Saturday, July 13, 2019

Find Palindrome Strings in an Array Using Java


In this article, we will see how to find all palindrome strings from an array. This is a very frequently asked questions in interview.  Also there is a question to find all palindrome number from an array.  Find few more collection interview question.  

PalindromeStrings.java

package com.techbyteslearn.lab.basic;

/**
 * This program illustrates how to find palindrome strings from an array.
 */
public class PalindromeStrings {

    public static void main(String[] args) {
        String[] stringArray = {
            "eye",
            "jdg",
            "javadevelopersguide",
            "aabaa",
            "hello",
            "pip"
        };

        for (int i = 0; i < stringArray.length; i++) {
            printOnlyPalindrome(stringArray[i]);
        }
    }

    private static void printOnlyPalindrome(String str) {
        String oldString = str;
        StringBuilder builder = new StringBuilder(str);

        if (builder.reverse().toString().equals(oldString)) {
            System.out.println(oldString + " is a Palindrome String.");
        }
    }
}

Output:

eye is a Palindrome String.
aabaa is a Palindrome String.
pip is a Palindrome String.
Happy Learning.

Find All Palindrome Numbers from an Array in Java

In this article, we will see how to find all palindrome from an array. This is a very basic questions in interview, the interviewer will ask the same questions in different way. So, its good to know all possible questions from palindrome. Also there is a question to find all palindrome number from a list.  Find few more collection interview question.  Today we will see how to check a number is palindrome or not. 

FindAllPalindrome.java

package com.techbyteslearn.lab.basic;

public class FindAllPalindrome {
public static void main(String[] args) {
int numberArray[] = { 120, 990, 121, 777, 808, 1280 };
for (int i = 0; i < numberArray.length; i++) {
printOnlyPalindrom(numberArray[i]);
}
}
private static void printOnlyPalindrom(int number) {
int finalNumber = 0;
int oldNumber = number;
// Repeat the loop until the number became zero.
while (number != 0) {
// Get the First Digit (i.e. 1)
int firstDigit = number % 10;
// Get the Result number.
finalNumber = (finalNumber * 10) + firstDigit;
// Now get the remaining digits , after finding the first digit
number = number / 10;
}
// Now compare the finalNumber and oldNumber both are same or not.
if (finalNumber == oldNumber)
System.out.println(finalNumber + " is a Palindrome.");
}
}

Output -

121 is a Palindrome.
777 is a Palindrome.
808 is a Palindrome.

How to Check Whether a Number Is a Palindrome in Java


In this article, we will see how to check if a number is palindrome or not. This is a very basic questions in interview. But, you never know about what kind of question the interviewer will ask. So, better you prepare for every certain questions. Also there is a question to find all palindrome number from a list.  Find few more collection interview question.  Today we will see how to check a number is palindrome or not. 


Palindrome.java

package com.techbyteslearn.lab.basic;

public class Palindrome {

    public static void main(String[] args) {
        int number = 121;
        int temp = number;
        int finalNumber = 0;

        // Repeat the loop until the number becomes zero.
        while (number != 0) {

            // Get the last digit.
            int lastDigit = number % 10;

            // Build the reversed number.
            finalNumber = (finalNumber * 10) + lastDigit;

            // Remove the last digit from the number.
            number = number / 10;
        }

        // Compare the reversed number with the original number.
        if (finalNumber == temp) {
            System.out.println("This number is a Palindrome.");
        } else {
            System.out.println("This number is not a Palindrome.");
        }
    }
}

Output:

This number is a Palindrome.
 
 

One important correction from the original comments: % 10 gets the last digit, not the first digit. Also, number has already become 0 by the time of the final comparison, so comparing with temp is the correct approach.



Happy Learning.

Thursday, July 11, 2019

Find Duplicate Values in a List Using Java

In this article, we will see how to find the duplicate values from an array or list using java.  This is one of important programming questions in technical interview. Each interviewer has different approach to access the candidate. But, the logic and the approach by candidate is really matter. In this program we have used Map and List both, so its a kind of collections interview questions. You can find few more collection interview question.  Today we will see how to find the duplicate values from array. 


The logic is very simple here, see the below.

