Freshers preparing for Java interviews should master core programs like Fibonacci series, prime number checking, string palindrome, bubble sort, and Armstrong number. These questions test fundamental logic, loop control, and problem-solving skills that interviewers consistently evaluate in entry-level technical rounds.<\/p>\n\n\n\n
Top 5 Must-Know Java Interview Programs for Freshers<\/strong><\/h2>\n\n\n\n![\"Top](\"https:\/\/www.guvi.in\/blog\/wp-content\/uploads\/2026\/03\/1-9.png\" "\\"\\"")
<\/figure>\n\n\n\nThese 5 must-know Java Interview programs are designed to test your problem-solving skills and your ability to apply Java concepts to real-world scenarios.<\/p>\n\n\n\n
1. Fibonacci Series<\/strong><\/h3>\n\n\n\n![\"\"](\"https:\/\/www.guvi.in\/blog\/wp-content\/uploads\/2026\/03\/2-9.png\" "\\"\\"")
<\/figure>\n\n\n\nThe Fibonacci series is one of the most common starter questions in Java interviews, but what it really checks is how well you understand iteration and value flow. The series begins with two base numbers, usually 0 and 1. Every number that follows is simply the sum of the previous two.<\/p>\n\n\n\n
When interviewers ask this, they are less interested in the series itself and more interested in how you manage changing values inside a loop. You need to remember the previous two numbers at every step and update them correctly as the loop progresses.<\/p>\n\n\n\n
What the question actually tests<\/strong><\/p>\n\n\n\nThis is not about Fibonacci numbers. It\u2019s about whether you understand state<\/strong>, iteration<\/strong>, and how values evolve over time<\/strong>.<\/p>\n\n\n\nInterviewers want to see:<\/p>\n\n\n\n
\n- Can you track previous values correctly?<\/li>\n\n\n\n
- Do you understand loop-based sequence generation?<\/li>\n\n\n\n
- Do you know when recursion is overkill?<\/li>\n<\/ul>\n\n\n\n
The core idea<\/strong><\/p>\n\n\n\nEach number is the sum of the previous two<\/strong>.<\/p>\n\n\n\nSo if you start with:<\/p>\n\n\n\n
0, 1<\/code><\/p>\n\n\n\nThe sequence becomes:<\/p>\n\n\n\n
0, 1, 1, 2, 3, 5, 8, ...<\/code><\/p>\n\n\n\nYou are always carrying two values forward<\/strong>.<\/p>\n\n\n\nHow to think before coding?<\/strong><\/p>\n\n\n\nAsk yourself:<\/p>\n\n\n\n
\n- What are my starting values?<\/li>\n\n\n\n
- How many times should the loop run?<\/li>\n\n\n\n
- Which variables change every iteration?<\/li>\n<\/ul>\n\n\n\n
You usually need:<\/p>\n\n\n\n
\n- one variable for the previous value<\/li>\n\n\n\n
- one for the current value<\/li>\n\n\n\n
- one temporary variable for the next value<\/li>\n<\/ul>\n\n\n\n
Logic walkthrough<\/strong><\/p>\n\n\n\n\n- Initialize first two numbers.<\/li>\n\n\n\n
- Print them.<\/li>\n\n\n\n
- Loop from the 3rd position to n.<\/li>\n\n\n\n
- Calculate
next = previous + current<\/code>.<\/li>\n\n\n\n- Shift values forward.<\/li>\n<\/ol>\n\n\n\n
Common mistakes<\/strong><\/p>\n\n\n\n\n- Forgetting to update variables in the right order.<\/li>\n\n\n\n
- Using recursion without explaining its cost.<\/li>\n\n\n\n
- Off-by-one errors in loop conditions.<\/li>\n<\/ul>\n\n\n\n
Interview twist you might hear<\/strong><\/p>\n\n\n\n\n- \u201cCan you do this using recursion?\u201d<\/li>\n\n\n\n
- \u201cWhat\u2019s the time complexity?\u201d<\/li>\n\n\n\n
- \u201cHow would you generate only even Fibonacci numbers?\u201d<\/li>\n<\/ul>\n\n\n\n
If you say:<\/p>\n\n\n\n
Iterative solution is O(n) time and O(1) space, you instantly sound more senior.<\/em><\/p>\n\n\n\nHere is the code to Fibonacci Series:<\/p>\n\n\n\n
\nclass FibonacciSeries{ \npublic static void main(String args[]) \n{ \n int a1=0,b2=1,c3,i,count=10; \n System.out.println(a1+\" \"+b2); \/\/simply printing 0 and 1 <\/strong> \n \n for(i=2;i<count;++i) \/\/the loop starts from 2 because 0 and 1 have already been printed <\/strong> \n { \n c3=a1+b2; \n System.out.println(\" \"+c3); \n a1=b2; \n b2=c3; \n } \n \n}} \n<\/code><\/pre>\n\n\n\n2. Check if a Number is Prime<\/strong><\/h3>\n\n\n\nThe prime number program looks simple on the surface, but it\u2019s a great test of logical thinking and optimization. A prime number is a number greater than 1 that is divisible only by 1 and itself.<\/p>\n\n\n\n
Freshers often make the mistake of checking divisibility all the way up to the number itself. Interviewers want to see whether you know that you can stop much earlier. The key insight is that if a number has any factor greater than its square root, it must also have a factor smaller than the square root. So checking up to \u221an is enough.<\/p>\n\n\n\n
In an interview, your explanation matters as much as your code. Saying why you stop at the square root immediately signals strong fundamentals.<\/p>\n\n\n\n
What the question actually tests<\/strong><\/p>\n\n\n\nThis is about logical pruning<\/strong> and efficiency<\/strong>, not division.<\/p>\n\n\n\nInterviewers are checking:<\/p>\n\n\n\n
\n- Do you understand mathematical boundaries?<\/li>\n\n\n\n
- Can you optimize brute-force logic?<\/li>\n\n\n\n
- Do you handle edge cases?<\/li>\n<\/ul>\n\n\n\n
The core idea<\/strong><\/p>\n\n\n\nA prime number:<\/p>\n\n\n\n
\n- has exactly two divisors<\/strong>: 1 and itself<\/li>\n\n\n\n
- cannot be divided evenly by any other number<\/li>\n<\/ul>\n\n\n\n
How to think before coding?<\/strong><\/p>\n\n\n\nAsk:<\/p>\n\n\n\n
\n- What\u2019s the smallest number that is not prime? (1)<\/li>\n\n\n\n
- Do I need to check all numbers till n? (No)<\/li>\n\n\n\n
- When can I stop early?<\/li>\n<\/ul>\n\n\n\n
Key insight:
If a number has a factor greater than \u221an, it must also have one smaller than \u221an<\/strong>. So checking till \u221an is enough.<\/p>\n\n\n\nLogic walkthrough<\/strong><\/p>\n\n\n\n\n- If number \u2264 1 \u2192 not prime.<\/li>\n\n\n\n
- Loop from 2 to \u221an.<\/li>\n\n\n\n
