Java Coding Interview Questions for Freshers (2026): 25 Programs with Logic, Code, Output and Complexity

Which Java programs are asked in fresher interviews? The most common are reverse a string, palindrome, Fibonacci, prime number, factorial, Armstrong number, anagram check, count vowels, find the largest and second largest in an array, remove duplicates, swap two numbers without a third variable, word frequency with HashMap and star or number patterns. Interviewers care less about the answer and more about whether you can explain your logic, handle edge cases and state the time complexity.

This guide from VibrantMinds Technologies in Pune gives 25 Java coding interview questions for freshers in a fixed format: problem, logic, code, sample output and complexity. It is written for anyone searching Java coding interview questions for freshers, Java programs for interview, Java coding questions with solutions, Java string programs for interview, Java array programs for interview, Java pattern programs for freshers, Java coding round for freshers, Java logical programs for interview or a free IT job WhatsApp group for Java freshers in Pune.

A 5-STEP METHOD TO ANSWER ANY CODING QUESTION

CATEGORY A: NUMBER PROGRAMS (1 TO 7)

Program 1: Reverse a number

Logic: repeatedly take the last digit (n % 10), build the reverse, then drop the digit (n / 10).

java

int n = 1234, rev = 0;
while (n > 0) {
    rev = rev * 10 + n % 10;
    n /= 10;
}
System.out.println(rev);   // 4321

int n = 1234, rev = 0;
while (n > 0) {
    rev = rev * 10 + n % 10;
    n /= 10;
}
System.out.println(rev);   // 4321

int n = 1234, rev = 0;
while (n > 0) {
    rev = rev * 10 + n % 10;
    n /= 10;
}
System.out.println(rev);   // 4321

Complexity: O(d) time where d is the number of digits, O(1) space.

Program 2: Palindrome number

Logic: reverse the number and compare with the original.

java

static boolean isPalindrome(int num) {
    int original = num, rev = 0;
    while (num > 0) {
        rev = rev * 10 + num % 10;
        num /= 10;
    }
    return original == rev;
}
// isPalindrome(121) -> true, isPalindrome(123) -> false

static boolean isPalindrome(int num) {
    int original = num, rev = 0;
    while (num > 0) {
        rev = rev * 10 + num % 10;
        num /= 10;
    }
    return original == rev;
}
// isPalindrome(121) -> true, isPalindrome(123) -> false

static boolean isPalindrome(int num) {
    int original = num, rev = 0;
    while (num > 0) {
        rev = rev * 10 + num % 10;
        num /= 10;
    }
    return original == rev;
}
// isPalindrome(121) -> true, isPalindrome(123) -> false

Complexity: O(d), O(1).

Program 3: Factorial

Logic: multiply 1 to n. Use long since factorials grow quickly.

java

static long factorial(int n) {
    long result = 1;
    for (int i = 2; i <= n; i++) result *= i;
    return result;
}
// factorial(5) -> 120

static long factorial(int n) {
    long result = 1;
    for (int i = 2; i <= n; i++) result *= i;
    return result;
}
// factorial(5) -> 120

static long factorial(int n) {
    long result = 1;
    for (int i = 2; i <= n; i++) result *= i;
    return result;
}
// factorial(5) -> 120

Complexity: O(n), O(1). Mention: very large n overflows long, so BigInteger is needed.

Program 4: Fibonacci series

Logic: each number is the sum of the previous two.

java

int a = 0, b = 1;
for (int i = 0; i < 10; i++) {
    System.out.print(a + " ");
    int next = a + b;
    a = b;
    b = next;
}
// 0 1 1 2 3 5 8 13 21 34

int a = 0, b = 1;
for (int i = 0; i < 10; i++) {
    System.out.print(a + " ");
    int next = a + b;
    a = b;
    b = next;
}
// 0 1 1 2 3 5 8 13 21 34

int a = 0, b = 1;
for (int i = 0; i < 10; i++) {
    System.out.print(a + " ");
    int next = a + b;
    a = b;
    b = next;
}
// 0 1 1 2 3 5 8 13 21 34

Complexity: O(n), O(1). Follow-up: the simple recursive version is O(2^n), so mention memoization.