  • At first we need we need to create a Map to hold the key-value pair. Where key is the array element and value is the counter for number of time the array element repeats.
  • Then we will iterate the array and put into the map as per the above step. If the map contains the element earlier, then we will update the value +1.
  • Finally we will have the map , which holds the array elements with the counter for repentance. 
  • Now, we will iterate the Map , by checking the condition where the counter is more than 1 (i.e. its duplicated or repeated).


DuplicateFinder.java

package com.techbyteslearn.lab.basic;

import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.Map;
import java.util.Map.Entry;

public class DuplicateFinder {

    public static void main(String[] args) {

        ArrayList<Integer> list = new ArrayList<>(
                Arrays.asList(4, 3, 5, 25, 25, 25, 13, 5, 22, 4, 90));

        System.out.println("Input List Data = " + list);

        Map<Integer, Integer> map = new HashMap<Integer, Integer>();

        for (int i = 0; i < list.size(); i++) {

            if (map.isEmpty()) {
                map.put(list.get(i), 1);
            } else if (map.containsKey(list.get(i))) {
                map.put(list.get(i), map.get(list.get(i)) + 1);
            } else {
                map.put(list.get(i), 1);
            }
        }

        System.out.println("\nDuplicate values are: ");

        // Iterate the Map and display the duplicate values.
        for (Entry<Integer, Integer> entry : map.entrySet()) {

            if (entry.getValue() > 1) {
                System.out.println(entry.getKey());

                // TODO: We can now put these values into any list.
            }
        }
    }
}

Output:

Input List Data = [4, 3, 5, 25, 25, 25, 13, 5, 22, 4, 90]

Duplicate values are:
4
5
25

Note- One small point: because HashMap does not guarantee iteration order, the order of 4, 5, and 25 in the output can vary.



Happy Learning.

Find the Largest Number in an Array Using Java

In this article, we will see how to find the largest number from an array using java.  This is one of basic questions in technical interviews. Earlier post we had seen how to find the smallest element from array. Now we will see how to find the largest number from integer array using java. 

The logic is very simple here, see the below.

  • At first we need to assume any element as largest value. Example - 0th location.
  • Then iterate over the array and compare with each element , whether its larger than the assumed larger value or not. If array element is larger then assign the array element value to assumed variable. Repeat the entire until end. 


FindLargestNumberInArray.java

package com.techbyteslearn.lab.basic;

public class FindLargestNumberInArray {

    // Find the largest value from an array.
    public static void main(String[] args) {
        int[] arr = {200, 3, 4, 24, 33, 24, 22, 55, 90, 103, 150};

        // Assume the largest value is at the 0th index.
        int largest = arr[0];

        for (int i = 0; i < arr.length; i++) {
            if (arr[i] >= largest) {
                largest = arr[i];
            }
        }

        System.out.println("Largest Number is ::" + largest);
    }
}

Output:

Largest Number is ::200

The original i < arr.length - 1 skips the last element. Using i < arr.length checks the complete array.


Using Java 8

int largest = IntStream.of(arr).boxed().max(Comparator.naturalOrder()).get().intValue() ;


 Happy Learning.

Find the Smallest Number in an Array Using Java


In this article, we will see how to find the smallest number from an array using java.  This is one of basic questions in technical interviews. Earlier post we had seen how to use stream for finding the missing number. Now we will see how to find the smallest number from integer array using java. 


The logic is very simple here, see the below.

  • At first we need to assume the first smallest element.
  • Then iterate over the array and compare with each element , whether its smaller than the assumed value or not. If array element is smaller then assign the array element value to assumed variable. Repeat the entire until end. 

FindSmallestNumberInArray.java

package com.techbyteslearn.lab.basic;

public class FindSmallestNumberInArray {

    // Find the smallest value from an array.
    public static void main(String[] args) {
        int[] arr = {200, 3, 4, 24, 33, 24, 22, 55, 90, 103, 150};

        // Assign the 0th index as the first smallest number.
        int smallest = arr[0];

        for (int i = 0; i < arr.length; i++) {
            if (arr[i] <= smallest) {
                smallest = arr[i];
            }
        }

        System.out.println("Smallest Element is - " + smallest);
    }
}

Output:

Smallest Element is - 3

The original i < arr.length - 1 skips the last element. Using i < arr.length is the correct condition.