- If any number divides evenly \u2192 not prime.<\/li>\n\n\n\n
- If none do \u2192 prime.<\/li>\n<\/ol>\n\n\n\n
Common mistakes<\/strong><\/p>\n\n\n\n\n- Checking till
n-1<\/code> (wastes time).<\/li>\n\n\n\n- Forgetting that 1 is not prime.<\/li>\n\n\n\n
- Not breaking the loop early.<\/li>\n<\/ul>\n\n\n\n
Interview twist you might hear<\/strong><\/p>\n\n\n\n\n- \u201cHow would you check primes for very large numbers?\u201d<\/li>\n\n\n\n
- \u201cHow would you find all primes in a range?\u201d (Sieve of Eratosthenes)<\/li>\n\n\n\n
- \u201cWhat\u2019s the time complexity?\u201d<\/li>\n<\/ul>\n\n\n\n
Correct answer:<\/p>\n\n\n\n
O(\u221an) time, O(1) space<\/em><\/p>\n\n\n\nHere is the code for finding whether a number is prime or not!<\/p>\n\n\n\n
public class PrimeCheck{ \n public static void main(String args[]){ \n int a,b=0,flag=0; \n int num=3; \/\/this variable represents the number to be checked <\/strong> \n b=num\/2; \n if(num==0||num==1){ \n System.out.println(num+\" is not prime number\"); \n }else{ \n for(a=2;i<=b;i++){ \n if(num%a==0){ \n System.out.println(a+\" is not prime number\"); \n flag=1; \n break; \n } \n } \n if(flag==0) { System.out.println(a+\" is prime number\"); } \n }\n} \n} \n<\/code><\/pre>\n\n\n\n3. String Palindrome<\/strong><\/h3>\n\n\n\n![\"String](\"https:\/\/www.guvi.in\/blog\/wp-content\/uploads\/2026\/03\/3-8.png\" "\\"\\"")
<\/figure>\n\n\n\nA string palindrome question tests how well you handle strings, indexes, and comparisons in Java. A palindrome is a string that reads the same forwards and backwards, such as \u201clevel\u201d or \u201cmadam\u201d.<\/p>\n\n\n\n
Many beginners reverse the string and compare it with the original. While that works, interviewers often prefer a more efficient approach that compares characters from both ends moving inward. This shows better space awareness and algorithmic thinking.<\/p>\n\n\n\n
The idea is simple. You start with one pointer at the beginning of the string and another at the end. If the characters match, you move both pointers toward the center. If at any point they don\u2019t match, the string is not a palindrome.<\/p>\n\n\n\n
What the question actually tests<\/strong><\/p>\n\n\n\nThis checks string handling<\/strong>, indexes<\/strong>, and comparison logic<\/strong>.<\/p>\n\n\n\nMore importantly, it checks:<\/p>\n\n\n\n
\n- Do you understand immutability?<\/li>\n\n\n\n
- Can you reason with characters?<\/li>\n\n\n\n
- Do you choose the right approach?<\/li>\n<\/ul>\n\n\n\n
The core idea<\/strong><\/p>\n\n\n\nA string is a palindrome if:<\/p>\n\n\n\n
original == reversed<\/code><\/p>\n\n\n\nExamples:<\/p>\n\n\n\n
\n- level<\/li>\n\n\n\n
- madam<\/li>\n\n\n\n
- radar<\/li>\n<\/ul>\n\n\n\n
How to think before coding?<\/strong><\/p>\n\n\n\nYou have two main approaches:<\/p>\n\n\n\n
\n- Reverse the string and compare.<\/li>\n\n\n\n
- Compare characters from both ends moving inward.<\/li>\n<\/ol>\n\n\n\n
Approach 2 is more efficient and interview-friendly.<\/p>\n\n\n\n
Logic walkthrough (two-pointer method)<\/strong><\/p>\n\n\n\n\n- Set
left = 0<\/code>, right = length - 1<\/code><\/li>\n\n\n\n- Compare characters at both positions.<\/li>\n\n\n\n
- If mismatch \u2192 not a palindrome.<\/li>\n\n\n\n
- Move inward until pointers cross.<\/li>\n<\/ol>\n\n\n\n
Common mistakes<\/strong><\/p>\n\n\n\n\n- Using
==<\/code> instead of .equals()<\/code>.<\/li>\n\n\n\n- Ignoring case sensitivity.<\/li>\n\n\n\n
- Not handling spaces or punctuation when asked.<\/li>\n<\/ul>\n\n\n\n
Interview twist you might hear<\/strong><\/p>\n\n\n\n\n- \u201cIgnore spaces and special characters.\u201d<\/li>\n\n\n\n
- \u201cMake it case-insensitive.\u201d<\/li>\n\n\n\n
- \u201cCheck palindrome without extra memory.\u201d<\/li>\n<\/ul>\n\n\n\n
If you handle these calmly, you stand out.<\/p>\n\n\n\n
Here is the code for Palindrome checking:<\/p>\n\n\n\n
class PalindromePractice{ \n public static void main(String args[]){ \n int p,sum=0,tempo; \n int num=454; \/\/This variable represents the number to be checked for palindrome<\/strong> \n \n tempo=num; \n while(num>0){ \n p=num%10; \/\/getting the remainder <\/strong>\n sum=(sum*10)+p; \n num=num\/10; \n } \n if(tempo==sum) \n System.out.println(\"palindrome number \"); \n else \n System.out.println(\"not palindrome\"); \n} \n} <\/code><\/pre>\n\n\n\n\ud83d\udca1 Did You Know?<\/strong>
Many so-called \u201cbasic\u201d Java interview programs are actually used by interviewers to judge senior-level thinking. For example, a Fibonacci question can quickly turn into a discussion on time and space complexity, while a palindrome check reveals how well you understand memory usage and string handling. These problems aren\u2019t about memorization: they\u2019re designed to see how clearly you think under pressure and how well you explain your logic.<\/div>\n\n\n\n4. Bubble Sort <\/strong><\/h3>\n\n\n\n![\"Bubble](\"https:\/\/www.guvi.in\/blog\/wp-content\/uploads\/2026\/03\/4-6.png\" "\\"\\"")
<\/figure>\n\n\n\nBubble Sort<\/a> is often the first sorting algorithm <\/a>that freshers encounter in interviews. It\u2019s not used much in real-world systems, but it\u2019s excellent for testing your understanding of nested loops and array manipulation.<\/p>\n\n\n\nThe core idea is that you repeatedly compare adjacent elements and swap them if they are out of order. After each pass, the largest unsorted element \u201cbubbles\u201d to the end of the array. With every new pass, one more element is already sorted, reducing the number of comparisons.<\/p>\n\n\n\n
What does <\/strong>the question actually test?<\/strong><\/p>\n\n\n\nThis is not about sorting efficiency. It\u2019s about nested loops<\/strong>, comparisons<\/strong>, and swapping logic<\/strong>.<\/p>\n\n\n\nInterviewers want to see:<\/p>\n\n\n\n
\n- Can you reason step by step?<\/li>\n\n\n\n
