Java 20 years celebrations

Integer Literal - Octals

What is the Output of the below Program:

//OcatalSample.java
public class OcatalSample{
    public static void main(String []args){
      int i = 012;
      int j = 034;
      int k = 056;
      int m = 08; 
      System.out.println("i=" +i + ", j="+j+ ", k="+k+ ", m="+m);
    }
}

public class ReverseNumber {
   public static void main(String[] args) {
      int number = 987;
      System.out.println("Reverse of given number is: "+
               getReverseNumber(number));   
   }

   private static int getReverseNumber(int number) {
      int reverseNumber = 0;
      while (number > 0) {
          int reminder = number % 10;
          reverseNumber = reverseNumber * 10 + reminder;
          number = number / 10;
      }
      return reverseNumber;
   }
}

//Output:
//Reverse of given number is: 789

 Output: It won’t compile. It gives below compilation error
             Exception in thread "main" java.lang.Error: Unresolved compilation problem:
                       The literal 08 of type int is out of range
            at OcatalSample.main(OcatalSample.java:7)

Why??
An integer literal may be expressed in decimal (base 10), hexadecimal (base 16), octal (base 8), or binary (base 2).
Octal representation in java(base 8) :
An integer literal prefixed with 0 is treated as octal. It means when an integer literal starts with 0 in Java, it's assumed to be in octal notation. The digits 8 and 9 are illegal in octal—the digits can range only between 0 and 7.

In the above example #line7 gives an error, because as per above line range of octal is 0 to 7 only.

Note: Generally in octal representation, 7 + 1 is 10.

String comparison tip

In Java if you actually want to test whether two strings have the same value you should use .equals(). It checks value equality. I think most of the people knew it. 

Today I want to add a very valuable tip:

If you have strings or constants to compare, always put them first in the equals clause. 

//StringComparisonTip.java
public class StringComparisonTip {
   public static void main(String[] args) {
     // Recommended String comparison
     if("Venkatesh".equals(getMyName())){  
        System.out.println("We are equal");
     }
     // It is Correct but it may be throws NPE at runtime
     if(getMyName().equals("Venkatesh")){  
        System.out.println("We are equal");
     }
   }
   public static String getMyName(){
     return null;
   }
}

In the above code line #5 (if("Venkatesh".equals(getMyName()))) is much better than the line #9 (if(getMyName().equals("Venkatesh"))) because in the second case getMyName() may be null and throws an java.lang.NullPointerException while in the first case it doesn't matter if getMyName() is null. 

Note: I am not saying second one is wrong but I am saying it is not recommended by Java Guru's(Who has great knowledge in Java). Both are syntactically correct and you never get compilation error.

Count occurrence of an element

Write a program to count the total number of duplicated entries in a List.

For Example :
Input List of countries : japan,usa,uae,japan,uae,usa,australia
Expected output:
japan:2
usa:2
uae:2
australia:1

Solutions:

Solution 1:
1) Java API provides a class called Collections. This class contains static utility methods for manipulating collections. It provides a static method for counting the number of elements in specified collection, as you can see below for more information. 

static int

frequency(Collection<?> c, Object o)  Returns the number of elements in the specified collection equal to the specified object.      c - the collection in which to determine the frequency of o      o - the object whose frequency is to be determined

// CountDuplicates.java
import java.util.Arrays;
import java.util.Collections;
import java.util.LinkedHashSet;
import java.util.List;
import java.util.Set;

public class CountDuplicates {
   public static void main(String[] args) {
      String[] listOfArray = 
      { "japan", "usa", "uae", "japan", "uae", "usa", "australia" };
      List list = Arrays.asList(listOfArray);
      System.out.println("Frequency of each country ");
      Set uniqueSet = new LinkedHashSet(list);
      for (String countryName : uniqueSet) {
      System.out.println(countryName + ": " 
      + Collections.frequency(list, countryName));
      }
   }
}
//Output::
//Frequency of each country 
//japan: 2
//usa: 2
//uae: 2
//australia: 1

Solution 2:

// CountDuplicates1.java
import java.util.LinkedHashMap;
import java.util.Map.Entry;
import java.util.Map;

public class CountDuplicates1 {
   public static void main(String[] args) {
      String[] listOfArray = 
      { "japan", "usa", "uae", "japan", "uae", "usa", "australia" };
      Map<String, Integer> map = new LinkedHashMap<String,Integer>();
      for( String countryName : listOfArray){
          int count = 0;
          if(map.get(countryName) != null){
               count = map.get(countryName);
          }
          map.put(countryName, count+1);
      } 
      for (Entry countryEntry : map.entrySet()) {
          System.out.println(countryEntry.getKey()+ ": " 
         + countryEntry.getValue() );
      }
   }
}
//Output::
//Frequency of each country 
//japan: 2
//usa: 2
//uae: 2
//australia: 1