Using Java 8

IntStream.of(arr).boxed().min(Comparator.naturalOrder()).get().intValue();

Wednesday, July 10, 2019

Find One Missing Number from a List Using Java


In this article, we will see how to find the missing number from a list using java.  This is one of important common interview question asked in interview. You can see, how to find all missing numbers from a list. In this program we will use core java or the traditional way using for loop for finding the miss number from a list. Earlier post we had seen how to use stream for finding the missing number. Now we will see how to find one missing number using traditional core java style. 


The logic is very simple here, see the below.

  • At first we need to find the MAX number from the list. We need this MAX number because , we need to calculate the SUM of all natural number up to that max number. 
  • Then , we need to calculate the sum of all those natural number.
  • Then we will subtract each element from the given list from sumOfNaturalNumbers.
  • Now, the at the last  the value inside sumOfNaturalNumbers is the missing number.

FindOneMissingNumber.java

package com.techbyteslearn.lab.basic;

import java.util.ArrayList;
import java.util.Arrays;

public class FindOneMissingNumber {

    public static void main(String[] args) {
        ArrayList<Integer> numberList = new ArrayList<>(
                Arrays.asList(10, 3, 2, 4, 5, 6, 7, 9, 8, 14, 1, 11, 13));

        int sumOfNaturalNumbers = getSumUptoMax(findMax(numberList));

        for (int i = 0; i < numberList.size(); i++) {
            /*
             * Subtract each element from the list from sumOfNaturalNumbers.
             * The final value will be the missing number.
             */
            sumOfNaturalNumbers = sumOfNaturalNumbers - numberList.get(i);
        }

        int missingNumber = sumOfNaturalNumbers;
        System.out.println("Missing Number is :: " + missingNumber);
    }

    // Find the sum of all natural numbers up to limitNumber.
    private static int getSumUptoMax(int limitNumber) {
        int sum = 0;

        for (int i = 1; i <= limitNumber; i++) {
            sum = sum + i;
        }

        return sum;
    }

    // Find the greatest value from the list.
    private static int findMax(ArrayList<Integer> numberList) {
        int largest = numberList.get(0);

        for (int i = 1; i < numberList.size(); i++) {
            if (numberList.get(i) > largest) {
                largest = numberList.get(i);
            }
        }

        return largest;
    }
}

Output:

Missing Number is :: 12

 

This approach assumes the list contains the numbers from 1 through the maximum value, with exactly one number missing.


Happy Learning.

Find a Missing Number from a List Using Java 8 Streams

In this article, we will find the missing number from a list of numbers using java 8.  This is one of important questions asked in interview. This program is to find the only one missing number. Check how to find all missing numbers from a list. In this program we will use only java 8 stream to find the missing number. Earlier post we had seen how to use stream to sort the employee. Check more how find one missing number using traditional core java style


FindMissingNumber.java

package com.techbyteslearn.lab.basic;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Comparator;
import java.util.stream.IntStream;

public class FindMissingNumber {
public static void main(String[] args) {
ArrayList<Integer> numberList = new ArrayList<Integer>(Arrays.asList(1, 3, 2, 4, 5, 6, 7, 9, 10));
// Get the Max value from the List.
int maxValue = numberList.stream().max(Comparator.naturalOrder()).get().intValue();
// Get sum of all natural numbers - upto the above maxvalue
int sumOfAllNumber = IntStream.range(1, maxValue + 1).sum();
// Get the sum of all number inside List.
int sumofList = numberList.stream().mapToInt(Integer::intValue).sum();
// Now print the missing number.
System.out.println("The Missing Number is:: " + (sumOfAllNumber - sumofList));
}
}

Output - 

The Missing Number is:: 8

Tuesday, July 9, 2019

Sort Employees by Name and Age Using Java 8

This program for writing a program to sort the employee by name and age using JDK 8. Java 8 introduced many out of box features for developers. The comparator in java 8 is marked as @FunctionalInterface, and it provide a cleaner way to develop your code. We had already discussed how the Comparator<T> interface works before java 8. In this post, we will use the stream API with comparator. 


Employee.java


package com.techbyteslearn.lab.basic;

public class Employee {
private String name;
private int age;
private String department;
public Employee(String name, int age, String department) {
super();
this.name = name;
this.age = age;
this.department = department;
}
@Override
public String toString() {
return "\n Employee [name=" + name + ", age=" + age + ", department=" + department + "]";
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
public int getAge() {
return age;
}
public void setAge(int age) {
this.age = age;
}
public String getDepartment() {
return department;
}
public void setDepartment(String department) {
this.department = department;
}
}


Now we will create an action or main method to use the above POJO class for sorting.