Program 5: Prime number check

Logic: a number is prime if no number from 2 to its square root divides it.

java

static boolean isPrime(int n) {
    if (n < 2) return false;
    for (int i = 2; (long) i * i <= n; i++) {
        if (n % i == 0) return false;
    }
    return true;
}

static boolean isPrime(int n) {
    if (n < 2) return false;
    for (int i = 2; (long) i * i <= n; i++) {
        if (n % i == 0) return false;
    }
    return true;
}

static boolean isPrime(int n) {
    if (n < 2) return false;
    for (int i = 2; (long) i * i <= n; i++) {
        if (n % i == 0) return false;
    }
    return true;
}

Complexity: O(√n), O(1).

Program 6: Armstrong number

Logic: the sum of each digit raised to the power of the digit count equals the number (153 = 1³ + 5³ + 3³).

java

static boolean isArmstrong(int num) {
    int original = num, digits = String.valueOf(num).length(), sum = 0;
    while (num > 0) {
        sum += Math.pow(num % 10, digits);
        num /= 10;
    }
    return sum == original;
}

static boolean isArmstrong(int num) {
    int original = num, digits = String.valueOf(num).length(), sum = 0;
    while (num > 0) {
        sum += Math.pow(num % 10, digits);
        num /= 10;
    }
    return sum == original;
}

static boolean isArmstrong(int num) {
    int original = num, digits = String.valueOf(num).length(), sum = 0;
    while (num > 0) {
        sum += Math.pow(num % 10, digits);
        num /= 10;
    }
    return sum == original;
}

Complexity: O(d), O(1).

Program 7: Swap two numbers without a third variable

Logic: use arithmetic.

java

int a = 10, b = 20;
a = a + b;   // 30
b = a - b;   // 10
a = a - b;   // 20

int a = 10, b = 20;
a = a + b;   // 30
b = a - b;   // 10
a = a - b;   // 20

int a = 10, b = 20;
a = a + b;   // 30
b = a - b;   // 10
a = a - b;   // 20

Mention: readability matters, and a temp variable is perfectly fine in real code. Large values can overflow with addition.

CATEGORY B: STRING PROGRAMS (8 TO 13)

Program 8: Reverse a string (without built-in reverse)

Logic: loop from the end and append each character

java

String s = "VibrantMinds";
String rev = "";
for (int i = s.length() - 1; i >= 0; i--) {
    rev += s.charAt(i);
}
System.out.println(rev);   // sdniMtnarbiV

String s = "VibrantMinds";
String rev = "";
for (int i = s.length() - 1; i >= 0; i--) {
    rev += s.charAt(i);
}
System.out.println(rev);   // sdniMtnarbiV

String s = "VibrantMinds";
String rev = "";
for (int i = s.length() - 1; i >= 0; i--) {
    rev += s.charAt(i);
}
System.out.println(rev);   // sdniMtnarbiV

Better: new StringBuilder(s).reverse().toString(). Say why: += in a loop creates many String objects, as Strings are immutable. Complexity is O(n) with StringBuilder.

Program 9: Palindrome string

Logic: two pointers moving inward.

java

static boolean isPalindrome(String s) {
    int i = 0, j = s.length() - 1;
    while (i < j) {
        if (s.charAt(i++) != s.charAt(j--)) return false;
    }
    return true;
}

static boolean isPalindrome(String s) {
    int i = 0, j = s.length() - 1;
    while (i < j) {
        if (s.charAt(i++) != s.charAt(j--)) return false;
    }
    return true;
}

static boolean isPalindrome(String s) {
    int i = 0, j = s.length() - 1;
    while (i < j) {
        if (s.charAt(i++) != s.charAt(j--)) return false;
    }
    return true;
}

Complexity: O(n), O(1). Edge case: ask whether case and spaces matter.

