COLLECTION - Top 100 important interview OUTPUT questions and answers in java, Set-3 > Q75- Q100
These very important questions have been framed to test your basic and in depth knowledge of Collection, these questions covers vast variety of questions that touches almost all concepts of Collection in java. Questions ranges from easy to hard for fresher/beginner to experienced developers . Go, have a crack on these!!
Collection interview Question 51.
Collection Output interview question 1.
import java.util.ArrayList;
import java.util.List;
public class MyClass {
public static void main(String[] args) {
//INSET HERE
l.add("2");
l.add("21");
}
}
Insert the correct one from given 4 lines -
List<String> l=new ArrayList<String>();
List<String> l=new ArrayList<Integer>();
List<Integer> l=new ArrayList<Integer>();
List<Integer> l=new ArrayList<String>();
Answer.
List<String> l=new ArrayList<String>();
Collection interview Question 52.
Collection Output interview question 2.
import java.util.ArrayList;
import java.util.List;
/** Copyright (c), AnkitMittal JavaMadeSoEasy.com */
public class MyClass {
public static void main(String[] args) {
//INSET HERE
List<String> list=new ArrayList<String>();
list.add("a");
list.add("b");
l.add(list);
}
}
Insert the correct one from given 4 lines -
List<List<String>> l=new ArrayList<List<String>>();
List<List> l=new ArrayList<List<String>>();
List<List<String>> l=new ArrayList<ArrayList<String>>();
List<List,String> l=new ArrayList<List,String>();
Answer.
List<List<String>> l=new ArrayList<List<String>>();
Collection interview Question 53.
Collection Output interview question 3.
import java.util.ArrayList;
import java.util.List;
/** Copyright (c), AnkitMittal JavaMadeSoEasy.com */
public class ArrayListTest {
public static void main(String args[]) {
List<String> arrayList = new ArrayList<String>();
arrayList.add("a");
arrayList.add("b");
arrayList.add("c");
System.out.println();
arrayList.add(1,"d");
System.out.println(arrayList);
}
}
Answer.
/*OUTPUT
[a, d, b, c]
*/
ArrayList maintains insertion order and element can be added at specific index as well.
Collection interview Question 54.
Collection Output interview question 4.
import java.util.ArrayList;
import java.util.List;
import java.util.ListIterator;
/** Copyright (c), AnkitMittal JavaMadeSoEasy.com */
public class ArrayListTest {
public static void main(String args[]) {
List<String> arrayList = new ArrayList<String>();
arrayList.add("a");
arrayList.add("b");
ListIterator<String> listIterator = arrayList.listIterator();
while (listIterator.hasNext()) {
System.out.println(listIterator.next());
listIterator.previous();
}
}
}
Answer.
/*OUTPUT
a infinite times
*/
ArrayList provides listIterator for traversing in forward and backward direction, so program will compile and run infinitely.
Collection interview Question 55.
Collection Output interview question 5.
import java.util.Iterator;
import java.util.Set;
import java.util.TreeSet;
public class ConcurrentSkipListMapTest {
public static void main(String args[]) {
Set set = new TreeSet();
set.add(1);
set.add("2");
set.add(3);
Iterator iterator = set.iterator();
while (iterator.hasNext()) {
System.out.print(iterator.next() + " ");
}
}
}
Answer.
/*OUTPUT
java.lang.ClassCastException
*/
TreeSet maintains natural order of elements, 1 is a Integer while "2" is a String, so they cannot be compared for sorted, hence runtime exception - ClassCastException is thrown.
Read :
These very important questions have been framed to test your basic and in depth knowledge of Collection, these questions covers vast variety of questions that touches almost all concepts of Collection in java. Questions ranges from easy to hard for fresher/beginner to experienced developers . Go, have a crack on these!!
Collection interview Question 51.
Collection Output interview question 1.
import java.util.ArrayList;
import java.util.List;
public class MyClass {
public static void main(String[] args) {
//INSET HERE
l.add("2");
l.add("21");
}
}
|
Insert the correct one from given 4 lines -
List<String> l=new ArrayList<String>();
List<String> l=new ArrayList<Integer>();
List<Integer> l=new ArrayList<Integer>();
List<Integer> l=new ArrayList<String>();
Answer.
List<String> l=new ArrayList<String>();
Collection interview Question 52.
Collection Output interview question 2.
import java.util.ArrayList;
import java.util.List;
/** Copyright (c), AnkitMittal JavaMadeSoEasy.com */
public class MyClass {
public static void main(String[] args) {
//INSET HERE
List<String> list=new ArrayList<String>();
list.add("a");
list.add("b");
l.add(list);
}
}
|
Insert the correct one from given 4 lines -
List<List<String>> l=new ArrayList<List<String>>();
List<List> l=new ArrayList<List<String>>();
List<List<String>> l=new ArrayList<ArrayList<String>>();
List<List,String> l=new ArrayList<List,String>();
Answer.
List<List<String>> l=new ArrayList<List<String>>();
Collection interview Question 53.
Collection Output interview question 3.
import java.util.ArrayList;
import java.util.List;
/** Copyright (c), AnkitMittal JavaMadeSoEasy.com */
public class ArrayListTest {
public static void main(String args[]) {
List<String> arrayList = new ArrayList<String>();
arrayList.add("a");
arrayList.add("b");
arrayList.add("c");
System.out.println();
arrayList.add(1,"d");
System.out.println(arrayList);
}
}
|
Answer.
/*OUTPUT
[a, d, b, c]
*/
ArrayList maintains insertion order and element can be added at specific index as well.
Collection interview Question 54.
Collection Output interview question 4.
import java.util.ArrayList;
import java.util.List;
import java.util.ListIterator;
/** Copyright (c), AnkitMittal JavaMadeSoEasy.com */
public class ArrayListTest {
public static void main(String args[]) {
List<String> arrayList = new ArrayList<String>();
arrayList.add("a");
arrayList.add("b");
ListIterator<String> listIterator = arrayList.listIterator();
while (listIterator.hasNext()) {
System.out.println(listIterator.next());
listIterator.previous();
}
}
}
|
Answer.
