma99/multi-selectbox-browser-specific/blob/master/test.jsp

So what we have in the first select box is -

'ABC intDEF',

'XYZ intABC',

'qwerty',

'abcdefgh',

'blah blah intDUMMY',

'blah blah hello',

'hello int'

I need to filter only those values having ‘int’, by hiding the non-matching ones.

Now the box would be showing -

'ABC intDEF',

'XYZ intABC',

'blah blah intDUMMY',

'hello int'

Shift + select End-To-End and append in the second box.

So far, looks good.

Try this code in Chrome, it will look perfect.

But now comes the interesting part, try in Mozilla Firefox, and inspect the second box.

“BOOM”

All those values that were hidden due to filter in the first box, have also been included if present within the range. What it can lead to is the angry set of users who prefer Firefox. :P

Clearly, Chrome is able to understand not to include hidden options, while Firefox fails here, and being a browser issue, it becomes difficult to comprehend.

What can be done?

A neat and clean work-around!

function pushSelectedToRight(pushObj) {

pushObj = $(pushObj);

var leftBox = $("#tempLeft");

var rightBox = $("#tempRight");

if(leftBox.find("option:selected").length > 0){

var tmp = leftBox.find("option:selected");

tmp.each(function() {

        if($(this).css("display")=="block"){

    rightBox.append($(this));

    }

    });

  rightBox.find("option").removeAttr("selected");

}

else{

alert("Please select atleast one item to move");

}

}

Speaking in terms of jQuery, it selects out the values using selected flag - leftBox.find("option:selected")

Apart from this better use ‘each’ method as well for checking the visible elements.

It won’t cause any issue on the Chrome side, but will also work for Firefox -

var tmp = leftBox.find("option:selected");

tmp.each(function() {

        if($(this).css("display")=="block"){

    rightBox.append($(this));

    }

    });

And this is how a day is saved!

Happy Coding!!

O(n^2) vs O(n) techniques for finding the max count of a digit in the given numerical String.

Hello Buddies

Came across this question related Java String handling, so thought of sharing.

Question:
Given a string made up of n number of digits, find the digit having the maximum number of occurrence.
Eg. "12333123654"
Here, 3 is the digit being repeated for the maximum number of times.

Answer:
First approach can be to scan the string as a character array, maintain a map that can update the occurrence count. In the end find the max value and then the corresponding key.


Question:
What is the complexity of this solution?

Answer:
O(n^2) since we are iterating and using contains check.


Question:
Any better approach?

Answer:
Solution with O(n) complexity, i.e. having a lookup array.
We have digits 0 to 9, which can be related to each index of an array. For the values, update the occurrence count in the values.

For complete solution, refer to this link.

Happy Coding!!

Cloning in java - An interviewer's perspective

Wassup Buddies!

'Cloning' is the topic we will be cloning down for this blogpost.

To save time and memory, coders often prefer to re-use already existing objects. Commonly, a new reference pointing to an existing object, or talking about veterans of Java, techniques like cloning are used.

Digging deeper in this context, we will understand what are the repercussions and whats are the various ways of cloning. Core java interviewers find it an interesting topic to engulf the candidates by asking various pros and cons.

I have placed the sample Java class that I created for my understanding and the differentiation of types, here. 

Please feel free to get your hands dirty. Many of our readers might find the topic repetitive, but I hope they will find a new perspective of seeing or explaining the concepts or would love to share their thoughts and suggestions for improving the knowledge of others.

A brief primer for better understanding of the code is as follows:

Scenario 1: Assigning the reference.

Test ob1 = new Test();

Where Test.java is -

class Test {

 int var1, var2;

 Dummy dummy = new Dummy();

  Test() {

  var1 = 10;

  var2 = 20;

  dummy.aVar = 30;

  dummy.bVar = 40;

 }

}

Simply straight forward, an object is having its reference.

ob1 original -- 10 20 30 40

Scenario 2: Creating a new reference for the same object, using the earlier created reference.