Program 10: Count vowels and consonants

java

String s = "Interview";
int vowels = 0, consonants = 0;
for (char c : s.toLowerCase().toCharArray()) {
    if (c >= 'a' && c <= 'z') {
        if ("aeiou".indexOf(c) != -1) vowels++;
        else consonants++;
    }
}
// vowels = 4, consonants = 5

String s = "Interview";
int vowels = 0, consonants = 0;
for (char c : s.toLowerCase().toCharArray()) {
    if (c >= 'a' && c <= 'z') {
        if ("aeiou".indexOf(c) != -1) vowels++;
        else consonants++;
    }
}
// vowels = 4, consonants = 5

String s = "Interview";
int vowels = 0, consonants = 0;
for (char c : s.toLowerCase().toCharArray()) {
    if (c >= 'a' && c <= 'z') {
        if ("aeiou".indexOf(c) != -1) vowels++;
        else consonants++;
    }
}
// vowels = 4, consonants = 5

Complexity: O(n), O(1).

Program 11: Check if two strings are anagrams

Logic: two strings are anagrams if sorted characters match.

java

static boolean isAnagram(String a, String b) {
    if (a.length() != b.length()) return false;
    char[] x = a.toLowerCase().toCharArray();
    char[] y = b.toLowerCase().toCharArray();
    Arrays.sort(x);
    Arrays.sort(y);
    return Arrays.equals(x, y);
}
// isAnagram("listen", "silent") -> true

static boolean isAnagram(String a, String b) {
    if (a.length() != b.length()) return false;
    char[] x = a.toLowerCase().toCharArray();
    char[] y = b.toLowerCase().toCharArray();
    Arrays.sort(x);
    Arrays.sort(y);
    return Arrays.equals(x, y);
}
// isAnagram("listen", "silent") -> true

static boolean isAnagram(String a, String b) {
    if (a.length() != b.length()) return false;
    char[] x = a.toLowerCase().toCharArray();
    char[] y = b.toLowerCase().toCharArray();
    Arrays.sort(x);
    Arrays.sort(y);
    return Arrays.equals(x, y);
}
// isAnagram("listen", "silent") -> true

Complexity: O(n log n). Follow-up: a frequency array of size 26 gives O(n).

Program 12: First non-repeating character

Logic: count frequencies in a LinkedHashMap, then return the first with count 1.

java

static char firstUnique(String s) {
    Map<Character, Integer> map = new LinkedHashMap<>();
    for (char c : s.toCharArray()) map.put(c, map.getOrDefault(c, 0) + 1);
    for (Map.Entry<Character, Integer> e : map.entrySet()) {
        if (e.getValue() == 1) return e.getKey();
    }
    return '\0';   // none found
}
// firstUnique("swiss") -> 'w'

static char firstUnique(String s) {
    Map<Character, Integer> map = new LinkedHashMap<>();
    for (char c : s.toCharArray()) map.put(c, map.getOrDefault(c, 0) + 1);
    for (Map.Entry<Character, Integer> e : map.entrySet()) {
        if (e.getValue() == 1) return e.getKey();
    }
    return '\0';   // none found
}
// firstUnique("swiss") -> 'w'

static char firstUnique(String s) {
    Map<Character, Integer> map = new LinkedHashMap<>();
    for (char c : s.toCharArray()) map.put(c, map.getOrDefault(c, 0) + 1);
    for (Map.Entry<Character, Integer> e : map.entrySet()) {
        if (e.getValue() == 1) return e.getKey();
    }
    return '\0';   // none found
}
// firstUnique("swiss") -> 'w'

Complexity: O(n), O(k) for k unique characters.