/*OUTPUT
a infinite times
*/
ArrayList provides listIterator for traversing in forward and backward direction, so program will compile and run infinitely.
Collection interview Question 55.
Collection Output interview question 5.
import java.util.Iterator;
import java.util.Set;
import java.util.TreeSet;
public class ConcurrentSkipListMapTest {
public static void main(String args[]) {
Set set = new TreeSet();
set.add(1);
set.add("2");
set.add(3);
Iterator iterator = set.iterator();
while (iterator.hasNext()) {
System.out.print(iterator.next() + " ");
}
}
}
|
Answer.
/*OUTPUT
java.lang.ClassCastException
*/
TreeSet maintains natural order of elements, 1 is a Integer while "2" is a String, so they cannot be compared for sorted, hence runtime exception - ClassCastException is thrown.
Read :
COLLECTION - Top 100 important interview OUTPUT questions and answers in java, Set-3 > Q75- Q100
Collection interview Question 56.
Collection Output interview question 6.
import java.util.ArrayList;
import java.util.Iterator;
import java.util.List;
/** Copyright (c), AnkitMittal JavaMadeSoEasy.com */
public class ArrayListTest {
public static void main(String args[]) {
List<String> arrayList = new ArrayList<String>();
arrayList.add("a");
arrayList.add("b");
Iterator<String> iterator = arrayList.iterator();
while (iterator.hasNext()) {
System.out.println(iterator.next());
arrayList.add("c");
}
}
}
Answer.
/*OUTPUT
a
Exception in thread "main" java.util.ConcurrentModificationException
at java.util.ArrayList$Itr.checkForComodification(Unknown Source)
at java.util.ArrayList$Itr.next(Unknown Source)
at ArrayListTest.main(ArrayListTest.java:23)
*/
ConcurrentModificationException is thrown because ArrayList cannot modified during iteration.
Collection interview Question 57.
Collection Output interview question 7.
import java.util.Iterator;
import java.util.Map;
import java.util.concurrent.ConcurrentSkipListMap;
/** Copyright (c), AnkitMittal JavaMadeSoEasy.com */
public class ConcurrentSkipListMapTest {
public static void main(String args[]) {
Map<Integer, String> concurrentSkipListMap = new ConcurrentSkipListMap<Integer, String>();
concurrentSkipListMap.put(11, "audi");
Iterator<Integer> keyIterator = concurrentSkipListMap.keySet()
.iterator();
while (keyIterator.hasNext()) {
System.out.println(keyIterator.next());
concurrentSkipListMap.put(13, "bmw");
}
}
}
Answer.
/*OUTPUT
11
*/
ConcurrentSkipListMap can be modified during iteration- i.e. iteration is fail-safe, but newly added keys won’t be a part of that iteration
Collection interview Question 58.
Collection Output interview question 8.
import java.util.List;
import java.util.concurrent.CopyOnWriteArrayList;
/** Copyright (c), AnkitMittal JavaMadeSoEasy.com */
public class CopyOnWriteArrayListTest {
public static void main(String args[]) {
List<String> copyOnWriteArrayList = new CopyOnWriteArrayList<String>();
copyOnWriteArrayList.add("ind");
copyOnWriteArrayList.add("usa");
copyOnWriteArrayList.add(null);
for (String string : copyOnWriteArrayList) {
System.out.print(string+" ");
copyOnWriteArrayList.add("newEle3");
}
}
}
Answer.
/* OUTPUT
ind usa null
*/
Collection interview Question 59.
Collection Output interview question 9.
import java.util.EnumMap;
import java.util.Map;
/** Copyright (c), AnkitMittal JavaMadeSoEasy.com */
public class EnumMapExample{
public enum Days{
Monday ;
}
public static void main(String args[]) {
Map<Days, String> daysEnumMap = new EnumMap<Days, String>(Days.class);
daysEnumMap.put(Days.Monday, "Day1");
daysEnumMap.put(Days.Tuesday, "Day2");
System.out.println("daysEnumMap.get(Days.Monday) : " +
daysEnumMap.get(Days.Monday));
System.out.println("daysEnumMap.containsKey(Days.Monday) : " +
daysEnumMap.containsKey(Days.Tuesday));
}
}
Answer. Program will fail to compile because a EnumMap is specialized Map implementation for use with enum type keys. In EnumMap all keys comes from a single enum type that is specified when the Map is created.
Read :
Collection interview Question 56.
Collection Output interview question 6.
import java.util.ArrayList;
import java.util.Iterator;
import java.util.List;
/** Copyright (c), AnkitMittal JavaMadeSoEasy.com */
public class ArrayListTest {
public static void main(String args[]) {
List<String> arrayList = new ArrayList<String>();
arrayList.add("a");
arrayList.add("b");
Iterator<String> iterator = arrayList.iterator();
while (iterator.hasNext()) {
System.out.println(iterator.next());
arrayList.add("c");
}
}
}
|
Answer.
/*OUTPUT
a
Exception in thread "main" java.util.ConcurrentModificationException
at java.util.ArrayList$Itr.checkForComodification(Unknown Source)
at java.util.ArrayList$Itr.next(Unknown Source)
at ArrayListTest.main(ArrayListTest.java:23)
*/
ConcurrentModificationException is thrown because ArrayList cannot modified during iteration.
Collection interview Question 57.