- Do you understand how arrays mutate?<\/li>\n\n\n\n
- Can you explain each pass clearly?<\/li>\n<\/ul>\n\n\n\n
The core idea<\/strong><\/p>\n\n\n\nBubble Sort:<\/p>\n\n\n\n
\n- compares adjacent elements<\/li>\n\n\n\n
- swaps them if they\u2019re in the wrong order<\/li>\n\n\n\n
- pushes the largest element to the end in each pass<\/li>\n<\/ul>\n\n\n\n
How to think before coding?<\/strong><\/p>\n\n\n\nAsk:<\/p>\n\n\n\n
\n- How many passes do I need? (n-1)<\/li>\n\n\n\n
- After each pass, what is already sorted?<\/li>\n\n\n\n
- Can I reduce comparisons in later passes?<\/li>\n<\/ul>\n\n\n\n
Logic walkthrough<\/strong><\/p>\n\n\n\n\n- Outer loop controls passes.<\/li>\n\n\n\n
- Inner loop compares adjacent elements.<\/li>\n\n\n\n
- Swap if left > right.<\/li>\n\n\n\n
- Reduce inner loop range each pass.<\/li>\n\n\n\n
- <\/li>\n<\/ol>\n\n\n\n
Common mistakes<\/strong><\/p>\n\n\n\n\n- Wrong loop boundaries.<\/li>\n\n\n\n
- Forgetting to reduce inner loop range.<\/li>\n\n\n\n
- Messing up swap logic.<\/li>\n<\/ul>\n\n\n\n
Interview twist you might hear<\/strong><\/p>\n\n\n\n\n- \u201cWhat\u2019s the time complexity?\u201d \u2192 O(n\u00b2)<\/li>\n\n\n\n
- \u201cIs Bubble Sort stable?\u201d \u2192 Yes<\/li>\n\n\n\n
- \u201cWhy is it rarely used in real systems?\u201d<\/li>\n<\/ul>\n\n\n\n
Correct reasoning:<\/p>\n\n\n\n
It\u2019s simple but inefficient for large datasets.<\/em><\/p>\n\n\n\nHere is how you solve bubble sort:<\/p>\n\n\n\n
public class BubbleSort { \n static void bubbleSort(int[] arr) { \n int num = arr.length; \n int tempo = 0; \n for(int i=0; i < num; i++){ \n for(int j=1; j < (num-i); j++){ \n if(arr[j-1] > arr[j]){ \n \/\/swapping elements <\/strong> \n tempo = arr[j-1]; \n arr[j-1] = arr[j]; \n arr[j] = tempo; \n } \n \n } \n } \n \n } \n public static void main(String[] args) { \n int arr[] ={3,60,35,2,45,320,5}; \n \n System.out.println(\"The array before Bubble Sort:\"); \n for(int i=0; i < arr.length; i++){ \n System.out.println(arr[i] + \" \"); \n } \n System.out.println(); \n \n bubbleSort(arr); \/\/sorting the array elements using bubble sort method we wrote earlier <\/strong>\n \n System.out.println(\"The array after Bubble Sort:\"); \n for(int i=0; i < arr.length; i++){ \n System.out.println(arr[i] + \" \"); \n } \n \n } \n} <\/code><\/pre>\n\n\n\n5. Armstrong Number<\/strong><\/h3>\n\n\n\n![\"Armstrong](\"https:\/\/www.guvi.in\/blog\/wp-content\/uploads\/2026\/03\/5-4.png\" "\\"\\"")
<\/figure>\n\n\n\nThe Armstrong number program tests your ability to break a number into digits and apply mathematical logic. A number is called an Armstrong number if the sum of its digits raised to the power of the number of digits equals the original number.<\/p>\n\n\n\n
For example, 153 is an Armstrong number because:
1\u00b3 + 5\u00b3 + 3\u00b3 = 153<\/p>\n\n\n\n
To solve this, you first store the original number. Then you repeatedly extract the last digit using the modulus operator, raise it to the required power, and add it to a running sum. At the end, you compare the sum with the original number.<\/p>\n\n\n\n
What does <\/strong>the question actually test?<\/strong><\/p>\n\n\n\nThis checks digit extraction<\/strong>, math operations<\/strong>, and loop control<\/strong>.<\/p>\n\n\n\nIt also quietly checks:<\/p>\n\n\n\n
\n- Can you break a number into digits?<\/li>\n\n\n\n
- Can you reconstruct logic from a definition?<\/li>\n<\/ul>\n\n\n\n
The core idea<\/strong><\/p>\n\n\n\nA number is Armstrong if:<\/p>\n\n\n\n
sum of (each digit ^ number of digits) == original number<\/code><\/p>\n\n\n\nExample:<\/p>\n\n\n\n
153 1\u00b3 + 5\u00b3 + 3\u00b3 = 153<\/code><\/p>\n\n\n\nHow to think before coding?<\/strong><\/p>\n\n\n\nYou need to:<\/p>\n\n\n\n
\n- Preserve the original number.<\/li>\n\n\n\n
- Extract digits one by one.<\/li>\n\n\n\n
- Apply power logic.<\/li>\n\n\n\n
- Compare result with original.<\/li>\n<\/ol>\n\n\n\n
Logic walkthrough<\/strong><\/p>\n\n\n\n\n- Store original number.<\/li>\n\n\n\n
- While number > 0:\n
\n- digit = number % 10<\/li>\n\n\n\n
- add digit\u00b3 to sum<\/li>\n\n\n\n
- number = number \/ 10<\/li>\n<\/ul>\n<\/li>\n\n\n\n
- Compare sum with original.<\/li>\n<\/ol>\n\n\n\n
Common mistakes<\/strong><\/p>\n\n\n\n\n- Forgetting to reset the number.<\/li>\n\n\n\n
- Hardcoding power without explanation.<\/li>\n\n\n\n
- Using floating-point comparisons carelessly.<\/li>\n<\/ul>\n\n\n\n
Interview twist you might hear<\/strong><\/p>\n\n\n\n\n- \u201cWhat about numbers with different digit counts?\u201d<\/li>\n\n\n\n
- \u201cCan you make this generic?\u201d<\/li>\n\n\n\n
- \u201cWhat\u2019s the time complexity?\u201d<\/li>\n<\/ul>\n\n\n\n
A strong answer:<\/p>\n\n\n\n
Time complexity is O(d), where d is the number of digits.<\/em><\/p>\n\n\n\nHere is the code for checking Armstrong numbers:<\/p>\n\n\n\n
public class ArmstrongCheck {\n\n public static void main(String[] args) {\n\n int num = 1634, ogNum, remainder, final = 0, h = 0;\n\n ogNum = num;\n\n for (;ogNum != 0; ogNum \/= 10, ++h);\n\n ogNum = num;\n\n for (;ogNum != 0; ogNum \/= 10)\n {\n remainder = ogNum % 10;\n final += Math.pow(remainder, h);\n }\n\n if(final == num)\n System.out.println(num + \" is an Armstrong number.\");\n else\n System.out.println(num + \" is not an Armstrong number.\");\n }\n}\n<\/code><\/pre>\n\n\n\nTips for Solving Interview Questions Efficiently<\/strong><\/h2>\n\n\n\nNow that you are familiar with some of the most commonly asked Java interview programs, here are some tips to help you get through the interviews with ease:<\/p>\n\n\n\n
\n- Practice, practice, practice<\/strong> – the more you practice solving Java programs, the more comfortable and confident you will be during the interview.<\/li>\n\n\n\n
- Understand the logic behind the programs<\/strong> – it is important to not just memorize the solutions to the programming questions but to also understand the logic behind them.<\/li>\n\n\n\n