Program 13: Remove duplicate characters from a string

java

static String removeDuplicates(String s) {
    Set<Character> seen = new LinkedHashSet<>();
    for (char c : s.toCharArray()) seen.add(c);
    StringBuilder sb = new StringBuilder();
    for (char c : seen) sb.append(c);
    return sb.toString();
}
// removeDuplicates("programming") -> "progamin"

static String removeDuplicates(String s) {
    Set<Character> seen = new LinkedHashSet<>();
    for (char c : s.toCharArray()) seen.add(c);
    StringBuilder sb = new StringBuilder();
    for (char c : seen) sb.append(c);
    return sb.toString();
}
// removeDuplicates("programming") -> "progamin"

static String removeDuplicates(String s) {
    Set<Character> seen = new LinkedHashSet<>();
    for (char c : s.toCharArray()) seen.add(c);
    StringBuilder sb = new StringBuilder();
    for (char c : seen) sb.append(c);
    return sb.toString();
}
// removeDuplicates("programming") -> "progamin"

Complexity: O(n), O(k).

CATEGORY C: ARRAY PROGRAMS (14 TO 19)

Program 14: Find the largest element

java

int[] arr = {12, 45, 7, 89, 23};
int max = arr[0];
for (int i = 1; i < arr.length; i++) {
    if (arr[i] > max) max = arr[i];
}
// 89

int[] arr = {12, 45, 7, 89, 23};
int max = arr[0];
for (int i = 1; i < arr.length; i++) {
    if (arr[i] > max) max = arr[i];
}
// 89

int[] arr = {12, 45, 7, 89, 23};
int max = arr[0];
for (int i = 1; i < arr.length; i++) {
    if (arr[i] > max) max = arr[i];
}
// 89

Complexity: O(n), O(1).

Program 15: Find the second largest element

Logic: one pass, tracking the largest and second largest without sorting.

java

int first = Integer.MIN_VALUE, second = Integer.MIN_VALUE;
for (int n : arr) {
    if (n > first) {
        second = first;
        first = n;
    } else if (n > second && n != first) {
        second = n;
    }
}
// second = 45

int first = Integer.MIN_VALUE, second = Integer.MIN_VALUE;
for (int n : arr) {
    if (n > first) {
        second = first;
        first = n;
    } else if (n > second && n != first) {
        second = n;
    }
}
// second = 45

int first = Integer.MIN_VALUE, second = Integer.MIN_VALUE;
for (int n : arr) {
    if (n > first) {
        second = first;
        first = n;
    } else if (n > second && n != first) {
        second = n;
    }
}
// second = 45

Complexity: O(n), O(1). Edge case: arrays with all equal elements have no second largest.

Program 16: Reverse an array in place

java

int i = 0, j = arr.length - 1;
while (i < j) {
    int temp = arr[i];
    arr[i++] = arr[j];
    arr[j--] = temp;
}

int i = 0, j = arr.length - 1;
while (i < j) {
    int temp = arr[i];
    arr[i++] = arr[j];
    arr[j--] = temp;
}

int i = 0, j = arr.length - 1;
while (i < j) {
    int temp = arr[i];
    arr[i++] = arr[j];
    arr[j--] = temp;
}

Complexity: O(n), O(1).

Program 17: Remove duplicates from a sorted array (in place)

java

static int removeDuplicates(int[] a) {
    if (a.length == 0) return 0;
    int k = 1;
    for (int i = 1; i < a.length; i++) {
        if (a[i] != a[i - 1]) a[k++] = a[i];
    }
    return k;   // new length
}
// {1,1,2,2,3} -> k = 3, array starts {1,2,3}

static int removeDuplicates(int[] a) {
    if (a.length == 0) return 0;
    int k = 1;
    for (int i = 1; i < a.length; i++) {
        if (a[i] != a[i - 1]) a[k++] = a[i];
    }
    return k;   // new length
}
// {1,1,2,2,3} -> k = 3, array starts {1,2,3}

static int removeDuplicates(int[] a) {
    if (a.length == 0) return 0;
    int k = 1;
    for (int i = 1; i < a.length; i++) {
        if (a[i] != a[i - 1]) a[k++] = a[i];
    }
    return k;   // new length
}
// {1,1,2,2,3} -> k = 3, array starts {1,2,3}

Complexity: O(n), O(1).