SortEmployeeWithStream.java

package com.techbyteslearn.lab.basic;
import java.util.ArrayList;
import java.util.Comparator;
import java.util.stream.Collectors;
public class SortEmployeeWithStream {
public static void main(String[] args) {
Employee1 e1 = new Employee1("Ander Koli", 32, "Sales");
Employee1 e2 = new Employee1("Andrew Smith", 23, "Sales");
Employee1 e3 = new Employee1("David Jone", 52, "Sales");
Employee1 e4 = new Employee1("Cuba Station", 23, "Marketing");
Employee1 e5 = new Employee1("Bradley Head", 23, "Marketing");
Employee1 e6 = new Employee1("Peter Parker", 34, "Sales");
// Create an arraylist and add all the employee object into that list.
ArrayList<Employee1> employeeList = new ArrayList<Employee1>();
employeeList.add(e1);
employeeList.add(e2);
employeeList.add(e3);
employeeList.add(e4);
employeeList.add(e5);
employeeList.add(e6);
System.out.println("Employee list before sorting -\n" + employeeList);
ArrayList<Employee1> sortedList = (ArrayList) employeeList.stream()
.sorted(Comparator.comparing(Employee1::getName).thenComparing(Employee1::getAge))
.collect(Collectors.toList());
System.out.println("Employee list after sorting -\n" + sortedList);
}
}


Output-

Employee list before sorting -
[
 Employee1 [name=Ander Koli, age=32, department=Sales],
 Employee1 [name=Andrew Smith, age=23, department=Sales],
 Employee1 [name=David Jone, age=52, department=Sales],
 Employee1 [name=Cuba Station, age=23, department=Marketing],
 Employee1 [name=Bradley Head, age=23, department=Marketing],
 Employee1 [name=Peter Parker, age=34, department=Sales]]
Employee list after sorting -
[
 Employee1 [name=Ander Koli, age=32, department=Sales],
 Employee1 [name=Andrew Smith, age=23, department=Sales],
 Employee1 [name=Bradley Head, age=23, department=Marketing],
 Employee1 [name=Cuba Station, age=23, department=Marketing],
 Employee1 [name=David Jone, age=52, department=Sales],
 Employee1 [name=Peter Parker, age=34, department=Sales]]

The Comparator.comparing  and thenComparing  two static methods inside Comparator. As we know Comparator is a Functional interface which provides default and static methods along with implementation.  These functional interfaces are giving out of box functionality. See how before java 8 with Comparator interface examples.

Also we can achieve the above sorting by creating multiple different comparators , which we can use when we need. 

We have created two comparator as compareByAge, CompareByDept . Now we can use those comparators at any place we need along with stream. Below sample code snippet shows the usages. Read these reference documents for more about  stream, comparator, functions, jdk 8 features.


Comparator<Employee1> compareByAge = Comparator.comparing(Employee1::getAge);
Comparator<Employee1> compareByDept = Comparator.comparing(Employee1::getDepartment);
ArrayList<Employee1> sortedList = (ArrayList) employeeList.stream()
.sorted(compareByAge.thenComparing(compareByDept)).collect(Collectors.toList());


Java Comparator Example for custom sorting by employee age and department

This program illustrates , how to sort custom object using Comparator<T>. We have used Employee list to sort by age and department. As per the default ordering it will follow the natural ordering.

The below Employee class is our POJO , we will use this class to sort the employee list by age and department.

Employee.java

package com.javadevelopersguide.lab.basic;
/**
 * @author manoj.bardhan
 *
 */
public class Employee {
private String name;
private int age;
private String department;
public Employee(String name, int age, String department) {
super();
this.name = name;
this.age = age;
this.department = department;
}
@Override
public String toString() {
return "\n Employee [name=" + name + ", age=" + age + ", department=" + department + "]";
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
public int getAge() {
return age;
}
public void setAge(int age) {
this.age = age;
}
public String getDepartment() {
return department;
}
public void setDepartment(String department) {
this.department = department;
}
}

Now, we need to create a comparator to compare two same objects. We can create as many comparators by implementing each field/attributes of the POJO. Its purely business requirement, how you need your sorting functionality. Here we need the sorting based on the employee  age and employee department.