Test ob2 = ob1;

Now make some changes using the second reference to see the impact on the object created.

ob2.var1 = 100;

ob2.dummy.aVar = 1000;

What do we get?

ob1 original -- 10 20 30 40

ob1 changed -- 100 20 1000 40

ob2 original -- 100 20 1000 40

So clearly, as expected, reference will impact the original object.

Scenario 3: Shallow Cloning

CloneableTest cloneableTest = new CloneableTest();

Where CloneableTest.java is -

class CloneableTest implements Cloneable {

 int var1, var2;

 Dummy dummy = new Dummy();

  CloneableTest() {

  var1 = 10;

  var2 = 20;

  dummy.aVar = 30;

  dummy.bVar = 40;

 }

  public Object clone() 

throws CloneNotSupportedException {

  return super.clone();
}

}

Note: For using cloneableTest.clone(), we need to have the class implementing Cloneable interface, else we will get ''CloneNotSupportedException".

Bringing in a new reference for testing -

CloneableTest newCloneable = null;

try {

 newCloneable = (CloneableTest) cloneableTest.clone();

 newCloneable.var1 = 99;

 newCloneable.var2 = 99;

 newCloneable.dummy.aVar = 99;

 newCloneable.dummy.bVar = 99;

} catch (CloneNotSupportedException e) {

 // e.printStackTrace();

 System.err.println

("Exception since Cloneable interface is not implemented!!");

}

Changes have been made, but what about the original object? Let's check!

Original Before : 10 & 20 & 30 & 40

Cloned   : 99 & 99 & 99 & 99

Original After : 10 & 20 & 99 & 99

Hence take away from this example - original primitive attributes won't change, but non primitives will change. 

Still the cloned object is not yet completely change proof.

Scenario 4: Deep Cloning

CloneableDeepTest cloneableDeepTest = new CloneableDeepTest();

Where CloneableDeepTest.java is -

class CloneableDeepTest implements Cloneable {

 int var1, var2;

 Dummy dummy = new Dummy();

  CloneableDeepTest() {

  var1 = 10;

  var2 = 20;

  dummy.aVar = 30;

  dummy.bVar = 40;

    }

   public Object clone() 

throws CloneNotSupportedException {

 CloneableDeepTest cloneableDeepTest 

= (CloneableDeepTest) super.clone();

 cloneableDeepTest.dummy = new Dummy();

 return cloneableDeepTest;

   }

}

Bringing in new reference again -

CloneableDeepTest cloneableDeepTest2 = null;

try {

 cloneableDeepTest2 = 

(CloneableDeepTest) cloneableDeepTest.clone();

 cloneableDeepTest2.var1 = 55;

 cloneableDeepTest2.var2 = 66;

 cloneableDeepTest2.dummy.aVar = 77;

 cloneableDeepTest2.dummy.bVar = 88;

} catch (CloneNotSupportedException e) {

 // e.printStackTrace();

 System.err.println

("Exception since Cloneable interface is not implemented!!");

}

What should be the impact on the original object now? Let's check!

Original Before : 10 & 20 & 30 & 40

Cloned   : 55 & 66 & 77 & 88

Original After : 10 & 20 & 30 & 40

Sigh! The cloned object still retains the expected original values for both primitive and non primitive attributes.

Major difference between the implementation of shallow and deep clonings can be seen in the overridden clone() methods -

For shallow we simply call super.clone() as follows -

public Object clone() 

throws CloneNotSupportedException {

 return super.clone();

}

While for deep cloning, we need to go a bit further and control the way how inner objects are being generated as follows -

public Object clone() 

throws CloneNotSupportedException {

 CloneableDeepTest cloneableDeepTest = 

(CloneableDeepTest) super.clone();

 cloneableDeepTest.dummy = new Dummy();

 return cloneableDeepTest;

}

That's all I have for now.

Happy Coding!!

How to read and write a CSV file using core java?

Hello Buddies.

CSV: A CSV is a comma separated values file which allows data to be saved in a table structured format

Let's try to read a CSV file using java, filter out the required rows and create a new CSV per our own requirements.

