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Program should set the start and end index and then display this accordingly. For example the above pattern is for displayNumbersBetween(2,10) and following is for displayNumbersBetween(5,7)
Ans. public class BuggyBread {
public static void main(String args[]) {
String str = "we are what we repeatedly do excellence then is not an act but a haBit";
Set<String> wordSet = new TreeSet(); // Using Linked Hash Set as we would like to retrieve words in the insertion order
Ans.
public class BuggyBread {
public static void main(String args[]) {
String str = "we are what we repeatedly Do excellence, then, is not an act but a haBit";
Set<String> wordSet = new LinkedHashSet(); // Using Linked Hash Set as we would like to retrieve words in the insertion order
Ans. public class BuggyBread {
public static void main(String args[]) {
String str1 = "we are what we repeatedly Do excellence, then, is not an act but a haBit";
String str2 = "we are what we repeatedly Do is";
String[] str1Words = str1.split(" ");
String[] str2Words = str2.split(" ");
Ans. public class BuggyBread {
public static void main(String args[]) {
String str1 = "we are what we repeatedly Do excellence, then, is not an act but a haBit";
String str2 = "we are what we repeatedly Do is";
String[] str1Words = str1.split(" ");
String[] str2Words = str2.split(" ");
Ans. import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.io.*;
public class Common {
public static void main(String ar[])throws Exception
{
File f=new File("a.txt");
File f1=new File("x.java");
System.out.println(f.exists());
FileInputStream fin = new FileInputStream(f);
FileInputStream fin1 = new FileInputStream(f1);
Ans. public class BuggyBread{
public static void main (String args[]) {
Set<Character> set = new HashSet();
Set<Character> setWithDuplicateChar = new HashSet();
String str = "hello world";
for(char character: str.toCharArray()){
if(set.contains(character)){
setWithDuplicateChar.add(character);
} else {
set.add(character);
}
}
System.out.println(setWithDuplicateChar);
}
}
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Q48. Write a program to print all even numbers first and then all odd numbers till 100. For example the output should be
2
4
6
8
...... 100
and then
1
3
5
7
...... 99
using only 1 for loop ? Is it possible with just one loop ?
public class BuggyBread{
public static void main (String args[]) {
int x = 50;
for(int i=1;i <= 100;i++){
if(i<=50){
System.out.println(i*2);
} else {
System.out.println(i-x);
x = x - 1;
}
}
}
}
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Q49. Write a program in Java that prints the numbers from 1 to 100. But for multiples of three print Fizz instead of the number and for the multiples of five print Buz. For numbers which are multiples of both three and five print FizzBuzz
Q50. Write a Program to check if string is a Colidrome ?
(Colidrome is a word that has n alphabets followed by the reverse of the n alphabets, for ex - mallom)
Ans.
public class Class {
public static void main(String[] args) {
String str = "mallam";
String firstHalf = str.substring(0, str.length() / 2);
String secondHalf = str.substring(str.length() / 2);
if (firstHalf.equals(new StringBuilder(secondHalf).reverse().toString())) {
System.out.println("It's a Colidrome");
} else {
System.out.println("It's not a Colidrome");
}
}
}
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Ans. public class Class{
public static void main(String[] args){
String string1 = "Hello I am Jack. I live in United States. I live in california state.";
String string2 = "I live in";
int startIndex = 0;
int endIndex = string1.length()-1;
Ans. public class Class{
public static void main(String[] args){
String string1 = "Hello I am Jack. I live in United States. I live in california state.";
String string2 = "I live in";
if(string1.indexOf(string2) >= -1){
System.out.println("string2 is sub string of string1");
} else {
System.out.println("string2 is not sub string of string1");
}
}
}
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Ans. public class Class{
public static void main(String[] args){
List<Integer> collector = new ArrayList();
Scanner scanner = new Scanner(System.in);
int x = scanner.nextInt();
while(x != 0){
collector.add(x);
x = scanner.nextInt();
}
System.out.println(collector.stream().collect(Collectors.averagingInt(p->((Integer)p))));
}
}
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Ans. int duplicateArray[] = { 1, 2, 2, 3, 4, 5, 6, 8, 9}
Set unique = new HashSet();
for (int i = 0; i < duplicateArray.length; i) {
if (unique.contains(duplicateArray[i])) {
System.out.println(duplicateArray[i]);
} else {
unique.add(duplicateArray[i]);
}
}
Complexity O(n) = nHashSet contains and add has O(n) = 1
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