Collection Output interview question 7.
import java.util.Iterator;
import java.util.Map;
import java.util.concurrent.ConcurrentSkipListMap;
/** Copyright (c), AnkitMittal JavaMadeSoEasy.com */
public class ConcurrentSkipListMapTest {
public static void main(String args[]) {
Map<Integer, String> concurrentSkipListMap = new ConcurrentSkipListMap<Integer, String>();
concurrentSkipListMap.put(11, "audi");
Iterator<Integer> keyIterator = concurrentSkipListMap.keySet()
.iterator();
while (keyIterator.hasNext()) {
System.out.println(keyIterator.next());
concurrentSkipListMap.put(13, "bmw");
}
}
}
|
Answer.
/*OUTPUT
11
*/
ConcurrentSkipListMap can be modified during iteration- i.e. iteration is fail-safe, but newly added keys won’t be a part of that iteration
Collection interview Question 58.
Collection Output interview question 8.
import java.util.List;
import java.util.concurrent.CopyOnWriteArrayList;
/** Copyright (c), AnkitMittal JavaMadeSoEasy.com */
public class CopyOnWriteArrayListTest {
public static void main(String args[]) {
List<String> copyOnWriteArrayList = new CopyOnWriteArrayList<String>();
copyOnWriteArrayList.add("ind");
copyOnWriteArrayList.add("usa");
copyOnWriteArrayList.add(null);
for (String string : copyOnWriteArrayList) {
System.out.print(string+" ");
copyOnWriteArrayList.add("newEle3");
}
}
}
|
Answer.
/* OUTPUT
ind usa null
*/
Collection interview Question 59.
Collection Output interview question 9.
import java.util.EnumMap;
import java.util.Map;
/** Copyright (c), AnkitMittal JavaMadeSoEasy.com */
public class EnumMapExample{
public enum Days{
Monday ;
}
public static void main(String args[]) {
Map<Days, String> daysEnumMap = new EnumMap<Days, String>(Days.class);
daysEnumMap.put(Days.Monday, "Day1");
daysEnumMap.put(Days.Tuesday, "Day2");
System.out.println("daysEnumMap.get(Days.Monday) : " +
daysEnumMap.get(Days.Monday));
System.out.println("daysEnumMap.containsKey(Days.Monday) : " +
daysEnumMap.containsKey(Days.Tuesday));
}
}
|
Answer. Program will fail to compile because a EnumMap is specialized Map implementation for use with enum type keys. In EnumMap all keys comes from a single enum type that is specified when the Map is created.
Read :
COLLECTION - Top 100 important interview OUTPUT questions and answers in java, Set-3 > Q75- Q100
Collection interview Question 60.
Collection Output interview question 10.
import java.util.Iterator;
import java.util.Set;
import java.util.concurrent.CopyOnWriteArraySet;
/** Copyright (c), AnkitMittal JavaMadeSoEasy.com */
public class CopyOnWriteArraySetTest {
public static void main(String args[]) {
Set<String> copyOnWriteArraySet = new CopyOnWriteArraySet<String>();
copyOnWriteArraySet.add("a");
copyOnWriteArraySet.add("b");
Iterator<String> iterator = copyOnWriteArraySet.iterator();
while (iterator.hasNext()) {
copyOnWriteArraySet.add("c");
System.out.println(iterator.next());
}
}
}
Answer.
/*OUTPUT
a
b
*/
CopyOnWriteArraySet can be modified during iteration- i.e. iteration is fail-safe, but newly added elements won’t be a part of that iteration, and insertion order is not guaranteed.
Collection interview Question 61.
Collection Output interview question 11.
import java.util.LinkedHashMap;
import java.util.Map;
/** Copyright (c), AnkitMittal JavaMadeSoEasy.com */
public class LinkedHashMapTest {
public static void main(String args[]) {
Map<Integer, String> m = new LinkedHashMap<Integer, String>();
m.put(11, "audi");
m.put(null, null);
m.put(11, "bmw");
m.put(null, "fer");
System.out.println(m.size());
System.out.println(m);
}
}
Answer.
/*OUTPUT
2
{11=bmw, null=fer}
*/
LinkedHashMap maintains insertion order of keys, and allows one null key and many null values.
Collection interview Question 60.
Collection Output interview question 10.
import java.util.Iterator;
import java.util.Set;
import java.util.concurrent.CopyOnWriteArraySet;
/** Copyright (c), AnkitMittal JavaMadeSoEasy.com */
public class CopyOnWriteArraySetTest {
public static void main(String args[]) {
Set<String> copyOnWriteArraySet = new CopyOnWriteArraySet<String>();
copyOnWriteArraySet.add("a");
copyOnWriteArraySet.add("b");
Iterator<String> iterator = copyOnWriteArraySet.iterator();
while (iterator.hasNext()) {
copyOnWriteArraySet.add("c");
System.out.println(iterator.next());
}
}
}
|
Answer.
/*OUTPUT
a
b
*/
CopyOnWriteArraySet can be modified during iteration- i.e. iteration is fail-safe, but newly added elements won’t be a part of that iteration, and insertion order is not guaranteed.
Collection interview Question 61.
Collection Output interview question 11.
import java.util.LinkedHashMap;
import java.util.Map;
/** Copyright (c), AnkitMittal JavaMadeSoEasy.com */
public class LinkedHashMapTest {
public static void main(String args[]) {
Map<Integer, String> m = new LinkedHashMap<Integer, String>();
m.put(11, "audi");
m.put(null, null);
m.put(11, "bmw");
m.put(null, "fer");
System.out.println(m.size());
System.out.println(m);
}
}
|
Answer.
/*OUTPUT
2
{11=bmw, null=fer}
*/
LinkedHashMap maintains insertion order of keys, and allows one null key and many null values.
Must read: Differences between HashMap, Hashtable, LinkedHashMap and TreeMap in java
Collection interview Question 62.