- Keep it simple<\/strong> – while it is important to write efficient and optimized code, it is also important to keep it simple and easy to read.<\/li>\n\n\n\n
- Show your work<\/strong> – be sure to explain your thought process and the steps you took to arrive at your solution.<\/li>\n\n\n\n
- Test your code thoroughly<\/strong> – a must for all programmers: shows how serious you are about your code so get into a habit of testing it regularly.<\/li>\n<\/ol>\n\n\n\n
<\/figure>\n\n\n\nThese 5 must-know Java Interview programs are designed to test your problem-solving skills and your ability to apply Java concepts to real-world scenarios.<\/p>\n\n\n\n
1. Fibonacci Series<\/strong><\/h3>\n\n\n\n<\/figure>\n\n\n\nThe Fibonacci series is one of the most common starter questions in Java interviews, but what it really checks is how well you understand iteration and value flow. The series begins with two base numbers, usually 0 and 1. Every number that follows is simply the sum of the previous two.<\/p>\n\n\n\n
When interviewers ask this, they are less interested in the series itself and more interested in how you manage changing values inside a loop. You need to remember the previous two numbers at every step and update them correctly as the loop progresses.<\/p>\n\n\n\n
What the question actually tests<\/strong><\/p>\n\n\n\nThis is not about Fibonacci numbers. It\u2019s about whether you understand state<\/strong>, iteration<\/strong>, and how values evolve over time<\/strong>.<\/p>\n\n\n\nInterviewers want to see:<\/p>\n\n\n\n
\n- Can you track previous values correctly?<\/li>\n\n\n\n
- Do you understand loop-based sequence generation?<\/li>\n\n\n\n
- Do you know when recursion is overkill?<\/li>\n<\/ul>\n\n\n\n
The core idea<\/strong><\/p>\n\n\n\nEach number is the sum of the previous two<\/strong>.<\/p>\n\n\n\nSo if you start with:<\/p>\n\n\n\n
0, 1<\/code><\/p>\n\n\n\nThe sequence becomes:<\/p>\n\n\n\n
0, 1, 1, 2, 3, 5, 8, ...<\/code><\/p>\n\n\n\nYou are always carrying two values forward<\/strong>.<\/p>\n\n\n\nHow to think before coding?<\/strong><\/p>\n\n\n\nAsk yourself:<\/p>\n\n\n\n
\n- What are my starting values?<\/li>\n\n\n\n
- How many times should the loop run?<\/li>\n\n\n\n
- Which variables change every iteration?<\/li>\n<\/ul>\n\n\n\n
You usually need:<\/p>\n\n\n\n
\n- one variable for the previous value<\/li>\n\n\n\n
- one for the current value<\/li>\n\n\n\n
- one temporary variable for the next value<\/li>\n<\/ul>\n\n\n\n
Logic walkthrough<\/strong><\/p>\n\n\n\n\n- Initialize first two numbers.<\/li>\n\n\n\n
- Print them.<\/li>\n\n\n\n
- Loop from the 3rd position to n.<\/li>\n\n\n\n
- Calculate
next = previous + current<\/code>.<\/li>\n\n\n\n- Shift values forward.<\/li>\n<\/ol>\n\n\n\n
Common mistakes<\/strong><\/p>\n\n\n\n\n- Forgetting to update variables in the right order.<\/li>\n\n\n\n
- Using recursion without explaining its cost.<\/li>\n\n\n\n
- Off-by-one errors in loop conditions.<\/li>\n<\/ul>\n\n\n\n
Interview twist you might hear<\/strong><\/p>\n\n\n\n\n- \u201cCan you do this using recursion?\u201d<\/li>\n\n\n\n
- \u201cWhat\u2019s the time complexity?\u201d<\/li>\n\n\n\n
- \u201cHow would you generate only even Fibonacci numbers?\u201d<\/li>\n<\/ul>\n\n\n\n
If you say:<\/p>\n\n\n\n
Iterative solution is O(n) time and O(1) space, you instantly sound more senior.<\/em><\/p>\n\n\n\nHere is the code to Fibonacci Series:<\/p>\n\n\n\n
\nclass FibonacciSeries{ \npublic static void main(String args[]) \n{ \n int a1=0,b2=1,c3,i,count=10; \n System.out.println(a1+\" \"+b2); \/\/simply printing 0 and 1 <\/strong> \n \n for(i=2;i<count;++i) \/\/the loop starts from 2 because 0 and 1 have already been printed <\/strong> \n { \n c3=a1+b2; \n System.out.println(\" \"+c3); \n a1=b2; \n b2=c3; \n } \n \n}} \n<\/code><\/pre>\n\n\n\n2. Check if a Number is Prime<\/strong><\/h3>\n\n\n\nThe prime number program looks simple on the surface, but it\u2019s a great test of logical thinking and optimization. A prime number is a number greater than 1 that is divisible only by 1 and itself.<\/p>\n\n\n\n
Freshers often make the mistake of checking divisibility all the way up to the number itself. Interviewers want to see whether you know that you can stop much earlier. The key insight is that if a number has any factor greater than its square root, it must also have a factor smaller than the square root. So checking up to \u221an is enough.<\/p>\n\n\n\n
In an interview, your explanation matters as much as your code. Saying why you stop at the square root immediately signals strong fundamentals.<\/p>\n\n\n\n
What the question actually tests<\/strong><\/p>\n\n\n\nThis is about logical pruning<\/strong> and efficiency<\/strong>, not division.<\/p>\n\n\n\nInterviewers are checking:<\/p>\n\n\n\n
\n- Do you understand mathematical boundaries?<\/li>\n\n\n\n
- Can you optimize brute-force logic?<\/li>\n\n\n\n
- Do you handle edge cases?<\/li>\n<\/ul>\n\n\n\n
The core idea<\/strong><\/p>\n\n\n\nA prime number:<\/p>\n\n\n\n
\n- has exactly two divisors<\/strong>: 1 and itself<\/li>\n\n\n\n
- cannot be divided evenly by any other number<\/li>\n<\/ul>\n\n\n\n
How to think before coding?<\/strong><\/p>\n\n\n\nAsk:<\/p>\n\n\n\n
\n- What\u2019s the smallest number that is not prime? (1)<\/li>\n\n\n\n
- Do I need to check all numbers till n? (No)<\/li>\n\n\n\n
- When can I stop early?<\/li>\n<\/ul>\n\n\n\n
Key insight:
If a number has a factor greater than \u221an, it must also have one smaller than \u221an<\/strong>. So checking till \u221an is enough.<\/p>\n\n\n\nLogic walkthrough<\/strong><\/p>\n\n\n\n\n- If number \u2264 1 \u2192 not prime.<\/li>\n\n\n\n
- Loop from 2 to \u221an.<\/li>\n\n\n\n
- If any number divides evenly \u2192 not prime.<\/li>\n\n\n\n