Program 18: Find the missing number from 1 to n

Logic: expected sum is n(n+1)/2, so subtract the actual sum.

java

int[] arr = {1, 2, 4, 5};
int n = 5;
int expected = n * (n + 1) / 2;
int actual = 0;
for (int x : arr) actual += x;
System.out.println(expected - actual);   // 3

int[] arr = {1, 2, 4, 5};
int n = 5;
int expected = n * (n + 1) / 2;
int actual = 0;
for (int x : arr) actual += x;
System.out.println(expected - actual);   // 3

int[] arr = {1, 2, 4, 5};
int n = 5;
int expected = n * (n + 1) / 2;
int actual = 0;
for (int x : arr) actual += x;
System.out.println(expected - actual);   // 3

Complexity: O(n), O(1).

Program 19: Two sum (find a pair with a given sum)

Logic: store seen numbers in a HashMap and look for the complement.

java

static int[] twoSum(int[] nums, int target) {
    Map<Integer, Integer> seen = new HashMap<>();
    for (int i = 0; i < nums.length; i++) {
        int need = target - nums[i];
        if (seen.containsKey(need)) return new int[]{seen.get(need), i};
        seen.put(nums[i], i);
    }
    return new int[]{};
}
// twoSum({2,7,11,15}, 9) -> indices {0,1}

static int[] twoSum(int[] nums, int target) {
    Map<Integer, Integer> seen = new HashMap<>();
    for (int i = 0; i < nums.length; i++) {
        int need = target - nums[i];
        if (seen.containsKey(need)) return new int[]{seen.get(need), i};
        seen.put(nums[i], i);
    }
    return new int[]{};
}
// twoSum({2,7,11,15}, 9) -> indices {0,1}

static int[] twoSum(int[] nums, int target) {
    Map<Integer, Integer> seen = new HashMap<>();
    for (int i = 0; i < nums.length; i++) {
        int need = target - nums[i];
        if (seen.containsKey(need)) return new int[]{seen.get(need), i};
        seen.put(nums[i], i);
    }
    return new int[]{};
}
// twoSum({2,7,11,15}, 9) -> indices {0,1}

Complexity: O(n), O(n). The brute-force nested loop is O(n²), so mention the improvement.

CATEGORY D: PATTERN PROGRAMS (20 TO 22)

Program 20: Right-angled star triangle

java

int rows = 5;
for (int i = 1; i <= rows; i++) {
    for (int j = 1; j <= i; j++) System.out.print("* ");
    System.out.println();
}

int rows = 5;
for (int i = 1; i <= rows; i++) {
    for (int j = 1; j <= i; j++) System.out.print("* ");
    System.out.println();
}

int rows = 5;
for (int i = 1; i <= rows; i++) {
    for (int j = 1; j <= i; j++) System.out.print("* ");
    System.out.println();
}

Program 21: Pyramid

java

int rows = 5;
for (int i = 1; i <= rows; i++) {
    for (int s = 1; s <= rows - i; s++) System.out.print(" ");
    for (int j = 1; j <= 2 * i - 1; j++) System.out.print("*");
    System.out.println();
}

int rows = 5;
for (int i = 1; i <= rows; i++) {
    for (int s = 1; s <= rows - i; s++) System.out.print(" ");
    for (int j = 1; j <= 2 * i - 1; j++) System.out.print("*");
    System.out.println();
}

int rows = 5;
for (int i = 1; i <= rows; i++) {
    for (int s = 1; s <= rows - i; s++) System.out.print(" ");
    for (int j = 1; j <= 2 * i - 1; j++) System.out.print("*");
    System.out.println();
}

Program 22: Number triangle

java

for (int i = 1; i <= 5; i++) {
    for (int j = 1; j <= i; j++) System.out.print(j + " ");
    System.out.println();
}

for (int i = 1; i <= 5; i++) {
    for (int j = 1; j <= i; j++) System.out.print(j + " ");
    System.out.println();
}

for (int i = 1; i <= 5; i++) {
    for (int j = 1; j <= i; j++) System.out.print(j + " ");
    System.out.println();
}

Pattern tip: the outer loop controls rows, the inner loops control spaces and characters. Complexity is O(n²).