Collection Output interview question 12.
import java.util.Iterator;
import java.util.Map;
import java.util.concurrent.ConcurrentSkipListMap;
/** Copyright (c), AnkitMittal JavaMadeSoEasy.com */
public class ConcurrentSkipListMapTest {
public static void main(String args[]){
Map<Integer,String> concurrentSkipListMap=new ConcurrentSkipListMap<Integer,String>();
concurrentSkipListMap.put(11, "audi");
concurrentSkipListMap.put(44, null);
Iterator<Integer> keyIterator = concurrentSkipListMap.keySet().iterator();
while (keyIterator.hasNext()) {
System.out.println(keyIterator.next());
}
}
}
Answer.
/*OUTPUT
java.lang.NullPointerException
*/
Concurrentskiplistmap does not any null key or null value.
Collection interview Question 62.
Collection Output interview question 12.
import java.util.Iterator;
import java.util.Map;
import java.util.concurrent.ConcurrentSkipListMap;
/** Copyright (c), AnkitMittal JavaMadeSoEasy.com */
public class ConcurrentSkipListMapTest {
public static void main(String args[]){
Map<Integer,String> concurrentSkipListMap=new ConcurrentSkipListMap<Integer,String>();
concurrentSkipListMap.put(11, "audi");
concurrentSkipListMap.put(44, null);
Iterator<Integer> keyIterator = concurrentSkipListMap.keySet().iterator();
while (keyIterator.hasNext()) {
System.out.println(keyIterator.next());
}
}
}
|
Answer.
/*OUTPUT
java.lang.NullPointerException
*/
Concurrentskiplistmap does not any null key or null value.
Must read: Differences and Similarities between TreeMap and ConcurrentSkipListMap with program in java
Collection interview Question 63.
Collection Output interview question 13. How many buckets will be there and what will be size of HashMap?
package p1;
import java.util.HashMap;
class Employee {
private String name;
public Employee(String name) { // constructor
this.name = name;
}
//no hashCode() method
//no equals() method
@Override
public String toString() {
return "Employee[ name=" + name + "] ";
}
}
/** Copyright (c), AnkitMittal JavaMadeSoEasy.com */
public class Program1 {
public static void main(String...a){
HashMap<Employee, String> hm=new HashMap<Employee, String>();
hm.put(new Employee("a"), "emp1");
hm.put(new Employee("b"), "emp2");
hm.put(new Employee("a"), "emp1 OVERRIDDEN");
System.out.println("HashMap's data> "+hm);
System.out.println("HashMap's size> "+hm.size());
System.out.println(hm.get(new Employee("a")));
}
}
Answer.
/*OUTPUT
HashMap's data> {Employee[ name=a] =emp1 OVERRIDDEN, Employee[ name=a] =emp1, Employee[ name=b] =emp2}
HashMap's size> 3
null
*/
Buckets= As hashCode() method is not there, hashcode generated for 3 objects will be different and we will end up using 3 buckets.
Size= As equals() method is not their, size will be 3.
get()=we won’t be able to get object.
Collection interview Question 64.
Collection Output interview question 14. How many buckets will be there and what will be size of HashMap?
package p2;
import java.util.HashMap;
class Employee {
private String name;
public Employee(String name) { // constructor
this.name = name;
}
@Override
public int hashCode(){
return (this.name==null ? 0: this.name.hashCode() );
}
@Override
public boolean equals(Object obj){
Employee emp=(Employee)obj;
return (emp.name==this.name || emp.name.equals(this.name));
}
@Override
public String toString() {
return "Employee[ name=" + name + "] ";
}
}
/** Copyright (c), AnkitMittal JavaMadeSoEasy.com */
public class Program2 {
public static void main(String...a){
HashMap<Employee, String> hm=new HashMap<Employee, String>();
hm.put(new Employee("a"), "emp1");
hm.put(new Employee("b"), "emp2");
hm.put(new Employee("a"), "emp1 OVERRIDDEN");
System.out.println("HashMap's data> "+hm);
System.out.println("HashMap's size> "+hm.size());
System.out.println(hm.get(new Employee("a")));
}
}
Answer.
/*OUTPUT
HashMap's data> {Employee[ name=b] =emp2, Employee[ name=a] =emp1 OVERRIDDEN}
HashMap's size> 2
emp1 OVERRIDDEN
*/
Buckets= As hashCode() method is overridden perfectly, 2 bucket locations will be used.
Size= As equals() method is their, size will be 2,
value corresponding to Employee with id=1 and name=’sam’ was employee1 data
& was overridden by value employee1 data OVERRIDDEN
get()=we will be able to get object.
Collection interview Question 65.
Collection Output interview question 15. How many buckets will be there and what will be size of HashMap?
package p3;
import java.util.HashMap;
class Employee {
private String name;
public Employee(String name) { // constructor
this.name = name;
}
@Override
public int hashCode(){
return 1;
}
@Override
public boolean equals(Object obj){
return true;
}
@Override
public String toString() {
return "Employee[ name=" + name + "] ";
}
}
/** Copyright (c), AnkitMittal JavaMadeSoEasy.com */
public class Program3 {
public static void main(String...a){
HashMap<Employee, String> hm=new HashMap<Employee, String>();
hm.put(new Employee("a"), "emp1");
hm.put(new Employee("b"), "emp2");
hm.put(new Employee("a"), "emp1 OVERRIDDEN");
System.out.println("HashMap's data> "+hm);
System.out.println("HashMap's size> "+hm.size());
System.out.println(hm.get(new Employee("a")));
}
}
Answer.
/*OUTPUT
HashMap's data> {Employee[ name=a] =emp1 OVERRIDDEN}
HashMap's size> 1
emp1 OVERRIDDEN
*/
Buckets= As hashCode() method returns 1, only 1 bucket location will be used.
Size= As equals() method always returns true, size will be 1, all three employees will be stored on same bucket location in one Entry (new Entry will keep on overriding previous Entry). We will always get last stored key-value pair only.
get()=we will be able to get object.
Collection interview Question 66.