- If none do \u2192 prime.<\/li>\n<\/ol>\n\n\n\n
Common mistakes<\/strong><\/p>\n\n\n\n\n- Checking till
n-1<\/code> (wastes time).<\/li>\n\n\n\n- Forgetting that 1 is not prime.<\/li>\n\n\n\n
- Not breaking the loop early.<\/li>\n<\/ul>\n\n\n\n
Interview twist you might hear<\/strong><\/p>\n\n\n\n\n- \u201cHow would you check primes for very large numbers?\u201d<\/li>\n\n\n\n
- \u201cHow would you find all primes in a range?\u201d (Sieve of Eratosthenes)<\/li>\n\n\n\n
- \u201cWhat\u2019s the time complexity?\u201d<\/li>\n<\/ul>\n\n\n\n
Correct answer:<\/p>\n\n\n\n
O(\u221an) time, O(1) space<\/em><\/p>\n\n\n\nHere is the code for finding whether a number is prime or not!<\/p>\n\n\n\n
public class PrimeCheck{ \n public static void main(String args[]){ \n int a,b=0,flag=0; \n int num=3; \/\/this variable represents the number to be checked <\/strong> \n b=num\/2; \n if(num==0||num==1){ \n System.out.println(num+\" is not prime number\"); \n }else{ \n for(a=2;i<=b;i++){ \n if(num%a==0){ \n System.out.println(a+\" is not prime number\"); \n flag=1; \n break; \n } \n } \n if(flag==0) { System.out.println(a+\" is prime number\"); } \n }\n} \n} \n<\/code><\/pre>\n\n\n\n3. String Palindrome<\/strong><\/h3>\n\n\n\n<\/figure>\n\n\n\nA string palindrome question tests how well you handle strings, indexes, and comparisons in Java. A palindrome is a string that reads the same forwards and backwards, such as \u201clevel\u201d or \u201cmadam\u201d.<\/p>\n\n\n\n
Many beginners reverse the string and compare it with the original. While that works, interviewers often prefer a more efficient approach that compares characters from both ends moving inward. This shows better space awareness and algorithmic thinking.<\/p>\n\n\n\n
The idea is simple. You start with one pointer at the beginning of the string and another at the end. If the characters match, you move both pointers toward the center. If at any point they don\u2019t match, the string is not a palindrome.<\/p>\n\n\n\n
What the question actually tests<\/strong><\/p>\n\n\n\nThis checks string handling<\/strong>, indexes<\/strong>, and comparison logic<\/strong>.<\/p>\n\n\n\nMore importantly, it checks:<\/p>\n\n\n\n
\n- Do you understand immutability?<\/li>\n\n\n\n
- Can you reason with characters?<\/li>\n\n\n\n
- Do you choose the right approach?<\/li>\n<\/ul>\n\n\n\n
The core idea<\/strong><\/p>\n\n\n\nA string is a palindrome if:<\/p>\n\n\n\n
original == reversed<\/code><\/p>\n\n\n\nExamples:<\/p>\n\n\n\n
\n- level<\/li>\n\n\n\n
- madam<\/li>\n\n\n\n
- radar<\/li>\n<\/ul>\n\n\n\n
How to think before coding?<\/strong><\/p>\n\n\n\nYou have two main approaches:<\/p>\n\n\n\n
\n- Reverse the string and compare.<\/li>\n\n\n\n
- Compare characters from both ends moving inward.<\/li>\n<\/ol>\n\n\n\n
Approach 2 is more efficient and interview-friendly.<\/p>\n\n\n\n
Logic walkthrough (two-pointer method)<\/strong><\/p>\n\n\n\n\n- Set
left = 0<\/code>, right = length - 1<\/code><\/li>\n\n\n\n- Compare characters at both positions.<\/li>\n\n\n\n
- If mismatch \u2192 not a palindrome.<\/li>\n\n\n\n
- Move inward until pointers cross.<\/li>\n<\/ol>\n\n\n\n
Common mistakes<\/strong><\/p>\n\n\n\n\n- Using
==<\/code> instead of .equals()<\/code>.<\/li>\n\n\n\n- Ignoring case sensitivity.<\/li>\n\n\n\n
- Not handling spaces or punctuation when asked.<\/li>\n<\/ul>\n\n\n\n
Interview twist you might hear<\/strong><\/p>\n\n\n\n\n- \u201cIgnore spaces and special characters.\u201d<\/li>\n\n\n\n
- \u201cMake it case-insensitive.\u201d<\/li>\n\n\n\n
- \u201cCheck palindrome without extra memory.\u201d<\/li>\n<\/ul>\n\n\n\n
If you handle these calmly, you stand out.<\/p>\n\n\n\n
Here is the code for Palindrome checking:<\/p>\n\n\n\n
class PalindromePractice{ \n public static void main(String args[]){ \n int p,sum=0,tempo; \n int num=454; \/\/This variable represents the number to be checked for palindrome<\/strong> \n \n tempo=num; \n while(num>0){ \n p=num%10; \/\/getting the remainder <\/strong>\n sum=(sum*10)+p; \n num=num\/10; \n } \n if(tempo==sum) \n System.out.println(\"palindrome number \"); \n else \n System.out.println(\"not palindrome\"); \n} \n} <\/code><\/pre>\n\n\n\n\ud83d\udca1 Did You Know?<\/strong>
Many so-called \u201cbasic\u201d Java interview programs are actually used by interviewers to judge senior-level thinking. For example, a Fibonacci question can quickly turn into a discussion on time and space complexity, while a palindrome check reveals how well you understand memory usage and string handling. These problems aren\u2019t about memorization: they\u2019re designed to see how clearly you think under pressure and how well you explain your logic.<\/div>\n\n\n\n4. Bubble Sort <\/strong><\/h3>\n\n\n\n<\/figure>\n\n\n\nBubble Sort<\/a> is often the first sorting algorithm <\/a>that freshers encounter in interviews. It\u2019s not used much in real-world systems, but it\u2019s excellent for testing your understanding of nested loops and array manipulation.<\/p>\n\n\n\nThe core idea is that you repeatedly compare adjacent elements and swap them if they are out of order. After each pass, the largest unsorted element \u201cbubbles\u201d to the end of the array. With every new pass, one more element is already sorted, reducing the number of comparisons.<\/p>\n\n\n\n
What does <\/strong>the question actually test?<\/strong><\/p>\n\n\n\nThis is not about sorting efficiency. It\u2019s about nested loops<\/strong>, comparisons<\/strong>, and swapping logic<\/strong>.<\/p>\n\n\n\nInterviewers want to see:<\/p>\n\n\n\n
\n- Can you reason step by step?<\/li>\n\n\n\n
- Do you understand how arrays mutate?<\/li>\n\n\n\n
- Can you explain each pass clearly?<\/li>\n<\/ul>\n\n\n\n
The core idea<\/strong><\/p>\n\n\n\nBubble Sort:<\/p>\n\n\n\n
\n- compares adjacent elements<\/li>\n\n\n\n