Input - Provided as a path for the file having below details -

ID	NAME	AGE	GENDER	OCCUPATION	ADDRESS
1	ABC	22	M	SE	IN
2	DEF	32	F	SE	US
3	ABC	12	M	SE	UK
4	DEF	42	F	SE	IN
5	ABC	52	M	SE	US
6	DEF	53	F	SE	UK
7	ABC	60	M	SE	IN
8	DEF	67	F	SE	US
9	ABC	74	M	SE	UK
10	DEF	81	F	SE	IN
11	ABC	88	M	SE	US

We will filter out the people living in India and will have a new CSV only for those people.

Here's a code snippet -

import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.File;
import java.io.FileReader;
import java.io.FileWriter;
import java.io.IOException;

public class CSVReader {

 private static String mainPath = "C:\\\\Users\\\\namit.sharma\\\\Downloads\\\\";

 public static void main(String[] args) throws IOException {

  FileReader fileReader = new FileReader(new File(mainPath + "sampleExcel1.csv"));

  BufferedReader bufferedReader = new BufferedReader(fileReader);

  String string;
  do {
   string = bufferedReader.readLine();
   if (null != string) {
    System.out.println(string);
    filterForIndians(string);
   }
  } while (null != string);

  bufferedReader.close();

 }

 private static void filterForIndians(String string) throws IOException {
  File file = new File(mainPath + "fileredCsv.csv");
  if (file.exists()) {
   file.delete();
  }
  FileWriter fileWriter = new FileWriter(new File(mainPath + "fileredCsv.csv"),true);
  if ("IN".equals(string.split(",")[5])) {
   BufferedWriter bufferedWriter = new BufferedWriter(fileWriter);
   bufferedWriter.write(string+"\n");
   bufferedWriter.close();
  }
 }

}

Now we can check the generated csv. It will have the desired rows -

1	ABC	22	M	SE	IN
4	DEF	42	F	SE	IN
7	ABC	60	M	SE	IN
10	DEF	81	F	SE	IN

How to count the number of leaves in a binary tree?

Hello Buddies.

Here's a quick code snippet for finding the number of leaves in a 'Binary Tree', after a node gets deleted!

Have a look-

Input -

5
-1 0 0 1 1
1

Input description -

Line 1: No. of nodes in Binary Tree
Line 2: All the nodes' parents, where -1 means no parent - root, 0 means root as parent.
Line 3: Node at which this value exists, remove it.


Node to remove -

public class TreeProblem {
 private static int count = 0;

 public static void main(String args[]) throws Exception {

  Scanner scanner = new Scanner(System.in);
  int numOfNodes = scanner.nextInt();
  scanner.nextLine();
  String intStr = scanner.nextLine();
  String[] array = intStr.split(" ");
  int[] arr = new int[array.length];
  for (int i = 0; i < numOfNodes; i++) {
   arr[i] = Integer.valueOf(array[i]);
  }
  int nodeToDelete = scanner.nextInt();

  // create a tree
  Node root = null;
  int valueIndex = 0;
  Node current = null;
  for (int i = 0; i < numOfNodes; i++) {
   if (arr[i] == -1) {
    root = new Node(valueIndex++);
    current = root;
   } else {
    if (arr[i] == current.value) {
     if (null == current.left) {
      current.left = new Node(valueIndex++);
     } else if (null == current.right) {
      current.right = new Node(valueIndex++);
      current = current.left;
     }
    }
   }
  }

  // scan the tree for deleting the node
  // it clearly means we have to ignore the leaves at that node, for the remaining
  // keep the count going

  treetraversal(root, nodeToDelete);

  System.out.println("Remaining leaves = " + count);
  scanner.close();
 }

 private static void treetraversal(Node root, int nodeToDelete) {
  if (null == root) {
   return;
  }
  System.out.println("-- " + root.value);
  if (root.value == nodeToDelete) {
   return;
  }
  if (null != root.left) {
   treetraversal(root.left, nodeToDelete);
  }
  if (null != root.right) {
   treetraversal(root.right, nodeToDelete);
  }
  if (null == root.left && null == root.right) {
   count++;
  }
 }

}

class Node {

 int value;
 Node parent, left, right;

 Node(int value) {
  this.value = value;
 }
}