27.09.2020

Java LinkedList can be used both as Stack and Queue

While using LinkedList as Stack use push() method to add and pop() method to retrieve an element.

And for Queue, use add() to add and poll() to remove an element from the queue.

20.10.2020

Escaping twice inside a string

The problem:

I was trying to use querySelector() function to get an element by an attribute. However the attribute was not so standard css attribute. It was like row:id.

So, whenever I was trying to query like this way: document.querySelector('["row:id"]'), it threw a syntax error that row:id is not a valid selector.

The solution:

I needed to escape the “:” twice. Once for the javascript itself and another for the css. So, I ended up writing this way document.querySelector('["row\\:id"]'). This time it worked. :D

07.12.2020

The problem:

I needed a data structure that can identify duplicate while inserting and can be retrieved by using an index.

The solution:

First issue can be solved by using HashSet. I was aware of that. I just didn’t know that HashSet can be converted into an Object[] array. Like this: Object[] array = set.toArray()

After that, I can retrieve any value by simply indexing and casting to a data type. i.e. int val = (int) array[i]

17.12.2020

The problem:

Suppose you have two strings s1 = watercooler and s2 = ercoolerwat. Determine s2 is a rotated array of s1. You can only use isSubstring() method only once. This constraint makes this problem a bit difficult to solve.

The solution:

We can always divide a string into x and y. Like s1 = x + y. In this case, x = wat and y = cooler. For s2 it becomes s2 = y + x.

Now, what if we concatenate the s1 twice. s1s1 = x + y + x + y = x + s2 + y Now we can use isSubtring only once and get our desired result.

03.01.2021

The problem:

5/2 returns 2.0 I expected 2.5.

Solution:

(double)5 / 2 returns 2.5 Reason is: When java performs division, it assumes 5 as int and divides accordingly. So, we need to tell java that in this case 5 is not int rather double and divide it accordingly.

Problem:

Oftentimes, I have to disregard all non-letters in a sentence. I used to create a hashset containing all the punctuations. Then while traversing through the string, I would check in the hashset. Which is inefficient.

Better Way:

Replace all the non-letter characters with spaces. Then while traversing only look for the space. E.g.: "this is a very short sentence! I wonder, why?".replaceAll("[^a-zA-Z0-9]", " ");

Here we are using a regex which simply looks for everything except a-z, A-Z and 0-9.

14.03.2021

Problem: Converting an arryList of integer array to a 2D int array.

Solution:

    List<int[]> list = new ArrayList<int[]>();
    int[][] arr = list.toArray(new int[list.size()][list.get(0).length]);

17.03.2021

Problem: When you want to use wildcard matching but don’t want to go through the complexities of regex, what can you do?

Solution: You can use glob pattern. It’s not as powerful or flexible as RegEx, however, it’s quite simple to use. In many cases it’s good enough for wildcard searching. E.g. src/*.js will find all the javascript files inside the src directory.

28.05.2021

Problem: What is a reducer?

Solution: By the name of the function it seems like, this reducer function takes a bunch of data and by some rule, it reduces those input data and return them.

This is the case in reality. Let’s take real example from redux. In redux we can create a reducer function that takes two arguments: state and action. state is the input data and and action tells the function what kind of modification it should perform on the state. E.g.:

const reducer = (state, action) => {
    if(action.type === 'Increase') return state.time + 1
    else if(action.type === 'Decrease') return state.time - 1
    else return state.time
}