Swap two numbers

Solutions:
1) Using temporary variable / Using extra space
2) Without using temporary variable / Without using extra space
     a) Using arithmetic operators
     b) Using bitwise operator

Solution 1: Using temporary variable

//SwapUsingExtraVariable.java
public class SwapUsingExtraVariable {
    public static void main(String[] args) {
        int a = 10;
        int b = 20;
        System.out.println("Values before swapping");
        System.out.println("a= " +a + " b= "+b);
        int temp = a;
        a = b;
        b = temp;
        System.out.println("Values after swapping");
        System.out.println("a= " +a + " b= "+b);
    }
}
// Output:
//Values before swapping
//a= 10 b= 20
//Values after swapping
//a= 20 b= 10

Solution 2 a: Using arithmetic operators

//SwapUsingArthematic.java
public class SwapUsingArthematic {
	public static void main(String[] args) {
		int a = 10;
		int b = 20;
		
		System.out.println("Values before swapping");
		System.out.println("a= " +a + " b= "+b);
		
		a = a + b;
		b = a - b;
		a = a - b;
                
                // a = a * b;
                // b = a / b;
                // a = a / b;

		System.out.println("Values after swapping");
		System.out.println("a= " +a + " b= "+b);
	}
}
// Output:
//Values before swapping
//a= 10 b= 20
//Values after swapping
//a= 20 b= 10

   
We can use "*" & "/" operators also instead of "+" & "-". Check the above program lines 14, 15 and 16 for the reference. But problem with "/" operator is it gives java.lang.ArithmeticException: / by zero Exception for "b" value is zero.

Solution 2 b: Using bitwise operator

//SwapUsingBitWiseOperator.java
public class SwapUsingBitWiseOperator {

	public static void main(String[] args) {
		int a = 10;
		int b = 20;
		
		System.out.println("Values before swapping");
		System.out.println("a= " +a + " b= "+b);
		
		a = a ^ b;
		b = a ^ b;
		a = a ^ b;

		System.out.println("Values after swapping");
		System.out.println("a= " +a + " b= "+b);
	}

}
// Output:
//Values before swapping
//a= 10 b= 20
//Values after swapping
//a= 20 b= 10

Technical Interview Question 1

You have an array of numbers from 0 to N-1, one of the numbers is removed, and replaced with a number already in the array which makes a duplicate of that number. How can we detect this duplicate in time O(N).

For example an array of 6 integers - 0,1,2,3,4,5. one of the number i.e "3" removed and replaced with "2". now array would become 0,1,2,2,4,5

Solution :

// InterviewQuestion1.java
public class InterviewQuestion1 {
   public static void main(String[] args) {
      int[] testArray = { 0, 1, 2, 2, 4, 5 };
      boolean[] testBool = new boolean[testArray.length];
      for (int i = 0; i < testArray.length; i++) {
         testBool[testArray[i]] = true;
      }
      System.out.print("Duplicate Number is : ");
      for (int i = 0; i < testArray.length; i++) {
        if (!testBool[i]) {
         System.out.print(testArray[i]);
         break;
        }
      }
   }
}

// Output: Duplicate Number is : 2

Sample Code

    // Hello World.java
    public class HelloWorld{
      public static void main(String[] args){
        System.out.print("Hi");      
      }      
    }
   

Java Program1

Write a Java Program to concatenate all characters at first and sum of numbers at last from the given string. 

For Example: 
Input String ::  azch1234dhrk345
Output: azchdhrk22

Solutions :
1) Using java regex package
2) Using Character class

Solution 1:

import java.util.regex.Matcher;
import java.util.regex.Pattern;

public class SumOfNumbersInString {

   public static void main(String[] args) {
      String myString = "azch1234dhrk345";
      SumOfNumbersInString s = new SumOfNumbersInString();
      System.out.print(s.getResultString(myString));
   }
     
   public String getResultString(String myString){
      Pattern pattern = Pattern.compile("\\d");
      Matcher match = pattern.matcher(myString);
      int sum = 0, index = 0;
      String otherString = "";
      boolean matchFind = false;
      while (match.find()) {
        matchFind = true;
        otherString += myString.substring(index, match.start());
        sum += Integer.parseInt(match.group());
        index = match.end();
      }
      if(index < myString.length()){
        otherString += myString.substring(index);
      }
      if(matchFind){
        return otherString + sum;
      } 
      return myString;
   }
}
// Output : azchdhrk22
 

