1. String & Array Java Programs
Essential Java coding interview programs for string reversing, palindromes, anagrams, array sorting, and matrix operations.
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2. DSA & Algorithms in Java
Implementations of Binary Search, Bubble Sort, Quick Sort, LinkedList, Stack, Queue, and Tree Traversals in Java.
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3. OOPS & Java Core Concepts
Core Java object-oriented programming concepts: Inheritance, Polymorphism, Abstraction, Interfaces, and Multithreading.
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4. Spring Boot REST API Guide
Step-by-step Java Spring Boot REST API CRUD application guide with MySQL, JPA Hibernate, and Controller/Service architecture.
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Top Java Coding Programs Quick Code
Program 1: How to Reverse a String in Java
public class ReverseString {
public static void main(String[] args) {
String str = "Manish Kumar Java Full Stack Developer";
String reversed = new StringBuilder(str).reverse().toString();
System.out.println("Reversed String: " + reversed);
}
}
Explanation: Utilizes Java's built-in StringBuilder.reverse() method for O(N) time complexity reversal.
Program 2: Fibonacci Series in Java
public class Fibonacci {
public static void main(String[] args) {
int n = 10, t1 = 0, t2 = 1;
System.out.print("First " + n + " terms: ");
for (int i = 1; i <= n; ++i) {
System.out.print(t1 + " ");
int sum = t1 + t2;
t1 = t2;
t2 = sum;
}
}
}
Explanation: Generates the Fibonacci sequence up to N terms using dynamic variable swapping.
Program 3: Check Palindrome Number in Java
public class PalindromeCheck {
public static void main(String[] args) {
int num = 12321, reversedNum = 0, remainder, originalNum = num;
while (num != 0) {
remainder = num % 10;
reversedNum = reversedNum * 10 + remainder;
num /= 10;
}
System.out.println(originalNum + (originalNum == reversedNum ? " is Palindrome" : " is not Palindrome"));
}
}
Explanation: Reverses an integer using modulo math to verify if it reads the same backward and forward.
Program 4: Binary Search Algorithm in Java
public class BinarySearch {
public static int binarySearch(int[] arr, int target) {
int low = 0, high = arr.length - 1;
while (low <= high) {
int mid = low + (high - low) / 2;
if (arr[mid] == target) return mid;
if (arr[mid] < target) low = mid + 1;
else high = mid - 1;
}
return -1;
}
}
Explanation: Efficient searching algorithm in a sorted array with logarithmic O(log N) time complexity.