EmployeeAgeComparator.java

package com.javadevelopersguide.lab.basic;
import java.util.Comparator;
/**
 * EmployeeAgeComparator is a comparator by Age.
 *
 * @author manoj.bardhan
 *
 */
public class EmployeeAgeComparator implements Comparator<Employee> {
@Override
public int compare(Employee emp1, Employee emp2) {
return emp1.getAge() - emp2.getAge();
}
}


EmployeeDeptComparator.java

package com.javadevelopersguide.lab.basic;
import java.util.Comparator;
/**
 * EmployeeDeptComparator is comparator by department.
 *
 * @author manoj.bardhan
 *
 */
public class EmployeeDeptComparator implements Comparator<Employee> {
@Override
public int compare(Employee emp1, Employee emp2) {
// Internally for comparing String we need to use compareTo()
return emp1.getDepartment().compareTo(emp2.getDepartment());
}
}

In the above we created two comparator for age and department. Now, we need to action on our comparators. We will call our newly created comparator from main() and see the result. Below EmployeeComparatorExample class show the comparator in action.


EmployeeComparatorExample.java

package com.javadevelopersguide.lab.basic;
import java.util.ArrayList;
import java.util.Collections;
/**
 * This program illustrates the simple use of Comparator<t> interface.
 *
 * @author manoj.bardhan
 *
 */
public class EmployeeComparatorExample {
public static void main(String[] args) {
Employee e1 = new Employee("Matt Kuban", 32, "IT");
Employee e2 = new Employee("Andrew Smith", 42, "HR");
Employee e3 = new Employee("Butler Jason", 52, "HR");
Employee e4 = new Employee("Miss Linda", 35, "HR");
Employee e5 = new Employee("Bradley Head", 23, "IT");
Employee e6 = new Employee("Peter Parker", 34, "ADMIN");
// Create an arraylist and add all the employee object into that list.
ArrayList<Employee> employeeList = new ArrayList<Employee>();
employeeList.add(e1);
employeeList.add(e2);
employeeList.add(e3);
employeeList.add(e4);
employeeList.add(e5);
employeeList.add(e6);
System.out.println("Employee Before Sort ::" + employeeList);
// Using EmployeeAgeComparator - to sort the employee by Age
EmployeeAgeComparator ageComparator = new EmployeeAgeComparator();
Collections.sort(employeeList, ageComparator);
// Using EmployeeDeptComparator - to sort the employee by Department
EmployeeDeptComparator deptComparator = new EmployeeDeptComparator();
Collections.sort(employeeList, deptComparator);
System.out.println("Employee After Sort ::" + employeeList);
}
}

Output :- 

Employee Before Sort ::[
 Employee [name=Matt Kuban, age=32, department=IT],
 Employee [name=Andrew Smith, age=42, department=HR],
 Employee [name=Butler Jason, age=52, department=HR],
 Employee [name=Miss Linda, age=35, department=HR],
 Employee [name=Bradley Head, age=23, department=IT],
 Employee [name=Peter Parker, age=34, department=ADMIN]]

Employee After Sort ::[
 Employee [name=Peter Parker, age=34, department=ADMIN],
 Employee [name=Miss Linda, age=35, department=HR],
 Employee [name=Andrew Smith, age=42, department=HR],
 Employee [name=Butler Jason, age=52, department=HR],
 Employee [name=Bradley Head, age=23, department=IT],
 Employee [name=Matt Kuban, age=32, department=IT]]

Happy Learning.



Comparable vs Comparator in Java: Understanding the Comparable Interface


In this article we will see what is Comparable<T> interface ? How to use this interface with some sample examples ?

The Comparator interface is present inside java.util package. Its comparison function, which imposes a total ordering on some collection of objects. Its mostly used while sorting a collection of objects.  Means, it compares its two arguments for order.  Returns a negative integer,zero, or a positive integer as the first argument is less than, equal to, or greater than the second one.

The Comparators can be passed to a sort method (i.e. Collections.sort or Arrays.sort) to allow precise control over the sorting order. Comparators can also be used to control the order of certain data structures (such as sorted sets or sorted maps), or to provide an ordering for collections of objects that don't have a natural ordering.