Collection Output interview question 16.
import java.util.HashMap;
import java.util.Map;
/** Copyright (c), AnkitMittal JavaMadeSoEasy.com */
public class HashMapTest {
public static void main(String args[]) {
Map<Integer, String> hashMap = new HashMap<Integer, String>();
hashMap.put(11, "a");
hashMap.put(null, "c");
hashMap.put(null, null);
System.out.println(hashMap.size());
System.out.println(hashMap);
}
}
Answer.
/*OUTPUT
2
{null=null, 11=a}
*/
HashMap does not maintains insertion order of keys, and allows one null key and many null values.
Must read : HashMap methods in java
Collection interview Question 67.
Collection Output interview question 17.
import java.util.Iterator;
import java.util.Vector;
/** Copyright (c), AnkitMittal JavaMadeSoEasy.com */
public class VectorTest {
public static void main(String args[]) {
Vector<String> vector = new Vector<String>();
vector.add("1");
vector.add("2");
Iterator<String> iterator = vector.iterator();
while (iterator.hasNext()) {
vector.add("3");
System.out.println(iterator.next());
}
}
}
Answer. ConcurrentModificationException will be thrown, because Iterator returned by Vector is fail-fast.
Collection interview Question 68.
Collection Output interview question 18.
import java.util.Collections;
import java.util.HashMap;
import java.util.Map;
/** Copyright (c), AnkitMittal JavaMadeSoEasy.com */
public class HashMapTest {
public static void main(String args[]){
Map<Integer,String> hashMap=new HashMap<Integer,String>();
hashMap.put(11, "a");
Collections.unmodifiableMap(hashMap);
hashMap.put(12, "b");
System.out.println(hashMap);
}
}
Answer.
/*OUTPUT
{11=a, 12=b}
*/
Although map has been made unmodifiable but we haven’t stored that reference anywhere, so we can continue to modify map. Read making map unmodifiable using Collections.unmodifiableMap
Collection interview Question 69.
Collection Output interview question 19.
import java.util.Hashtable;
import java.util.Map;
/** Copyright (c), AnkitMittal JavaMadeSoEasy.com */
public class HashTableTest {
public static void main(String args[]){
Map<Integer, String> hashtable = new Hashtable<Integer, String>();
hashtable.put(11, "a");
hashtable.put(null, "c");
hashtable.put(null, null);
System.out.println(hashtable.size());
System.out.println(hashtable);
}
}
Answer. NullPointerException will be thrown, because hashtable does not allow to store null key or values.
Read :
Collection interview Question 63.
Collection Output interview question 13. How many buckets will be there and what will be size of HashMap?
package p1;
import java.util.HashMap;
class Employee {
private String name;
public Employee(String name) { // constructor
this.name = name;
}
//no hashCode() method
//no equals() method
@Override
public String toString() {
return "Employee[ name=" + name + "] ";
}
}
/** Copyright (c), AnkitMittal JavaMadeSoEasy.com */
public class Program1 {
public static void main(String...a){
HashMap<Employee, String> hm=new HashMap<Employee, String>();
hm.put(new Employee("a"), "emp1");
hm.put(new Employee("b"), "emp2");
hm.put(new Employee("a"), "emp1 OVERRIDDEN");
System.out.println("HashMap's data> "+hm);
System.out.println("HashMap's size> "+hm.size());
System.out.println(hm.get(new Employee("a")));
}
}
|
Answer.
/*OUTPUT
HashMap's data> {Employee[ name=a] =emp1 OVERRIDDEN, Employee[ name=a] =emp1, Employee[ name=b] =emp2}
HashMap's size> 3
null
*/
Buckets= As hashCode() method is not there, hashcode generated for 3 objects will be different and we will end up using 3 buckets.
Size= As equals() method is not their, size will be 3.
get()=we won’t be able to get object.
Collection interview Question 64.
Collection Output interview question 14. How many buckets will be there and what will be size of HashMap?
package p2;
import java.util.HashMap;
class Employee {
private String name;
public Employee(String name) { // constructor
this.name = name;
}
@Override
public int hashCode(){
return (this.name==null ? 0: this.name.hashCode() );
}
@Override
public boolean equals(Object obj){
Employee emp=(Employee)obj;
return (emp.name==this.name || emp.name.equals(this.name));
}
@Override
public String toString() {
return "Employee[ name=" + name + "] ";
}
}
/** Copyright (c), AnkitMittal JavaMadeSoEasy.com */
public class Program2 {
public static void main(String...a){
HashMap<Employee, String> hm=new HashMap<Employee, String>();
hm.put(new Employee("a"), "emp1");
hm.put(new Employee("b"), "emp2");
hm.put(new Employee("a"), "emp1 OVERRIDDEN");
System.out.println("HashMap's data> "+hm);
System.out.println("HashMap's size> "+hm.size());
System.out.println(hm.get(new Employee("a")));
}
}
|
Answer.
/*OUTPUT
HashMap's data> {Employee[ name=b] =emp2, Employee[ name=a] =emp1 OVERRIDDEN}
HashMap's size> 2
emp1 OVERRIDDEN
*/
Buckets= As hashCode() method is overridden perfectly, 2 bucket locations will be used.
Size= As equals() method is their, size will be 2,
value corresponding to Employee with id=1 and name=’sam’ was employee1 data
& was overridden by value employee1 data OVERRIDDEN
get()=we will be able to get object.
Collection interview Question 65.
Collection Output interview question 15. How many buckets will be there and what will be size of HashMap?
package p3;
import java.util.HashMap;
class Employee {
private String name;
public Employee(String name) { // constructor
this.name = name;
}
@Override
public int hashCode(){
return 1;
}
@Override
public boolean equals(Object obj){
return true;
}
@Override
public String toString() {
return "Employee[ name=" + name + "] ";
}
}
/** Copyright (c), AnkitMittal JavaMadeSoEasy.com */
public class Program3 {
public static void main(String...a){
HashMap<Employee, String> hm=new HashMap<Employee, String>();
hm.put(new Employee("a"), "emp1");
hm.put(new Employee("b"), "emp2");
hm.put(new Employee("a"), "emp1 OVERRIDDEN");
System.out.println("HashMap's data> "+hm);
System.out.println("HashMap's size> "+hm.size());
System.out.println(hm.get(new Employee("a")));
}
}
|
Answer.
