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Ans. As Class B has been declared abstract , we can either implement any of these methods and just declare rest of them abstract. | ||||
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Like Discuss Correct / Improve  interfaces  abstract classes  code  coding | ||||
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Ans. false false | ||||
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Ans. We creates separate branches for each project if development work is going on parallel and they are to be released at different times. Once the first release is done, we merge the branch changes into trunk. If they all have to go at one time, we usually would merge everything in the trunk itself. | ||||
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Ans. If it's just smaller change, and single person is working, then this approach is fine. Otherwise there are risk on loosing it on your machine. Moreover , If there are multiple people working , it makes it hard to share code. It's better to create a separate branch and then merge it later to trunk. | ||||
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Ans. If the branch is to be created from Trunk and we are using Eclipse. Go to the Trunk Copy of the Project Right Click the project and then Click Branch/Tag In the Create Branch / Tag Dialog, Add the Destination Branch Url Check whether we want to make copy from the Head Revision or some specific revision Number | ||||
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Ans. There are many problems with the code 1. The method returns void and hence we cannot return any integer value. 2. We cannot return more than one value from a method. 3. The code after 1st Return statement is unreachable. | ||||
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Ans. Yes, It's an instance initialization block. | ||||
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Ans. x==y means that both references have same type and are pointing to same memory location and hence would always mean that they have same value. x.equals(y) is not required in this case. | ||||
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Ans. if x==y turns out to be true x.equals(y) will be true too. If x.equals(y) could be true even if x==y is true or not. So the only possible outcomes are 1 || 1 = 1 0 || 1 = 1 0 || 0 = 0 i.e the outcome of x.equals(y) check for x==y is not required in this if statement. | ||||
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Ans. 404 means the resource is not found on the server.The resource might not be deployed correctly. 500 means internal server error which means that resource was located but then it resulted in some exception or error. 200 means Ok. | ||||
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Ans. http://www.programmingsimplified.com/java/source-code/java-program-armstrong-number | ||||
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Ans. package com.string; import java.util.Scanner; public class String13 { public static void main(String[] args) { System.out.println("Enter Sentence"); Scanner sc=new Scanner(System.in); String sentence=sc.nextLine(); String[] words=sentence.split(" "); int count=0; for (String string : words) { string.trim(); if(!string.equals("")){ count++; System.out.println(string " " count); } } } } | ||||
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Ans. int[] arr = {1,-1,2,-3,3,-4,4,5,6,-5,-6,-7,-8,8,9,-9}; List positiveNumbers = new ArrayList<>(); List negativeNumbers = new ArrayList<>(); for(int i = 0; i < arr.length(); i ){ if(I < 0){ negativeNumbers.add(i); } else { positiveNumbers.add(i); } } System.out.println("Positive Numbers:" + positiveNumbers); System.out.println("Negative Numbers:" + negativeNumbers); | ||||
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Ans. 500 is Internal Server Error 404 is resource not found 400 is Bad Request 403 is Forbidden 401 is Unauthorized 200 is OK | ||||
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Ans. https://www.tutorialspoint.com/javaexamples/thread_procon.htm | ||||
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Ans. General contract of hashCode is: 1.Whenever it is invoked on the same object more than once during an execution of a Java application, the hashCode method must consistently return the same integer, 2.If two objects are equal according to the equals(Object) method, then calling the hashCode method on each of the two objects must produce the same integer result. 3.It is not required that if two objects are unequal according to the equals(java.lang.Object) method, then calling the hashCode method on each of the two objects must produce distinct integer results. | ||||
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Ans. https://www.geeksforgeeks.org/lru-cache-implementation/ | ||||
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Ans. Yes its useless if we are not going to use its objects within Hash collection, For example - HashSet , HashMap. HashCode is used internally by these collections for Search. | ||||
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Ans. It will print 4 because member elements of an interface are implicitly static and hence the concept of overriding doesn't work. | ||||
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Ans. 10 | ||||
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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. 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){ 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; int countNoOfOccurences = 0; String remainingString = string1; while(startIndex < endIndex){ if(remainingString.indexOf(string2) != -1){ countNoOfOccurences++; startIndex = remainingString.indexOf(string2) + string2.length(); remainingString = remainingString.substring(startIndex); } else { break; } } System.out.println(countNoOfOccurences); } } | ||||
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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){ for(int i=1;i<= 100;i++){ if(i%3 == 0){ if(i%5 == 0){ System.out.println("FizzBuzz"); } System.out.println("Fizz"); } else if(i%5 == 0){ System.out.println("Buzz"); } else { System.out.println(i); } } } } | ||||
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Ans. public class BuggyBread { public static void main(String args[]) { Map<Character, Integer> countMap = new HashMap(); String str = "hello world"; for (char character : str.toCharArray()) { if (countMap.containsKey(character)) { countMap.put(character, countMap.get(character) + 1); } else { countMap.put(character, 1); } } System.out.println(countMap); } } | ||||
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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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This question was recently asked at 'Compro Technologies'.This question is still unanswered. Can you please provide an answer. | ||||
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Ans. 1. Object Level Reuse - private methods 2. Class Level Reuse - static methods 3. package level Reuse - default methods 4. Application Level Reuse - Classes 5. Multiple Applications Level Reuse - Libraries , Frameworks | ||||
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Ans. int arr[]={1,3,5,6,4,8,9,2,10}; Arrays.sort(); System.out.println(arr[arr.length-1]); | ||||
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