Easy Java Interview Questions

Check divisibility only up to the square root, which makes it O(sqrt n). public static boolean isPrime(int n) { if (n < 2) return false; if (n % 2 == 0) return n == 2; for (int i = 3; (long) i * i <= n; i += 2) { if (n % i == 0) return false; } return true; } The long cast avoids overflow for large n.

Take the last digit with % 10 and build the result, checking for overflow. public static int reverse(int x) { long rev = 0; while (x != 0) { rev = rev * 10 + x % 10; x /= 10; } if (rev > Integer.MAX_VALUE || rev < Integer.MIN_VALUE) return 0; return (int) rev; } Negative numbers work because % keeps the sign in Java.

A parent reference calls the child's overridden method at runtime. abstract class Shape { abstract double area(); } class Circle extends Shape { double r = 2; double area() { return Math.PI * r * r; } } class Square extends Shape { double s = 3; double area() { return s * s; } } Shape sh = new Circle(); System.out.println(sh.area()); // Circle area

Use two pointers from both ends. public static boolean isPalindrome(String s) { int i = 0, j = s.length() - 1; while (i < j) { char a = s.charAt(i), b = s.charAt(j); if (!Character.isLetterOrDigit(a)) { i++; continue; } if (!Character.isLetterOrDigit(b)) { j--; continue; } if (Character.toLowerCase(a) != Character.toLowerCase(b)) return false; i++; j--; } return true; } Time O(n), space O(1).

A map with merge() is simple; streams also work. Map<Character, Integer> freq = new LinkedHashMap<>(); for (char c : s.toCharArray()) { freq.merge(c, 1, Integer::sum); } // Streams version Map<Character, Long> f = s.chars() .mapToObj(c -> (char) c) .collect(Collectors.groupingBy(c -> c, LinkedHashMap::new, Collectors.counting())); LinkedHashMap keeps first-seen order.

Count characters with an array for lowercase letters. public static boolean isAnagram(String a, String b) { if (a.length() != b.length()) return false; int[] count = new int[26]; for (int i = 0; i < a.length(); i++) { count[a.charAt(i) - 'a']++; count[b.charAt(i) - 'a']--; } for (int c : count) if (c != 0) return false; return true; } O(n) time; use a HashMap for Unicode input.

Chain comparators. employees.sort( Comparator.comparing(Employee::department) .thenComparing(Employee::salary, Comparator.reverseOrder())); // Streams version returning a new list List<Employee> sorted = employees.stream() .sorted(Comparator.comparing(Employee::department) .thenComparing(Employee::salary, Comparator.reverseOrder())) .toList(); List.sort is stable, so equal keys keep their order.

Store each number's index and look up its complement. public int[] twoSum(int[] nums, int target) { Map<Integer, Integer> seen = new HashMap<>(); for (int i = 0; i < nums.length; i++) { Integer j = seen.get(target - nums[i]); if (j != null) return new int[]{j, i}; seen.put(nums[i], i); } return new int[0]; } One pass, O(n) time and O(n) space.

A LinkedHashSet keeps first-seen order. List<String> unique = new ArrayList<>(new LinkedHashSet<>(names)); // or with streams List<String> unique2 = names.stream() .distinct() .toList(); Both rely on equals() and hashCode(), and both run in O(n) on average.

Sort distinct salaries in descending order and skip one. Optional<Integer> second = employees.stream() .map(Employee::salary) .distinct() .sorted(Comparator.reverseOrder()) .skip(1) .findFirst(); This returns empty if fewer than two distinct salaries exist. For huge data a single pass tracking top two is O(n).

partitioningBy splits into a map with true and false keys. Map<Boolean, List<Integer>> parts = nums.stream() .collect(Collectors.partitioningBy(n -> n % 2 == 0)); List<Integer> evens = parts.get(true); List<Integer> odds = parts.get(false); Both keys are always present, even if one list is empty, unlike groupingBy.

Collectors.joining takes a delimiter, prefix and suffix. String result = names.stream() .filter(n -> !n.isBlank()) .map(String::trim) .collect(Collectors.joining(", ", "[", "]")); // [Asha, Ravi, Neha] For a simple delimiter only, String.join(", ", names) is enough.

Use a PreparedStatement and try-with-resources. public Optional<User> findByEmail(String email) throws SQLException { String sql = "SELECT id, name FROM users WHERE email = ?"; try (Connection c = dataSource.getConnection(); PreparedStatement ps = c.prepareStatement(sql)) { ps.setString(1, email); try (ResultSet rs = ps.executeQuery()) { if (rs.next()) return Optional.of(new User(rs.getLong(1), rs.getString(2))); } } return Optional.empty(); }

Use constructor injection and validation. @RestController @RequestMapping("/api/products") class ProductController { private final ProductService service; ProductController(ProductService service) { this.service = service; } @PostMapping ResponseEntity<Product> create(@Valid @RequestBody ProductRequest req) { Product p = service.create(req); return ResponseEntity.created(URI.create("/api/products/" + p.id())).body(p); } @GetMapping("/{id}") Product get(@PathVariable Long id) { return service.get(id); } }