CATEGORY E: COLLECTIONS-BASED PROGRAMS (23 TO 25)

Program 23: Word frequency count

java

String text = "java is fun and java is powerful";
Map<String, Integer> freq = new HashMap<>();
for (String w : text.split(" ")) {
    freq.put(w, freq.getOrDefault(w, 0) + 1);
}
System.out.println(freq);   // {java=2, is=2, fun=1, and=1, powerful=1} (order may vary)

String text = "java is fun and java is powerful";
Map<String, Integer> freq = new HashMap<>();
for (String w : text.split(" ")) {
    freq.put(w, freq.getOrDefault(w, 0) + 1);
}
System.out.println(freq);   // {java=2, is=2, fun=1, and=1, powerful=1} (order may vary)

String text = "java is fun and java is powerful";
Map<String, Integer> freq = new HashMap<>();
for (String w : text.split(" ")) {
    freq.put(w, freq.getOrDefault(w, 0) + 1);
}
System.out.println(freq);   // {java=2, is=2, fun=1, and=1, powerful=1} (order may vary)

Complexity: O(n), O(k).

Program 24: Find duplicates in a list

java

List<Integer> list = Arrays.asList(1, 2, 3, 2, 4, 1);
Set<Integer> seen = new HashSet<>();
Set<Integer> duplicates = new HashSet<>();
for (int x : list) {
    if (!seen.add(x)) duplicates.add(x);
}
System.out.println(duplicates);   // [1, 2]

List<Integer> list = Arrays.asList(1, 2, 3, 2, 4, 1);
Set<Integer> seen = new HashSet<>();
Set<Integer> duplicates = new HashSet<>();
for (int x : list) {
    if (!seen.add(x)) duplicates.add(x);
}
System.out.println(duplicates);   // [1, 2]

List<Integer> list = Arrays.asList(1, 2, 3, 2, 4, 1);
Set<Integer> seen = new HashSet<>();
Set<Integer> duplicates = new HashSet<>();
for (int x : list) {
    if (!seen.add(x)) duplicates.add(x);
}
System.out.println(duplicates);   // [1, 2]

Complexity: O(n), O(n). add() returns false if the element already exists.

Program 25: Sort a list of objects by a field

java

class Student {
    String name; int marks;
    Student(String name, int marks) { this.name = name; this.marks = marks; }
}
List<Student> list = new ArrayList<>();
list.add(new Student("Asha", 82));
list.add(new Student("Ravi", 91));
list.sort(Comparator.comparingInt((Student s) -> s.marks).reversed());

class Student {
    String name; int marks;
    Student(String name, int marks) { this.name = name; this.marks = marks; }
}
List<Student> list = new ArrayList<>();
list.add(new Student("Asha", 82));
list.add(new Student("Ravi", 91));
list.sort(Comparator.comparingInt((Student s) -> s.marks).reversed());

class Student {
    String name; int marks;
    Student(String name, int marks) { this.name = name; this.marks = marks; }
}
List<Student> list = new ArrayList<>();
list.add(new Student("Asha", 82));
list.add(new Student("Ravi", 91));
list.sort(Comparator.comparingInt((Student s) -> s.marks).reversed());

Complexity: O(n log n). Mention: Comparable for natural order and Comparator for custom order.

COMPLEXITY CHEAT SHEET

EDGE CASES CHECKLIST (TEST EVERY SOLUTION)

MISTAKES FRESHERS MAKE IN THE JAVA CODING ROUND

14-DAY JAVA CODING PRACTICE PLAN

MYTH VS REALITY

HOW VIBRANTMINDS HELPS FRESHERS CRACK JAVA CODING ROUNDS

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• Interview preparation: Coding practice, technical rounds, project rounds, HR rounds, aptitude, communication and mock interviews

• Placement support: Screening, drive coordination and interview support. Check https://candidates.vibrantminds.in/ for current openings.