This interface has one important method (Before JDK 8)  -

int compare(T o1, T o2);



The below Employee class is our POJO , we will use this class to sort the employee list by age and department.

Employee.java

package com.techbyteslearn.lab.basic;

public class Employee {
private String name;
private int age;
private String department;
public Employee(String name, int age, String department) {
super();
this.name = name;
this.age = age;
this.department = department;
}
@Override
public String toString() {
return "\n Employee [name=" + name + ", age=" + age + ", department=" + department + "]";
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
public int getAge() {
return age;
}
public void setAge(int age) {
this.age = age;
}
public String getDepartment() {
return department;
}
public void setDepartment(String department) {
this.department = department;
}
}

Now, we need to create a comparator to compare two same objects. We can create as many comparators by implementing each field/attributes of the pojo. Its purely business requirement, how you need your sorting functionality. Here we need the sorting based on the employee  age and employee department.

EmployeeAgeComparator.java

package com.techbyteslearn.lab.basic;
import java.util.Comparator;

public class EmployeeAgeComparator implements Comparator<Employee> {
@Override
public int compare(Employee emp1, Employee emp2) {
return emp1.getAge() - emp2.getAge();
}
}


EmployeeDeptComparator.java

package com.techbyteslearn.lab.basic;
import java.util.Comparator;

public class EmployeeDeptComparator implements Comparator<Employee> {
@Override
public int compare(Employee emp1, Employee emp2) {
// Internally for comparing String we need to use compareTo()
return emp1.getDepartment().compareTo(emp2.getDepartment());
}
}

In the above we created two comparator for age and department. Now, we need to action on our comparators. We will call our newly created comparator from main() and see the result. Below EmployeeComparatorExample class show the comparator in action.


EmployeeComparatorExample.java

package com.techbyteslearn.lab.basic;
import java.util.ArrayList;
import java.util.Collections;

public class EmployeeComparatorExample {
public static void main(String[] args) {
Employee e1 = new Employee("Matt Kuban", 32, "IT");
Employee e2 = new Employee("Andrew Smith", 42, "HR");
Employee e3 = new Employee("Butler Jason", 52, "HR");
Employee e4 = new Employee("Miss Linda", 35, "HR");
Employee e5 = new Employee("Bradley Head", 23, "IT");
Employee e6 = new Employee("Peter Parker", 34, "ADMIN");
// Create an arraylist and add all the employee object into that list.
ArrayList<Employee> employeeList = new ArrayList<Employee>();
employeeList.add(e1);
employeeList.add(e2);
employeeList.add(e3);
employeeList.add(e4);
employeeList.add(e5);
employeeList.add(e6);
System.out.println("Employee Before Sort ::" + employeeList);
// Using EmployeeAgeComparator - to sort the employee by Age
EmployeeAgeComparator ageComparator = new EmployeeAgeComparator();
Collections.sort(employeeList, ageComparator);
// Using EmployeeDeptComparator - to sort the employee by Department
EmployeeDeptComparator deptComparator = new EmployeeDeptComparator();
Collections.sort(employeeList, deptComparator);
System.out.println("Employee After Sort ::" + employeeList);
}
}

Output :- 

Employee Before Sort ::[
 Employee [name=Matt Kuban, age=32, department=IT],
 Employee [name=Andrew Smith, age=42, department=HR],
 Employee [name=Butler Jason, age=52, department=HR],
 Employee [name=Miss Linda, age=35, department=HR],
 Employee [name=Bradley Head, age=23, department=IT],
 Employee [name=Peter Parker, age=34, department=ADMIN]]

Employee After Sort ::[
 Employee [name=Peter Parker, age=34, department=ADMIN],
 Employee [name=Miss Linda, age=35, department=HR],
 Employee [name=Andrew Smith, age=42, department=HR],
 Employee [name=Butler Jason, age=52, department=HR],
 Employee [name=Bradley Head, age=23, department=IT],
 Employee [name=Matt Kuban, age=32, department=IT]]


JAVA 8 

In JDK 8,  Comparator<T> is functional Interface. Its annotated with @FunctionalInterface and this comparator interface has added few more default & static methods. As its a functional interface therefore its will be used as the assignment target for a lambda expression or method reference.

