Introduction
Java is still a powerful and relevant language, appealing to programmers of all skill levels due to its great demand, platform freedom, variety, robust tool ecosystem, and emphasis on scalability and performance. Yet, there are many challenges for beginners, and in this article, we discuss advanced Java coding challenges with solutions. Explore our advanced Java course syllabus.
Advanced Java Coding Challenges and Solutions for Beginners
These are a few complex concept-based Java coding challenges:
Reverse a String:
Create a Java application that can reverse a string without utilizing the String class’s reverse function.
Challenge
- Create a Java application that can reverse a string without utilizing the String class’s reverse function.
- You can learn how Java handles strings by completing this exercise.
- This challenge requires you to build code that reverses the sequencing of a string that you input.
Solution
The string might or might not have any spaces in it. Avoid utilizing the Java string class’s reverse function while implementing this code.
- This logic to reverse the string can be created in the main method.
- You can begin by taking strings and assigning the same typecase to each character.
- You can gradually increase the difficulty of your writing by using different typecases, punctuation, and symbols.
Java Code for Java String
import java.util.Scanner;
public class StringReversalClass {
public static String ReverseString(String str) {
// create a new string here using StringBuilder
StringBuilder sb = new StringBuilder();
for (int i = str.length() – 1; i >= 0; i–) {
sb.append(str.charAt(i));
}
return sb.toString();
}
// main function
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
System.out.print(“Enter the string that you want to reverse: “);
String original = scanner.nextLine();
String reversed = ReverseString(original);
System.out.println(“Original String: ” + original);
System.out.println(“Reversed String: ” + reversed);
}
}
You can begin by using strings that have the same typecase for every character. You can gradually push yourself by using varied typecases, punctuation, and symbols. Begin your learning journey with our Java course in Chennai.
Prime Number Checker
Make a function that, in the case that an input number is prime, returns TRUE. A whole number larger than one that can only be divided by one and itself is called a prime number.
Challenge
- A string that remains the same whether read forward or backward is called a palindrome. The task at hand is determining if a given string is a palindrome or not.
- If it’s a palindrome, you can answer “yes” or “true.” Return “false” or “no” if it isn’t.
Solution
- For instance, if the supplied string is “level,” it is interpreted the same from both ends. As a result, the string will be reported as a palindrome in your output.
- Examine the schedule provided below. This may be one method of determining if the string is a palindrome or not.
Java Code for Prime Number Check
public class PrimeExample{
public static void main(String args[]){
int i,m=0,flag=0;
int n=3;//it is the number to be checked
m=n/2;
if(n==0||n==1){
System.out.println(n+” is not prime number”);
}else{
for(i=2;i<=m;i++){
if(n%i==0){
System.out.println(n+” is not prime number”);
flag=1;
break;
}
}
if(flag==0) { System.out.println(n+” is prime number”); }
}//end of else
}
}
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Greedy Algorithms
By selecting the optimal solution at each stage, an understanding of greedy algorithms aids in the solution of challenging issues.
Challenge
In this case, the scheduling problem will be resolved by a greedy method.
Solution
The problem’s objective is to determine the best timetable for a group of tasks with varying start, end, and duration times.
Java Code for Greedy Algorithm:
// Class to represent a job
class Job {
int start;
int end;
int length;
// Constructor
public Job(int start, int end, int length)
{
this.start = start;
this.end = end;
this.length = length;
}
}
// Function to find the optimal schedule
public void findOptimalSchedule(Job[] jobs)
{
// Sort jobs according to their end times
Arrays.sort(jobs, new Comparator<Job>() {
public int compare(Job j1, Job j2)
{
return j1.end – j2.end;
}
});
// Initialize result
int result = 0;
// Iterate over all jobs
for (int i = 0; i < jobs.length; i++) {
// Check if current job can be included
// in the optimal solution
if (jobs[i].start >= result) {
result = jobs[i].end;
}
}
// Print result
System.out.println(“Optimal Schedule: ” + result);
}
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Error Handling
Challenge: In programming, error management is crucial since it facilitates error detection. If that code fails, you can add a throws clause to your method to cause a particular error.
