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Programming Articles
Page 1480 of 2547
Can I define more than one public class in a Java package?
No, while defining multiple classes in a single Java file you need to make sure that only one class among them is public. If you have more than one public classes a single file a compile-time error will be generated.ExampleIn the following example we have two classes Student and AccessData we are having both of them in the same class and declared both public.import java.util.Scanner; public class Student { private String name; private int age; Student(){ this.name = "Rama"; this.age = 29; } Student(String name, int age){ ...
Read MoreWhile overriding can the subclass choose not to throw an exception in java?
If the super-class method throws certain exception, you can override it without throwing any exception.ExampleIn the following example the sampleMethod() method of the super-class throws FileNotFoundException exception and, the sampleMethod() method does not throw any exception at all. Still this program gets compiled and executed without any errors.import java.io.File; import java.io.FileNotFoundException; import java.io.IOException; import java.util.Scanner; abstract class Super { public void sampleMethod()throws FileNotFoundException { System.out.println("Method of superclass"); } } public class ExceptionsExample extends Super { public void sampleMethod() { System.out.println("Method of Subclass"); } public static void main(String args[]) { ...
Read MoreFind if nCr is divisible by the given prime in C++
Suppose there are three variables N, R and P. The N and R are used to get the NCR and P is a prime. We have to find whether NCR is divisible by P. Suppose we have some numbers N = 7, R = 2 and P = 3, then 7C2 = 21, this is divisible by 3, so the output will be true.We know that NCR = N! / (R! * (N – R)! ). We will use Legendre Formula to largest power of P, which divides any N!, R! and (N – R)! in order to NCR to ...
Read MoreHow can we split a string by sentence as a delimiter in Java?
The split() method of the String class accepts a String value representing the delimiter and splits into an array of tokens (words), treating the string between the occurrence of two delimiters as one token.For example, if you pass single space “ ” as a delimiter to this method and try to split a String. This method considers the word between two spaces as one token and returns an array of words (between spaces) in the current String.If the String does not contain the specified delimiter this method returns an array containing the whole string as an element.Examplepublic class SplitExample { ...
Read MoreWhat is loose coupling how do we achieve it using Java?
Coupling refers to the dependency of one object type on another, if two objects are completely independent of each other and the changes done in one doesn’t affect the other both are said to be loosely coupled.You can achieve loose coupling in Java using interfaces -Exampleinterface Animal { void child(); } class Cat implements Animal { public void child() { System.out.println("kitten"); } } class Dog implements Animal { public void child() { System.out.println("puppy"); } } public class LooseCoupling { public static void main(String args[]) { Animal obj = new Cat(); obj.child(); } }Outputkitten
Read MoreHow can we check if specific string occurs multiple times in another string in Java?
You can find whether a String contains a specified sequence of characters using any of the methods −The indexOf() method − The indexOf() method of the String class accepts a string value and finds the (starting) index of it in the current String and returns it. This method returns -1 if it doesn’t find the given string in the current one.The contains() method − The contains a () method of the String class accepts a sequence of characters value and verifies whether it exists in the current String. If found it returns true else it returns false.In addition to these, you ...
Read MoreFind LCM of rational number in C++
Here we will see how to find the LCM of Rational numbers. We have a list of rational numbers. Suppose the list is like {2/7, 3/14, 5/3}, then the LCM will be 30/1.To solve this problem, we have to calculate LCM of all numerators, then gcd of all denominators, then the LCM of rational numbers, will be like −$$LCM =\frac{LCM\:of\:all\:𝑛𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟𝑠}{GCD\:of\:all\:𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟𝑠}$$Example#include #include #include using namespace std; int LCM(int a, int b) { return (a * b) / (__gcd(a, b)); } int numeratorLCM(vector vect) { int result = vect[0].first; for (int i = 1; i < ...
Read MoreHow to trim white space in StringBuffer in Java?
The String class of the java.lang package represents a set of characters. All string literals in Java programs, such as "abc", are implemented as instances of this class. Strings objects are immutable, once you create a String object you cannot change their values, if you try to do so instead of changing the value a new object is created with the required value and the reference shifts to the newly created one leaving the previous object unused.The StringBuffer (and StringBuilder) class is used when there is a necessity to make a lot of modifications to a String.Unlike Strings, objects of ...
Read MoreFind nth term of the Dragon Curve Sequence in C++
Here we will see a program, that can find nth term of the Dragon Curve sequence. The Dragon curve sequence is an infinite binary sequence. It starts with 1, and in each step, it alternatively adds 1s and 0s before and after each element of the previous term, to form the next term.Term 1 : 1Term 2 : 110Term 3 : 1101100Term 4 : 110110011100100We will start with 1, then add 1 and 0, alternatively after each element of the preceding term. When the new term obtained becomes the current term, then repeat the steps from 1 to n to ...
Read MoreFind Surpasser Count of each element in array in C++
Suppose one array A is given. We have to find a number of surpasser of each element in that array. The surpassers are greater elements which are present at the right side of the array of the current element. Suppose A = {2, 7, 5, 3, 0, 8, 1}, the surpassers are {4, 1, 1, 1, 2, 0, 0}, so 2 has 4 numbers at right side, which are greater than 4, and the same rule for others. The solution is very simple, two nested loops will be there, for each element, it will count surpassers, then store them into ...
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