I am sure you must be learning a lot through tutorials and then forgetting about things easily. And again watching those tutorials over and over again or may be some different one and forget again !!! It becomes easier to get lost in the tutorials hell and without practice your learning gets wasted because when the real problem arrives you don't know when to use it. So don't learn the basics only, solve small problems to practice the fundamentals.

In this tutorial we will learn few basic problem solving using javascript build in functions of string, array, etc.

This is the most favourite question of interviewers to reverse a string. Well you can solve it in different ways:

String.split() : Divides a string into substrings with the separator passed and converts into an array, 

Array.reverse() : Reverses the elements of an array, 

Array.join() : Creates and returns a new concatenated string of all the elements in the array specified by a string separator.

     return str.split('') // split each character of the string and return array of characters

               .reverse() // reverse the array of characters

               .join(''); // join the array into string again

Now the interviewer says don't use the reverse() function !! Now what ??

No worries, we have for loop for the rescue, here is how you can use it

  for(let i = str.length - 1; i >= 0; i--) { 

    reverseString = reverseString + str[i]; 

Show your smartness that you know the latest javascript patterns and use reduce

Reduce is a function that executes on each element in the array and results in a single output. It takes two parameters - first: a reducer function to execute on each array element and second - an initial value

              .reduce((reverseString, character) => character + reverseString, '');

Problem Solving 2: Count Vowels in a String

Out of the 26 alphabets in english 5 of them are considered as vowels: A, E, I, O, U. (Bonus knowledge : Y is considered as a sixth vowel. One vowel can have many sounds and sometimes can be silent without any sound at all).

Solution 1:  Using a old school for loop (Understand the concept)

Solution 2: Using for of loop (Build upon your fundamental and apply latest trends to get smarter)

Comment below for more things you would like to know about Or suggest a different way of solving 

This blog will help you understand the concept of solving very basic algorithms using javascript.

Solving Algorithms and Problems with Javascript

I have string from which I am getting data in array but I am not getting way to get so that I get related data in array like my string is,

[{"Class Room": ["Windows",  "Windows1"],  "Staffroom": ["Windows",  "Windows1"]}],  "


"Class Room" having two values in dropdown Windows and Windows1

"Staffroom" having two values in dropdown Windows and Windows1

How would i get dropdown related value in array i can use split operation but cant getting logic

Above given string is in list, my code is ,

String[] data = list.toString().split(",");


But it split in array then I would not have related data

I want it to split in way so that I would get relative dropdown data in array,

 "Class Room": ["Windows",  "Windows1"], value in aray index 0
   "Staffroom": ["Windows",  "Windows1"]}] value in array index 1

How to split a string to get related data JAVA

Is it possible to convert a string to an io.Writer type in Golang?

I will be using this string in fmt.Fprintf() but I am unable to convert the type.

Golang Convert String to io.Writer?

whats the difference between char string[] vs char *string[]?

Fisrt you have to define a state and then what ever you typed in the input field that value you have to put in that state and send that state value to the api.based on your search api will return values

 In react.js  how to Search a string value from the data table based on api?

    cout<< "hello welcome to Jacob's\n\n";

    cout << "Enter your username: " << endl;

    cout << "Enter your password: " << endl;

if ( username != "Jacob" && password != "welcome" )

    cout << "Incorrect username/password. Please try again." << "\n" <<

		//**********************menu price variables******************************

    float pizza=23.00, wings = 9.50, chilliFries = 7.75,exitout = 4, amount, pick,total;

     cout<<"**************Restaurant menu********************\n";

     cout<<"please make a selection from the menu\n";

     cout<<"************************************************\n\n";

     cout<<"how much "<<pick<<" would you like?";//type iteam.

     cout<<"is that all you want to order today?\ntype y for yes and n for no ";

why isn't the code reading my "pick" variable? 

    printf("Subject:%s\n", stu.score.subject);

    printf("Marks: %d\n", stu.score.marks);

can anyone tell me why i am not getting name it's show 'b' instead of Anik ?

Why "else"  condition is not working ?

But when I compile and run, the compiler says " Anik was not declared in this scope"....

if (condition) {
  // body of if statement
}

The if statement evaluates the condition inside the parentheses ( ).

If the condition evaluates to true, the code inside the body of if is executed.

If the condition evaluates to false, the code inside the body of if is skipped.

// Program to print positive number entered by the user
// If the user enters a negative number, it is skipped

#include <iostream>
using namespace std;

int main() {

  int number;

  cout << "Enter an integer: ";
  cin >> number;

  // checks if the number is positive
  if (number > 0) {
    cout << "You entered a positive integer: " << number << endl;
  }

  cout << "This statement is always executed.";

  return 0;
}

Enter an integer: 5
You entered a positive number: 5
This statement is always executed.

When the user enters 5, the condition number > 0 is evaluated to true and the statement inside the body of if is executed.

Enter a number: -5
This statement is always executed.

When the user enters -5, the condition number > 0 is evaluated to false and the statement inside the body of if is not executed.

    cout << "You entered a positive integer: " << number << endl;

  cout << "This statement is always executed.";

How can I include strings in conditional statements...in C++ ?

Read string with spaces by using the "%[^\n]" format specifier. The format specifier "%[^\n]" tells the compiler that read the characters until "\n" is not found. See the output, now the program can read complete strings with white space.

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openHow can I include strings in conditional statements...in C++ ?

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openwhat function can be used to read a string containing spaces from the keyboard ?