Solution 2:

//SumOfNumbersInString2.java
public class SumOfNumbersInString2 {
   public static void main(String[] args) {
     String myString = "azch1234dhrk345";
     SumOfNumbersInString2 s = new SumOfNumbersInString2();
     System.out.print(s.getResultString(myString));
   }

   public String getResultString(String myString) {
     StringBuffer sb = new StringBuffer();
     int sum = 0;
     // Iterate over the string from 0th index to length()-1 characters
     // Check for the digit
     // If it is a digit, add this value to sum
     // If it is not digit, append to String Buffer Object
     for (int i = 0; i < myString.length(); i++) {
        char ch = myString.charAt(i);
        if (Character.isDigit(ch)) {
           sum += Character.getNumericValue(ch);
        } else {
           sb.append(ch);
        }
     }
     return sb.toString() + sum;
   }
}
// Output : azchdhrk22

String Concatenation Operator

String name = "Venkatesh";
int age = 5;
int salary = 6;

System.out.println(name + age + salary);

What is the Output?

Answer:- 

Rule:
If either operand is a String, the + operator becomes a String concatenation operator. If both operands are numbers, the + operator is the addition operator.

Note: The operands are the things on the left or right side of the operator.
Example: the expression a+b --> here a & b are the operands and "+" is operator.


Finally the answer is : Venkatesh56

Sort Month Names

1) Sorting months by natural sort order or alphabetical order or not chronological order.
2) Sorting months by Chronological order.

Solution for Question 1 ::

import java.util.Arrays;
import java.util.Collections;
import java.util.List;

// Natural Order : Month name by Lexicographic order
public class SortSample {

   public static void main(String[] args) {

    List<String> wordList = Arrays.asList( "Feb","Jul",Jan", "Apr","Dec", "Aug", "Oct","May","Sep","Nov", "Jun","Mar");
     
    System.out.println(wordList);
  
    Collections.sort(wordList);
  
    System.out.println(wordList);     
  }
}

Solution for Question 2 ::

import java.text.ParseException;
import java.text.SimpleDateFormat;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.Date;
import java.util.List;

public class SortSample1 {

    public static void main(String[] args) {
   
      List<String> months = Arrays.asList( "Feb", "Jul", "Jan", "Apr", "Dec",       "Aug", "Oct", "May", "Sep", "Nov", "Jun","Mar");
  
      System.out.println(months);
  
      final Comparator<String> dateCompare = new Comparator<String>() {
 
         public int compare(String o1, String o2) {
     
           SimpleDateFormat s = new SimpleDateFormat("MMM");
           Date s1 = null;
           Date s2 = null;
           try {
             s1 = s.parse(o1);
             s2 = s.parse(o2);
           } catch (ParseException e) {
               e.printStackTrace();
           }
           return s1.compareTo(s2);
         }
      };
  
     Collections.sort(months, dateCompare);
  
     System.out.print(months);
 } 
}

CodingBat > roundSum

For this problem, we'll round an int value up to the next multiple of 10 if its rightmost digit is 5 or more, so 15 rounds up to 20. Alternately, round down to the previous multiple of 10 if its rightmost digit is less than 5, so 12 rounds down to 10. Given 3 ints, a b c, return the sum of their rounded values. To avoid code repetition, write a separate helper "public int round10(int num) {" and call it 3 times. Write the helper entirely below and at the same indent level as roundSum().

roundSum(16, 17, 18) → 60
roundSum(12, 13, 14) → 30
roundSum(6, 4, 4) → 10

Solution ::

 public int roundSum(int a, int b, int c) {
    return round10(a) + round10(b) + round10(c);
 }
 public int round10(int num) {
    int remainder = num % 10;
    num -= remainder;
    if (remainder >= 5) {
       num += 10;
    }
    return num;
 }

CodingBat > firstLast6

Given an array of ints, return true if 6 appears as either the first or last element in the array. The array will be length 1 or more.

firstLast6({1, 2, 6}) → true

firstLast6({6, 1, 2, 3}) → true

firstLast6({13, 6, 1, 2, 3}) → false

Solution ::

public class FirstLast6 {
 public static void main(String[] args) {
  FirstLast6 fl = new FirstLast6();
  int[] testArray = new int[] { 1, 2, 6 };
  System.out.print(fl.firstLast6(testArray));
 }

 public boolean firstLast6(int[] nums) {
  if (nums[0] == 6 || nums[nums.length - 1] == 6)
   return true;
  return false;
 }
}