/*OUTPUT
HashMap's data> {Employee[ name=a] =emp1 OVERRIDDEN}
HashMap's size> 1
emp1 OVERRIDDEN
*/
Buckets= As hashCode() method returns 1, only 1 bucket location will be used.
Size= As equals() method always returns true, size will be 1, all three employees will be stored on same bucket location in one Entry (new Entry will keep on overriding previous Entry). We will always get last stored key-value pair only.
get()=we will be able to get object.
Collection interview Question 66.
Collection Output interview question 16.
import java.util.HashMap;
import java.util.Map;
/** Copyright (c), AnkitMittal JavaMadeSoEasy.com */
public class HashMapTest {
public static void main(String args[]) {
Map<Integer, String> hashMap = new HashMap<Integer, String>();
hashMap.put(11, "a");
hashMap.put(null, "c");
hashMap.put(null, null);
System.out.println(hashMap.size());
System.out.println(hashMap);
}
}
|
Answer.
/*OUTPUT
2
{null=null, 11=a}
*/
HashMap does not maintains insertion order of keys, and allows one null key and many null values.
Must read : HashMap methods in java
Collection interview Question 67.
Collection Output interview question 17.
import java.util.Iterator;
import java.util.Vector;
/** Copyright (c), AnkitMittal JavaMadeSoEasy.com */
public class VectorTest {
public static void main(String args[]) {
Vector<String> vector = new Vector<String>();
vector.add("1");
vector.add("2");
Iterator<String> iterator = vector.iterator();
while (iterator.hasNext()) {
vector.add("3");
System.out.println(iterator.next());
}
}
}
|
Answer. ConcurrentModificationException will be thrown, because Iterator returned by Vector is fail-fast.
Collection interview Question 68.
Collection Output interview question 18.
import java.util.Collections;
import java.util.HashMap;
import java.util.Map;
/** Copyright (c), AnkitMittal JavaMadeSoEasy.com */
public class HashMapTest {
public static void main(String args[]){
Map<Integer,String> hashMap=new HashMap<Integer,String>();
hashMap.put(11, "a");
Collections.unmodifiableMap(hashMap);
hashMap.put(12, "b");
System.out.println(hashMap);
}
}
|
Answer.
/*OUTPUT
{11=a, 12=b}
*/
Although map has been made unmodifiable but we haven’t stored that reference anywhere, so we can continue to modify map. Read making map unmodifiable using Collections.unmodifiableMap
Collection interview Question 69.
Collection Output interview question 19.
import java.util.Hashtable;
import java.util.Map;
/** Copyright (c), AnkitMittal JavaMadeSoEasy.com */
public class HashTableTest {
public static void main(String args[]){
Map<Integer, String> hashtable = new Hashtable<Integer, String>();
hashtable.put(11, "a");
hashtable.put(null, "c");
hashtable.put(null, null);
System.out.println(hashtable.size());
System.out.println(hashtable);
}
}
|
Answer. NullPointerException will be thrown, because hashtable does not allow to store null key or values.
Read :
COLLECTION - Top 100 important interview OUTPUT questions and answers in java, Set-3 > Q75- Q100
Collection interview Question 70.
Collection Output interview question 20. How many buckets will be there and what will be size of HashMap?
package p4;
import java.util.HashMap;
class Employee {
private String name;
public Employee(String name) { // constructor
this.name = name;
}
@Override
public int hashCode(){
return 1;
}
//no equals() method
@Override
public String toString() {
return "Employee[ name=" + name + "] ";
}
}
/** Copyright (c), AnkitMittal JavaMadeSoEasy.com */
public class Program4 {
public static void main(String...a){
HashMap<Employee, String> hm=new HashMap<Employee, String>();
hm.put(new Employee("a"), "emp1");
hm.put(new Employee("b"), "emp2");
hm.put(new Employee("a"), "emp1 OVERRIDDEN");
System.out.println("HashMap's data> "+hm);
System.out.println("HashMap's size> "+hm.size());
System.out.println(hm.get(new Employee("a")));
}
}
Answer.
/*OUTPUT
HashMap's data> {Employee[ name=a] =emp1 OVERRIDDEN, Employee[ name=b] =emp2, Employee[ name=a] =emp1}
HashMap's size> 3
null
*/
Buckets= As hashCode() method returns 1, only 1 bucket location will be used.
Size= As equals() method doesn’t exist, size will be 3, all three employees will be stored on same bucket location but in different Entry.
get()=we won’t be able to get object.
Collection interview Question 72.
Collection Output interview question 22. How many buckets will be there and what will be size of HashMap?
package p5;
import java.util.HashMap;
class Employee {
private String name;
public Employee(String name) { // constructor
this.name = name;
}
//no hashCode() method
@Override
public boolean equals(Object obj){
return true;
}
@Override
public String toString() {
return "Employee[ name=" + name + "] ";
}
}
/** Copyright (c), AnkitMittal JavaMadeSoEasy.com */
public class Program5 {
public static void main(String...a){
HashMap<Employee, String> hm=new HashMap<Employee, String>();
hm.put(new Employee("a"), "emp1");
hm.put(new Employee("b"), "emp2");
hm.put(new Employee("a"), "emp1 OVERRIDDEN");
System.out.println("HashMap's data> "+hm);
System.out.println("HashMap's size> "+hm.size());
System.out.println(hm.get(new Employee("a")));
}
}
Answer.