Solution
Let’s look at an example of Java exception handling where the exception is handled with a try-catch sentence.
Java Code Sample for Error Handling
public class JavaExceptionExample{
public static void main(String args[]){
try{
//code that may raise exception
int data=100/0;
}catch(ArithmeticException e){System.out.println(e);}
//rest code of the program
System.out.println(“rest of the code…”);
}
}
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Merge Sort
Challenge: One of the most effective sorting algorithms is merge sort; write a Java program to demonstrate it.
Solution: The divide and conquer technique, which splits an issue into several smaller problems, is the foundation of the merge sort algorithm. Subproblems are solved one at a time until the final solutions are formed by combining subproblems.
Java Code for Merge Sort
import java.util.Arrays;
class Main {
void merge(int array[], int p, int q, int r) {
int n1 = q – p + 1;
int n2 = r – q;
int L[] = new int[n1];
int M[] = new int[n2];
for (int i = 0; i < n1; i++)
L[i] = array[p + i];
for (int j = 0; j < n2; j++)
M[j] = array[q + 1 + j];
int i, j, k;
i = 0;
j = 0;
k = p;
while (i < n1 && j < n2) {
if (L[i] <= M[j]) {
array[k] = L[i];
i++;
} else {
array[k] = M[j];
j++;
}
k++;
}
while (i < n1) {
array[k] = L[i];
i++;
k++;
}
while (j < n2) {
array[k] = M[j];
j++;
k++;
}
}
void mergeSort(int array[], int left, int right) {
if (left < right) {
int mid = (left + right) / 2;
mergeSort(array, left, mid);
mergeSort(array, mid + 1, right);
merge(array, left, mid, right);
}
}
public static void main(String args[]) {
int[] array = { 6, 5, 12, 10, 9, 1 };
Main ob = new Main();
ob.mergeSort(array, 0, array.length – 1);
System.out.println(“Sorted Array:”);
System.out.println(Arrays.toString(array));
}
}
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ArrayList
Challenge: In Java, the ArrayList class stores its elements in a dynamic array. It has no size restriction, much like an array.
Solution: In Java, ArrayList offers us dynamic arrays. Even though it might operate more slowly than conventional arrays, it can still be useful in systems that require a lot of array manipulation. The java.util package contains this class.
Java Code for ArrayList
import java.io.*;
import java.util.*;
class ArrayListExample {
public static void main(String[] args)
{
// Size of the
// ArrayList
int n = 5;
// Declaring the ArrayList with
// initial size n
ArrayList<Integer> arr1 = new ArrayList<Integer>(n);
// Declaring the ArrayList
ArrayList<Integer> arr2 = new ArrayList<Integer>();
// Printing the ArrayList
System.out.println(“Array 1:” + arr1);
System.out.println(“Array 2:” + arr2);
// Appending new elements at
// the end of the list
for (int i = 1; i <= n; i++) {
arr1.add(i);
arr2.add(i);
}
// Printing the ArrayList
System.out.println(“Array 1:” + arr1);
System.out.println(“Array 2:” + arr2);
}
}
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Recursion
Challenge: To grasp the idea, write codes with recursion. In dynamic programming, recursion is utilized.
Solution: The process of having a function call itself is known as recursion. This method offers a way to deconstruct complex issues into simpler, more manageable issues.
Java Code for Recursion
public class Main {
public static void main(String[] args) {
int result = sum(10);
System.out.println(result);
}
public static int sum(int k) {
if (k > 0) {
return k + sum(k – 1);
} else {
return 0;
}
}
}
Conclusion
For beginners in programming, Java may be difficult to understand because to its complex and verbose syntax, object-oriented paradigm, and sophisticated ideas like memory management, multithreading, and exception handling. Therefore, we have discussed advanced Java coding challenges with solutions. Hone your skills with our advanced Java training in Chennai.