- swaps them if they\u2019re in the wrong order<\/li>\n\n\n\n
- pushes the largest element to the end in each pass<\/li>\n<\/ul>\n\n\n\n
How to think before coding?<\/strong><\/p>\n\n\n\nAsk:<\/p>\n\n\n\n
\n- How many passes do I need? (n-1)<\/li>\n\n\n\n
- After each pass, what is already sorted?<\/li>\n\n\n\n
- Can I reduce comparisons in later passes?<\/li>\n<\/ul>\n\n\n\n
Logic walkthrough<\/strong><\/p>\n\n\n\n\n- Outer loop controls passes.<\/li>\n\n\n\n
- Inner loop compares adjacent elements.<\/li>\n\n\n\n
- Swap if left > right.<\/li>\n\n\n\n
- Reduce inner loop range each pass.<\/li>\n\n\n\n
- <\/li>\n<\/ol>\n\n\n\n
Common mistakes<\/strong><\/p>\n\n\n\n\n- Wrong loop boundaries.<\/li>\n\n\n\n
- Forgetting to reduce inner loop range.<\/li>\n\n\n\n
- Messing up swap logic.<\/li>\n<\/ul>\n\n\n\n
Interview twist you might hear<\/strong><\/p>\n\n\n\n\n- \u201cWhat\u2019s the time complexity?\u201d \u2192 O(n\u00b2)<\/li>\n\n\n\n
- \u201cIs Bubble Sort stable?\u201d \u2192 Yes<\/li>\n\n\n\n
- \u201cWhy is it rarely used in real systems?\u201d<\/li>\n<\/ul>\n\n\n\n
Correct reasoning:<\/p>\n\n\n\n
It\u2019s simple but inefficient for large datasets.<\/em><\/p>\n\n\n\nHere is how you solve bubble sort:<\/p>\n\n\n\n
public class BubbleSort { \n static void bubbleSort(int[] arr) { \n int num = arr.length; \n int tempo = 0; \n for(int i=0; i < num; i++){ \n for(int j=1; j < (num-i); j++){ \n if(arr[j-1] > arr[j]){ \n \/\/swapping elements <\/strong> \n tempo = arr[j-1]; \n arr[j-1] = arr[j]; \n arr[j] = tempo; \n } \n \n } \n } \n \n } \n public static void main(String[] args) { \n int arr[] ={3,60,35,2,45,320,5}; \n \n System.out.println(\"The array before Bubble Sort:\"); \n for(int i=0; i < arr.length; i++){ \n System.out.println(arr[i] + \" \"); \n } \n System.out.println(); \n \n bubbleSort(arr); \/\/sorting the array elements using bubble sort method we wrote earlier <\/strong>\n \n System.out.println(\"The array after Bubble Sort:\"); \n for(int i=0; i < arr.length; i++){ \n System.out.println(arr[i] + \" \"); \n } \n \n } \n} <\/code><\/pre>\n\n\n\n5. Armstrong Number<\/strong><\/h3>\n\n\n\n<\/figure>\n\n\n\nThe Armstrong number program tests your ability to break a number into digits and apply mathematical logic. A number is called an Armstrong number if the sum of its digits raised to the power of the number of digits equals the original number.<\/p>\n\n\n\n
For example, 153 is an Armstrong number because:
1\u00b3 + 5\u00b3 + 3\u00b3 = 153<\/p>\n\n\n\n
To solve this, you first store the original number. Then you repeatedly extract the last digit using the modulus operator, raise it to the required power, and add it to a running sum. At the end, you compare the sum with the original number.<\/p>\n\n\n\n
What does <\/strong>the question actually test?<\/strong><\/p>\n\n\n\nThis checks digit extraction<\/strong>, math operations<\/strong>, and loop control<\/strong>.<\/p>\n\n\n\nIt also quietly checks:<\/p>\n\n\n\n
\n- Can you break a number into digits?<\/li>\n\n\n\n
- Can you reconstruct logic from a definition?<\/li>\n<\/ul>\n\n\n\n
The core idea<\/strong><\/p>\n\n\n\nA number is Armstrong if:<\/p>\n\n\n\n
sum of (each digit ^ number of digits) == original number<\/code><\/p>\n\n\n\nExample:<\/p>\n\n\n\n
153 1\u00b3 + 5\u00b3 + 3\u00b3 = 153<\/code><\/p>\n\n\n\nHow to think before coding?<\/strong><\/p>\n\n\n\nYou need to:<\/p>\n\n\n\n
\n- Preserve the original number.<\/li>\n\n\n\n
- Extract digits one by one.<\/li>\n\n\n\n
- Apply power logic.<\/li>\n\n\n\n
- Compare result with original.<\/li>\n<\/ol>\n\n\n\n
Logic walkthrough<\/strong><\/p>\n\n\n\n\n- Store original number.<\/li>\n\n\n\n
- While number > 0:\n
\n- digit = number % 10<\/li>\n\n\n\n
- add digit\u00b3 to sum<\/li>\n\n\n\n
- number = number \/ 10<\/li>\n<\/ul>\n<\/li>\n\n\n\n
- Compare sum with original.<\/li>\n<\/ol>\n\n\n\n
Common mistakes<\/strong><\/p>\n\n\n\n\n- Forgetting to reset the number.<\/li>\n\n\n\n
- Hardcoding power without explanation.<\/li>\n\n\n\n
- Using floating-point comparisons carelessly.<\/li>\n<\/ul>\n\n\n\n
Interview twist you might hear<\/strong><\/p>\n\n\n\n\n- \u201cWhat about numbers with different digit counts?\u201d<\/li>\n\n\n\n
- \u201cCan you make this generic?\u201d<\/li>\n\n\n\n
- \u201cWhat\u2019s the time complexity?\u201d<\/li>\n<\/ul>\n\n\n\n
A strong answer:<\/p>\n\n\n\n
Time complexity is O(d), where d is the number of digits.<\/em><\/p>\n\n\n\nHere is the code for checking Armstrong numbers:<\/p>\n\n\n\n
public class ArmstrongCheck {\n\n public static void main(String[] args) {\n\n int num = 1634, ogNum, remainder, final = 0, h = 0;\n\n ogNum = num;\n\n for (;ogNum != 0; ogNum \/= 10, ++h);\n\n ogNum = num;\n\n for (;ogNum != 0; ogNum \/= 10)\n {\n remainder = ogNum % 10;\n final += Math.pow(remainder, h);\n }\n\n if(final == num)\n System.out.println(num + \" is an Armstrong number.\");\n else\n System.out.println(num + \" is not an Armstrong number.\");\n }\n}\n<\/code><\/pre>\n\n\n\nTips for Solving Interview Questions Efficiently<\/strong><\/h2>\n\n\n\nNow that you are familiar with some of the most commonly asked Java interview programs, here are some tips to help you get through the interviews with ease:<\/p>\n\n\n\n
\n- Practice, practice, practice<\/strong> – the more you practice solving Java programs, the more comfortable and confident you will be during the interview.<\/li>\n\n\n\n
- Understand the logic behind the programs<\/strong> – it is important to not just memorize the solutions to the programming questions but to also understand the logic behind them.<\/li>\n\n\n\n
- Keep it simple<\/strong> – while it is important to write efficient and optimized code, it is also important to keep it simple and easy to read.<\/li>\n\n\n\n