/*OUTPUT
HashMap's data> {Employee[ name=b] =emp2, Employee[ name=a] =emp1 OVERRIDDEN, Employee[ name=a] =emp1}
HashMap's size> 3
null
*/
Buckets= As hashCode() method is not there, hashcode generated for 3 objects will be different and we will end up using 3 buckets.
Size= Though equals() method is their(but because of hashCode() method’s absence) which always returns true, we won’t be able to locate correct bucket location for calling equals() method, so, size will be 3.
get()=we won’t be able to get object.
Collection interview Question 72.
Collection Output interview question 22. How many buckets will be there and what will be size of HashMap?
package p6;
import java.util.HashMap;
class Employee {
private String name;
public Employee(String name) { // constructor
this.name = name;
}
@Override
public int hashCode(){
return (this.name==null ? 0: this.name.hashCode() );
}
//no equals() method
@Override
public String toString() {
return "Employee[ name=" + name + "] ";
}
}
/** Copyright (c), AnkitMittal JavaMadeSoEasy.com */
public class Program6 {
public static void main(String...a){
HashMap<Employee, String> hm=new HashMap<Employee, String>();
hm.put(new Employee("a"), "emp1");
hm.put(new Employee("b"), "emp2");
hm.put(new Employee("a"), "emp1 OVERRIDDEN");
System.out.println("HashMap's data> "+hm);
System.out.println("HashMap's size> "+hm.size());
System.out.println(hm.get(new Employee("a")));
}
}
Answer.
/*OUTPUT
HashMap's data> {Employee[ name=b] =emp2, Employee[ name=a] =emp1 OVERRIDDEN, Employee[ name=a] =emp1}
HashMap's size> 3
null
*/
Buckets= As hashCode() method is overridden perfectly, 2 bucket locations will be used.
Size= As equals() method is not their, size will be 3,
get()=we won’t be able to get object.
Collection interview Question 73.
Collection Output interview question 23. How many buckets will be there and what will be size of HashMap?
package p7;
import java.util.HashMap;
class Employee {
private String name;
public Employee(String name) { // constructor
this.name = name;
}
//no hashCode() method
@Override
public boolean equals(Object obj){
Employee emp=(Employee)obj;
return (emp.name==this.name || emp.name.equals(this.name));
}
@Override
public String toString() {
return "Employee[ name=" + name + "] ";
}
}
/** Copyright (c), AnkitMittal JavaMadeSoEasy.com */
public class Program7 {
public static void main(String...a){
HashMap<Employee, String> hm=new HashMap<Employee, String>();
hm.put(new Employee("a"), "emp1");
hm.put(new Employee("b"), "emp2");
hm.put(new Employee("a"), "emp1 OVERRIDDEN");
System.out.println("HashMap's data> "+hm);
System.out.println("HashMap's size> "+hm.size());
System.out.println(hm.get(new Employee("a")));
}
}
Answer.
/*OUTPUT
HashMap's data> {Employee[ name=a] =emp1, Employee[ name=a] =emp1 OVERRIDDEN, Employee[ name=b] =emp2}
HashMap's size> 3
null
*/
Buckets= As hashCode() method is not there, hashcode generated for 3 objects will be different and we will end up using 3 buckets.
Size= Though equals() method is their(but because of hashCode() method’s absence), we won’t be able to locate correct bucket location for calling equals() method, so, size will be 3.
get()=we won’t be able to get object.
Collection interview Question 74.
Collection Output interview question 24.
import java.util.ArrayList;
import java.util.List;
/** Copyright (c), AnkitMittal JavaMadeSoEasy.com */
public class ArrayListTest {
public static void main(String args[]) {
List<String> arrayList = new ArrayList<String>();
arrayList.add("a");
arrayList.add("a");
arrayList.clear();
arrayList.add("b");
arrayList.add("b");
System.out.println(arrayList.size());
}
}
Answer.
/*OUTPUT
2
*/
ArrayList allows duplicates, clear() method removes all elements from list.
Collection interview Question 75.
Collection Output interview question 25.
import java.util.HashSet;
import java.util.Set;
/** Copyright (c), AnkitMittal JavaMadeSoEasy.com */
public class HashSetTest {
public static void main(String args[]) {
Set hashSet = new HashSet();
hashSet.add("1");
hashSet.add(1);
hashSet.add(null);
hashSet.add("null");
System.out.println(hashSet);
}
}
Answer.
/* OUTPUT
[null, 1, 1, null]
*/
HashSet does not store duplicates but >
“1” is a String, while 1 is Integer &
null is nothing, while “null” is a String
Also HashSet does not maintain insertion order and allows null.
Having any doubt? or you you liked the tutorial! Please comment in below section.
Please express your love by liking JavaMadeSoEasy.com (JMSE) on facebook, following on google+ or Twitter.
Related >>
Collection interview Question 70.
Collection Output interview question 20. How many buckets will be there and what will be size of HashMap?
package p4;
import java.util.HashMap;
class Employee {
private String name;
public Employee(String name) { // constructor
this.name = name;
}
@Override
public int hashCode(){
return 1;
}
//no equals() method
@Override
public String toString() {
return "Employee[ name=" + name + "] ";
}
}
/** Copyright (c), AnkitMittal JavaMadeSoEasy.com */
public class Program4 {
public static void main(String...a){
HashMap<Employee, String> hm=new HashMap<Employee, String>();
hm.put(new Employee("a"), "emp1");
hm.put(new Employee("b"), "emp2");
hm.put(new Employee("a"), "emp1 OVERRIDDEN");
System.out.println("HashMap's data> "+hm);
System.out.println("HashMap's size> "+hm.size());
System.out.println(hm.get(new Employee("a")));
}
}
|
Answer.
/*OUTPUT
HashMap's data> {Employee[ name=a] =emp1 OVERRIDDEN, Employee[ name=b] =emp2, Employee[ name=a] =emp1}
HashMap's size> 3
null
*/
Buckets= As hashCode() method returns 1, only 1 bucket location will be used.
