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Ans. matcher.find() should have been used instead of matcher.next() within while. | ||||
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Ans. It will print 4567,5678 and 6789 but Order cannot be predicted. | ||||
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Ans. This will give compile time error as we cannot retrieve the element from a specified index using Set. Set doesn't maintain elements in any order. | ||||
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Ans. WEDNESDAY FRIDAY SATURDAY Only one FRIDAY will be printed as Set doesn't allow duplicates.Elements will be printed in the order in which constants are declared in the Enum. TreeSet maintains the elements in the ascending order which is identified by the defined compareTo method. compareTo method in Enum has been defined such that the constant declared later are greater than the constants declared prior. | ||||
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Ans. int findMax(int[] items){ int maxNumber = 0; for(int x:items){ if(x > maxNumber){ maxNumber = x; } } return maxNumber; } | ||||
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Ans. Yes , It can be done using single for 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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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); byte b[]=new byte[10000]; byte b1[]=new byte[10000]; fin.read(b); fin1.read(b1); String s1 = new String(b); String s2 =new String(b1); String words1[] = s1.trim().split(" "); String words2[] = s2.trim().split(" "); Listlist1 = new ArrayList<>(Arrays.asList(words1)); Listlist2 = new ArrayList<>(Arrays.asList(words2)); list1.retainAll(list2); System.out.println(list1); } } | ||||
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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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Ans. private void permutation(String prefix, String sufix) { int ln = sufix.length(); if(ln == 0) { System.out.println(prefix); } else { IntStream.range(0, ln).forEach(i->permutation(prefix sufix.charAt(i), sufix.substring(0,i) sufix.substring(i 1, ln))); } } call:permutation("", "abcdef"); | ||||
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Ans. File folder = new File(path); if(!folder.exists()){ try { folder.mkdir(); } catch (Exception e) {} } | ||||
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Ans. A way of encoding characters as binary numbers. The Unicode character set includes characters used in many languages, not just English. Unicode is the character set that is used internally by Java. | ||||
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Ans. public class BuggyBread1{ public static void main (String args[]) { String str = "hheello world"; char[] charArray = str.toCharArray(); char selectedChar = 'a'; for(char char1: charArray){ if(!str.contains(Character.toString(char1).concat(Character.toString(char1)))){ selectedChar = char1; break; } } System.out.println(str.indexOf(Character.toString(selectedChar))); } } | ||||
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Ans. public class TotalUsingThreads extends Thread{ private static List<TotalUsingThreads> collector = new ArrayList<TotalUsingThreads>(); private int startFrom = 0; private Integer threadTotal = null; Test(int startFrom){ this.startFrom = startFrom; } public static void main(String[] args) throws InterruptedException{ int totalSum = 0; // Create all the threads and set the count starting point for all of them for(int count=0;count<10;count++){ TotalUsingThreads newThread = new TotalUsingThreads(count*100); newThread.start(); collector.add(newThread); } boolean allCollected = false; // Make sure that all thread totals are collected and all threads have completed their tasks while(!allCollected){ for(int count=0;count<10;count++){ if(collector.get(count).threadTotal == null){ Thread.sleep(100); break; } } allCollected = true; } // sum totals of all threads for(int count=0;count<10;count++){ totalSum += collector.get(count).threadTotal; } System.out.println(totalSum); } public void run(){ threadTotal = 0; for(int count=startFrom;count<startFrom+100;count++){ threadTotal += count; } } } | ||||
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Ans. convert(int binaryInt) { int sumValue=0; int multiple = 1; while(binaryInt > 0){ binaryDigit = binaryInt%10; binaryInt = binaryInt /10; sumValue = sumValue + (binaryDigit * multiple); multiple = multiple * 2; } return sumValue; } | ||||
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Ans. equal 1 | ||||
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Ans. BuggyBread | ||||
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Ans. Buggy | ||||
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Ans. Buggy | ||||
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Ans. Compile Time Error: Unhandled exception type Exception | ||||
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Ans. file.txtC:file.txtC:file.txt | ||||
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Ans. NullPointerException at line:File file = new File("/folder", child); | ||||
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Ans. ..file.txt C:WorkspaceProject..file.txt C:Workspacefile.txt | ||||
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Ans. c. File file = new File("../file.txt"); FileWriter fileWriter = new FileWriter(file); BufferedWriter bufferedOutputWriter = new BufferedWriter(fileWriter); | ||||
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Ans. IOException | ||||
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Ans. FileNotFoundException | ||||
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Ans. Though we are trying to serialize BuggyBread1 object but we haven't declared the class to implement Serializable. This will throw java.io.NotSerializableException upon execution. | ||||
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Ans. No, It will throw java.io.NotSerializableException. | ||||
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Ans. Yes. | ||||
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Ans. Date and Time after 5 months from now. | ||||
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Ans. http://www.c4learn.com/c-programs/program-to-check-whether-number-is.html | ||||
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