- Show your work<\/strong> – be sure to explain your thought process and the steps you took to arrive at your solution.<\/li>\n\n\n\n
- Test your code thoroughly<\/strong> – a must for all programmers: shows how serious you are about your code so get into a habit of testing it regularly.<\/li>\n<\/ol>\n\n\n\n
<\/figure>\n\n\n\nThe Fibonacci series is one of the most common starter questions in Java interviews, but what it really checks is how well you understand iteration and value flow. The series begins with two base numbers, usually 0 and 1. Every number that follows is simply the sum of the previous two.<\/p>\n\n\n\n
When interviewers ask this, they are less interested in the series itself and more interested in how you manage changing values inside a loop. You need to remember the previous two numbers at every step and update them correctly as the loop progresses.<\/p>\n\n\n\n
What the question actually tests<\/strong><\/p>\n\n\n\n This is not about Fibonacci numbers. It\u2019s about whether you understand state<\/strong>, iteration<\/strong>, and how values evolve over time<\/strong>.<\/p>\n\n\n\n Interviewers want to see:<\/p>\n\n\n\n The core idea<\/strong><\/p>\n\n\n\n Each number is the sum of the previous two<\/strong>.<\/p>\n\n\n\n So if you start with:<\/p>\n\n\n\n The sequence becomes:<\/p>\n\n\n\n You are always carrying two values forward<\/strong>.<\/p>\n\n\n\n How to think before coding?<\/strong><\/p>\n\n\n\n Ask yourself:<\/p>\n\n\n\n You usually need:<\/p>\n\n\n\n Logic walkthrough<\/strong><\/p>\n\n\n\n Common mistakes<\/strong><\/p>\n\n\n\n Interview twist you might hear<\/strong><\/p>\n\n\n\n If you say:<\/p>\n\n\n\n Iterative solution is O(n) time and O(1) space, you instantly sound more senior.<\/em><\/p>\n\n\n\n Here is the code to Fibonacci Series:<\/p>\n\n\n\n The prime number program looks simple on the surface, but it\u2019s a great test of logical thinking and optimization. A prime number is a number greater than 1 that is divisible only by 1 and itself.<\/p>\n\n\n\n Freshers often make the mistake of checking divisibility all the way up to the number itself. Interviewers want to see whether you know that you can stop much earlier. The key insight is that if a number has any factor greater than its square root, it must also have a factor smaller than the square root. So checking up to \u221an is enough.<\/p>\n\n\n\n In an interview, your explanation matters as much as your code. Saying why you stop at the square root immediately signals strong fundamentals.<\/p>\n\n\n\n What the question actually tests<\/strong><\/p>\n\n\n\n This is about logical pruning<\/strong> and efficiency<\/strong>, not division.<\/p>\n\n\n\n Interviewers are checking:<\/p>\n\n\n\n The core idea<\/strong><\/p>\n\n\n\n A prime number:<\/p>\n\n\n\n How to think before coding?<\/strong><\/p>\n\n\n\n Ask:<\/p>\n\n\n\n Key insight: Logic walkthrough<\/strong><\/p>\n\n\n\n Common mistakes<\/strong><\/p>\n\n\n\n Interview twist you might hear<\/strong><\/p>\n\n\n\n Correct answer:<\/p>\n\n\n\n O(\u221an) time, O(1) space<\/em><\/p>\n\n\n\n Here is the code for finding whether a number is prime or not!<\/p>\n\n\n\n A string palindrome question tests how well you handle strings, indexes, and comparisons in Java. A palindrome is a string that reads the same forwards and backwards, such as \u201clevel\u201d or \u201cmadam\u201d.<\/p>\n\n\n\n Many beginners reverse the string and compare it with the original. While that works, interviewers often prefer a more efficient approach that compares characters from both ends moving inward. This shows better space awareness and algorithmic thinking.<\/p>\n\n\n\n The idea is simple. You start with one pointer at the beginning of the string and another at the end. If the characters match, you move both pointers toward the center. If at any point they don\u2019t match, the string is not a palindrome.<\/p>\n\n\n\n What the question actually tests<\/strong><\/p>\n\n\n\n This checks string handling<\/strong>, indexes<\/strong>, and comparison logic<\/strong>.<\/p>\n\n\n\n More importantly, it checks:<\/p>\n\n\n\n The core idea<\/strong><\/p>\n\n\n\n A string is a palindrome if:<\/p>\n\n\n\n Examples:<\/p>\n\n\n\n How to think before coding?<\/strong><\/p>\n\n\n\n You have two main approaches:<\/p>\n\n\n\n Approach 2 is more efficient and interview-friendly.<\/p>\n\n\n\n Logic walkthrough (two-pointer method)<\/strong><\/p>\n\n\n\n Common mistakes<\/strong><\/p>\n\n\n\n Interview twist you might hear<\/strong><\/p>\n\n\n\n If you handle these calmly, you stand out.<\/p>\n\n\n\n Here is the code for Palindrome checking:<\/p>\n\n\n\n Bubble Sort<\/a> is often the first sorting algorithm <\/a>that freshers encounter in interviews. It\u2019s not used much in real-world systems, but it\u2019s excellent for testing your understanding of nested loops and array manipulation.<\/p>\n\n\n\n The core idea is that you repeatedly compare adjacent elements and swap them if they are out of order. After each pass, the largest unsorted element \u201cbubbles\u201d to the end of the array. With every new pass, one more element is already sorted, reducing the number of comparisons.<\/p>\n\n\n\n What does <\/strong>the question actually test?<\/strong><\/p>\n\n\n\n This is not about sorting efficiency. It\u2019s about nested loops<\/strong>, comparisons<\/strong>, and swapping logic<\/strong>.<\/p>\n\n\n\n Interviewers want to see:<\/p>\n\n\n\n The core idea<\/strong><\/p>\n\n\n\n Bubble Sort:<\/p>\n\n\n\n How to think before coding?<\/strong><\/p>\n\n\n\n Ask:<\/p>\n\n\n\n Logic walkthrough<\/strong><\/p>\n\n\n\n Common mistakes<\/strong><\/p>\n\n\n\n Interview twist you might hear<\/strong><\/p>\n\n\n\n Correct reasoning:<\/p>\n\n\n\n It\u2019s simple but inefficient for large datasets.