Size= As equals() method doesn’t exist, size will be 3, all three employees will be stored on same bucket location but in different Entry.
get()=we won’t be able to get object.
Collection interview Question 72.
Collection Output interview question 22. How many buckets will be there and what will be size of HashMap?
package p5;
import java.util.HashMap;
class Employee {
private String name;
public Employee(String name) { // constructor
this.name = name;
}
//no hashCode() method
@Override
public boolean equals(Object obj){
return true;
}
@Override
public String toString() {
return "Employee[ name=" + name + "] ";
}
}
/** Copyright (c), AnkitMittal JavaMadeSoEasy.com */
public class Program5 {
public static void main(String...a){
HashMap<Employee, String> hm=new HashMap<Employee, String>();
hm.put(new Employee("a"), "emp1");
hm.put(new Employee("b"), "emp2");
hm.put(new Employee("a"), "emp1 OVERRIDDEN");
System.out.println("HashMap's data> "+hm);
System.out.println("HashMap's size> "+hm.size());
System.out.println(hm.get(new Employee("a")));
}
}
|
Answer.
/*OUTPUT
HashMap's data> {Employee[ name=b] =emp2, Employee[ name=a] =emp1 OVERRIDDEN, Employee[ name=a] =emp1}
HashMap's size> 3
null
*/
Buckets= As hashCode() method is not there, hashcode generated for 3 objects will be different and we will end up using 3 buckets.
Size= Though equals() method is their(but because of hashCode() method’s absence) which always returns true, we won’t be able to locate correct bucket location for calling equals() method, so, size will be 3.
get()=we won’t be able to get object.
Collection interview Question 72.
Collection Output interview question 22. How many buckets will be there and what will be size of HashMap?
package p6;
import java.util.HashMap;
class Employee {
private String name;
public Employee(String name) { // constructor
this.name = name;
}
@Override
public int hashCode(){
return (this.name==null ? 0: this.name.hashCode() );
}
//no equals() method
@Override
public String toString() {
return "Employee[ name=" + name + "] ";
}
}
/** Copyright (c), AnkitMittal JavaMadeSoEasy.com */
public class Program6 {
public static void main(String...a){
HashMap<Employee, String> hm=new HashMap<Employee, String>();
hm.put(new Employee("a"), "emp1");
hm.put(new Employee("b"), "emp2");
hm.put(new Employee("a"), "emp1 OVERRIDDEN");
System.out.println("HashMap's data> "+hm);
System.out.println("HashMap's size> "+hm.size());
System.out.println(hm.get(new Employee("a")));
}
}
|
Answer.
/*OUTPUT
HashMap's data> {Employee[ name=b] =emp2, Employee[ name=a] =emp1 OVERRIDDEN, Employee[ name=a] =emp1}
HashMap's size> 3
null
*/
Buckets= As hashCode() method is overridden perfectly, 2 bucket locations will be used.
Size= As equals() method is not their, size will be 3,
get()=we won’t be able to get object.
Collection interview Question 73.
Collection Output interview question 23. How many buckets will be there and what will be size of HashMap?
package p7;
import java.util.HashMap;
class Employee {
private String name;
public Employee(String name) { // constructor
this.name = name;
}
//no hashCode() method
@Override
public boolean equals(Object obj){
Employee emp=(Employee)obj;
return (emp.name==this.name || emp.name.equals(this.name));
}
@Override
public String toString() {
return "Employee[ name=" + name + "] ";
}
}
/** Copyright (c), AnkitMittal JavaMadeSoEasy.com */
public class Program7 {
public static void main(String...a){
HashMap<Employee, String> hm=new HashMap<Employee, String>();
hm.put(new Employee("a"), "emp1");
hm.put(new Employee("b"), "emp2");
hm.put(new Employee("a"), "emp1 OVERRIDDEN");
System.out.println("HashMap's data> "+hm);
System.out.println("HashMap's size> "+hm.size());
System.out.println(hm.get(new Employee("a")));
}
}
|
Answer.
/*OUTPUT
HashMap's data> {Employee[ name=a] =emp1, Employee[ name=a] =emp1 OVERRIDDEN, Employee[ name=b] =emp2}
HashMap's size> 3
null
*/
Buckets= As hashCode() method is not there, hashcode generated for 3 objects will be different and we will end up using 3 buckets.
Size= Though equals() method is their(but because of hashCode() method’s absence), we won’t be able to locate correct bucket location for calling equals() method, so, size will be 3.
get()=we won’t be able to get object.
Collection interview Question 74.
Collection Output interview question 24.
import java.util.ArrayList;
import java.util.List;
/** Copyright (c), AnkitMittal JavaMadeSoEasy.com */
public class ArrayListTest {
public static void main(String args[]) {
List<String> arrayList = new ArrayList<String>();
arrayList.add("a");
arrayList.add("a");
arrayList.clear();
arrayList.add("b");
arrayList.add("b");
System.out.println(arrayList.size());
}
}
|
Answer.
/*OUTPUT
2
*/
ArrayList allows duplicates, clear() method removes all elements from list.
Collection interview Question 75.
Collection Output interview question 25.
import java.util.HashSet;
import java.util.Set;
/** Copyright (c), AnkitMittal JavaMadeSoEasy.com */
public class HashSetTest {
public static void main(String args[]) {
Set hashSet = new HashSet();
hashSet.add("1");
hashSet.add(1);
hashSet.add(null);
hashSet.add("null");
System.out.println(hashSet);
}
}
|
Answer.
/* OUTPUT
[null, 1, 1, null]
*/
HashSet does not store duplicates but >
“1” is a String, while 1 is Integer &
null is nothing, while “null” is a String
Also HashSet does not maintain insertion order and allows null.
Having any doubt? or you you liked the tutorial! Please comment in below section.
Please express your love by liking JavaMadeSoEasy.com (JMSE) on facebook, following on google+ or Twitter.
Related >>