• Free WhatsApp job alert groups for fresher IT jobs and drives

• Honesty first: VibrantMinds offers placement support, not a guaranteed job, and no legitimate recruiter should ask you to pay to be interviewed.

VibrantMinds works as a talent ecosystem connecting candidates, colleges and companies, with 1900+ partner colleges and 600+ hiring companies. More fresher job guidance: https://candidates.vibrantminds.in/blog/fresher-jobs-pune-2026

VIBRANTMINDS COURSES AT A GLANCE

Fees, syllabus and batch dates are on https://candidates.vibrantminds.in/. Please confirm them directly before enrolling.

THE VIBRANTMINDS PLACEMENT PIPELINE: SIX STEPS

REAL OUTCOMES: WHAT TO EXPECT AND WHAT NOT TO

Past students have shared positive feedback about the training and support, and have gone on to roles at companies such as Aress Software and Education Technologies, Saama Technologies and Three M Software. Results vary by candidate, skill level, effort and market conditions, so please confirm details directly with the VibrantMinds team. No one can honestly promise a guaranteed job.

JOIN FREE VIBRANTMINDS WHATSAPP GROUPS: JAVA AND IT JOB UPDATES FOR FRESHERS IN PUNE AND INDIA

Looking for free WhatsApp groups, free IT job WhatsApp groups, a Java WhatsApp group for freshers or a job updates WhatsApp group in India? The VibrantMinds groups share fresher IT openings, campus drives, walk-in interview alerts and interview preparation updates, free of charge. Join whichever group has space. All three carry the same type of updates.

Never pay to apply. Legitimate recruiters do not charge candidates for tests, interviews or offer letters. Do not share sensitive personal or financial details in any group chat.

FREQUENTLY ASKED QUESTIONS

1. What are the most asked Java coding interview questions for freshers?
Reverse a string, palindrome, Fibonacci, prime number, factorial, Armstrong number, anagram check, second largest in an array, remove duplicates, missing number, two sum, word frequency using HashMap and star or number patterns.

2. How do I reverse a string in Java without using a built-in method?
Loop from the last index to the first and append each character to a result, or swap characters from both ends using two pointers on a char[]. For efficiency, build the result with a StringBuilder instead of +=, because Strings are immutable.

3. How do I find the second largest number in an array in Java?
Track two variables, first and second, in a single pass. Update second to the old first when a bigger number appears, and update second when a number lies between them. This is O(n) and needs no sorting.

4. How do I check if two strings are anagrams in Java?
Check that the lengths match, then either sort both character arrays and compare them (O(n log n)), or count character frequencies in an array of size 26 (O(n)).

5. Is there a free WhatsApp group for Java freshers in Pune?
Yes. VibrantMinds runs free fresher IT job WhatsApp groups that share Java and other IT openings, campus drives and walk-in alerts. Join through the links above, and always verify a job post and never pay to apply.

CONCLUSION: PRACTISE THE PATTERNS, EXPLAIN THE LOGIC

The best way to handle Java coding interview questions for freshers is to learn the core 25 patterns, type each program yourself, say your logic aloud, mention complexity and test edge cases. Follow the 14-day plan, then try small variations so you aren’t relying on memory alone.

If you’re in Pune and want structured Java training with coding practice, interview preparation and placement support, VibrantMinds Technologies is here to help.

Call to action:

• Website: https://vibrantminds.in/

• Candidate portal and courses: https://candidates.vibrantminds.in/

• Fresher jobs guide: https://candidates.vibrantminds.in/blog/fresher-jobs-pune-2026

• Join the free WhatsApp groups: the three links above

VISIT OR CONTACT VIBRANTMINDS IN PUNE

• Company: VibrantMinds Technologies Pvt. Ltd.

• Address: Viva Academy Building, Warje, Pune, Maharashtra 411052

• Phone: +91 95035 79517

• Email: support@vibrantmindstech.com

• Website: https://vibrantminds.in/

• Candidate portal and courses: https://candidates.vibrantminds.in/