<\/em><\/p>\n\n\n\n Here is how you solve bubble sort:<\/p>\n\n\n\n The Armstrong number program tests your ability to break a number into digits and apply mathematical logic. A number is called an Armstrong number if the sum of its digits raised to the power of the number of digits equals the original number.<\/p>\n\n\n\n For example, 153 is an Armstrong number because: To solve this, you first store the original number. Then you repeatedly extract the last digit using the modulus operator, raise it to the required power, and add it to a running sum. At the end, you compare the sum with the original number.<\/p>\n\n\n\n What does <\/strong>the question actually test?<\/strong><\/p>\n\n\n\n This checks digit extraction<\/strong>, math operations<\/strong>, and loop control<\/strong>.<\/p>\n\n\n\n It also quietly checks:<\/p>\n\n\n\n The core idea<\/strong><\/p>\n\n\n\n A number is Armstrong if:<\/p>\n\n\n\n Example:<\/p>\n\n\n\n How to think before coding?<\/strong><\/p>\n\n\n\n You need to:<\/p>\n\n\n\n Logic walkthrough<\/strong><\/p>\n\n\n\n Common mistakes<\/strong><\/p>\n\n\n\n Interview twist you might hear<\/strong><\/p>\n\n\n\n A strong answer:<\/p>\n\n\n\n Time complexity is O(d), where d is the number of digits.<\/em><\/p>\n\n\n\n Here is the code for checking Armstrong numbers:<\/p>\n\n\n\n Now that you are familiar with some of the most commonly asked Java interview programs, here are some tips to help you get through the interviews with ease:<\/p>\n\n\n\n\n
0, 1<\/code><\/p>\n\n\n\n0, 1, 1, 2, 3, 5, 8, ...<\/code><\/p>\n\n\n\n\n
\n
\n
next = previous + current<\/code>.<\/li>\n\n\n\n\n
\n
\nclass FibonacciSeries{ \npublic static void main(String args[]) \n{ \n int a1=0,b2=1,c3,i,count=10; \n System.out.println(a1+\" \"+b2); \/\/simply printing 0 and 1 <\/strong> \n \n for(i=2;i<count;++i) \/\/the loop starts from 2 because 0 and 1 have already been printed <\/strong> \n { \n c3=a1+b2; \n System.out.println(\" \"+c3); \n a1=b2; \n b2=c3; \n } \n \n}} \n<\/code><\/pre>\n\n\n\n2. Check if a Number is Prime<\/strong><\/h3>\n\n\n\n
\n
\n
\n
If a number has a factor greater than \u221an, it must also have one smaller than \u221an<\/strong>. So checking till \u221an is enough.<\/p>\n\n\n\n\n
\n
n-1<\/code> (wastes time).<\/li>\n\n\n\n\n
public class PrimeCheck{ \n public static void main(String args[]){ \n int a,b=0,flag=0; \n int num=3; \/\/this variable represents the number to be checked <\/strong> \n b=num\/2; \n if(num==0||num==1){ \n System.out.println(num+\" is not prime number\"); \n }else{ \n for(a=2;i<=b;i++){ \n if(num%a==0){ \n System.out.println(a+\" is not prime number\"); \n flag=1; \n break; \n } \n } \n if(flag==0) { System.out.println(a+\" is prime number\"); } \n }\n} \n} \n<\/code><\/pre>\n\n\n\n3. String Palindrome<\/strong><\/h3>\n\n\n\n
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original == reversed<\/code><\/p>\n\n\n\n\n
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left = 0<\/code>, right = length - 1<\/code><\/li>\n\n\n\n\n
==<\/code> instead of .equals()<\/code>.<\/li>\n\n\n\n\n
class PalindromePractice{ \n public static void main(String args[]){ \n int p,sum=0,tempo; \n int num=454; \/\/This variable represents the number to be checked for palindrome<\/strong> \n \n tempo=num; \n while(num>0){ \n p=num%10; \/\/getting the remainder <\/strong>\n sum=(sum*10)+p; \n num=num\/10; \n } \n if(tempo==sum) \n System.out.println(\"palindrome number \"); \n else \n System.out.println(\"not palindrome\"); \n} \n} <\/code><\/pre>\n\n\n\n
Many so-called \u201cbasic\u201d Java interview programs are actually used by interviewers to judge senior-level thinking. For example, a Fibonacci question can quickly turn into a discussion on time and space complexity, while a palindrome check reveals how well you understand memory usage and string handling. These problems aren\u2019t about memorization: they\u2019re designed to see how clearly you think under pressure and how well you explain your logic.<\/div>\n\n\n\n4. Bubble Sort <\/strong><\/h3>\n\n\n\n
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public class BubbleSort { \n static void bubbleSort(int[] arr) { \n int num = arr.length; \n int tempo = 0; \n for(int i=0; i < num; i++){ \n for(int j=1; j < (num-i); j++){ \n if(arr[j-1] > arr[j]){ \n \/\/swapping elements <\/strong> \n tempo = arr[j-1]; \n arr[j-1] = arr[j]; \n arr[j] = tempo; \n } \n \n } \n } \n \n } \n public static void main(String[] args) { \n int arr[] ={3,60,35,2,45,320,5}; \n \n System.out.println(\"The array before Bubble Sort:\"); \n for(int i=0; i < arr.length; i++){ \n System.out.println(arr[i] + \" \"); \n } \n System.out.println(); \n \n bubbleSort(arr); \/\/sorting the array elements using bubble sort method we wrote earlier <\/strong>\n \n System.out.println(\"The array after Bubble Sort:\"); \n for(int i=0; i < arr.length; i++){ \n System.out.println(arr[i] + \" \"); \n } \n \n } \n} <\/code><\/pre>\n\n\n\n5. Armstrong Number<\/strong><\/h3>\n\n\n\n
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1\u00b3 + 5\u00b3 + 3\u00b3 = 153<\/p>\n\n\n\n\n
sum of (each digit ^ number of digits) == original number<\/code><\/p>\n\n\n\n153 1\u00b3 + 5\u00b3 + 3\u00b3 = 153<\/code><\/p>\n\n\n\n\n
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public class ArmstrongCheck {\n\n public static void main(String[] args) {\n\n int num = 1634, ogNum, remainder, final = 0, h = 0;\n\n ogNum = num;\n\n for (;ogNum != 0; ogNum \/= 10, ++h);\n\n ogNum = num;\n\n for (;ogNum != 0; ogNum \/= 10)\n {\n remainder = ogNum % 10;\n final += Math.pow(remainder, h);\n }\n\n if(final == num)\n System.out.println(num + \" is an Armstrong number.\");\n else\n System.out.println(num + \" is not an Armstrong number.\");\n }\n}\n<\/code><\/pre>\n\n\n\nTips for Solving Interview Questions Efficiently<\/strong><\/h2>\n\n\n\n
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