Below are few examples :-

default Comparator<T> reversed()
default Comparator<T> thenComparing(Comparator<? super T> other)
default <U> Comparator<T> thenComparing(
            Function<? super T, ? extends U> keyExtractor,
            Comparator<? super U> keyComparator)
default <U extends Comparable<? super U>> Comparator<T> thenComparing(
            Function<? super T, ? extends U> keyExtractor)
default Comparator<T> thenComparingInt(ToIntFunction<? super T> keyExtractor)



Here is the full list of methods added into Comparator<T> . We can use stream api of java 8 with lambda expression to implements comparator interface. We will see in the next subsequent posts on lambda expression and stream api.



Happy Learning.

Tuesday, July 2, 2019

Find Nearby Restaurants Using Google Places API

In the previous post we had seen how to get the current location (latitude and longitude) using java script.  Today we will use that functionality to get the nearby search using Google API.  We usually  do the search the near by restaurants , ATM, airport, book store , medicine store ,etc.  By the way we need this everyday in our daily life. 

We sometimes search the specific place or location by using the zipcode or pincode. Google API provide many features for Places, Routes and Map. Today we will find the nearest "resturants" by using a zip code or postcode. 

We have used google.maps.places and google.maps.geocoder in our example above. We have used geocoder , it is the process of converting addresses (like a street address) into geographic coordinates (like latitude and longitude), which you can use to place markers on a map, or position the map.


zipcodefind.html

<html>
<head>
        <style>
            html,
            body,
            #map-canvas {
                height: 100%;
                margin: 0px;
                padding: 0px
            }
        </style>
    <script
        src="https://maps.googleapis.com/maps/api/js?libraries=places&key=YOUR_API_KEY"></script>
   
   
   
    <script language="javascript">
        var map;
        var infowindow;
        function initialize() {
            var geocoder = new google.maps.Geocoder();
            var zipcode = document.getElementById('zipcode').value;
            geocoder.geocode({
                'address': zipcode, componentRestrictions: { country: 'IN' }
            }, function (results, status) {
                if (status === 'OK') {
                    var latLong = new google.maps.LatLng(results[0].geometry.location.lat(), results[0].geometry.location.lng());
                    console.log("Co-ordinates are::" + latLong);
                    map = new google.maps.Map(document.getElementById('map-canvas'), {
                        center: latLong,
                        zoom: 15
                    });
                    var request = {
                        location: latLong,
                        radius: 1000,
                        types: ['car_repair']
                    };
                    infowindow = new google.maps.InfoWindow();
                    var service = new google.maps.places.PlacesService(map);
                    service.nearbySearch(request, callback);
                } else {
                    alert('Search was not successful for the following reason: ' + status);
                }
            });
        }
        function callback(results, status) {
            if (status == google.maps.places.PlacesServiceStatus.OK) {
                for (var i = 0; i < results.length; i++) {
                    createMarker(results[i]);
                }
            }
        }
        function createMarker(place) {
            var placeLoc = place.geometry.location;
            var marker = new google.maps.Marker({
                map: map,
                position: place.geometry.location
            });
            google.maps.event.addListener(marker, 'click', function () {
                infowindow.setContent(place.name);
                infowindow.open(map, this);
            });
        }       
    </script>
</head>
<body>
    <div id="map-canvas" style="width: 50%; float:right"></div>
    <div style="width: 50%; float:left;padding-top: 20px;">
        <input id="zipcode" type="textbox" value="560078">
        <input id="submit" type="button" value="Get Nearby Search by Postcode" onclick="initialize()">
        <br>
        <span style="font-size: small">In this example we have used country <b>India</b>, and search type is <b>Resturants</b> </span>
    </div>
</body>
</html>




Now we will open this html file on browser as below screenshot. We have used marker for marking the near by place "resturants" for the given postcode. There are many supported place type google provides.
















In this example we have restricted the search inside India. But there are other country codes i.e. AU which is also supported. 

componentRestrictions: { country: 'IN' }


Find few reference documents below:-



Monday, July 1, 2019

Producer Consumer Example Using BlockingQueue in Java

Threading is a very tricky and interesting concept in java programming language. There are many problems we face in technology out of which producer-consumer is one. Today we will write a java program for showing producer consumer problem and its solution by using BlockingQueue implementation. 

In this program we will use ArrayBlockingQueue

FoodProducer.java

package com.techbyteslearn.lab.concurrent; import java.util.concurrent.BlockingQueue; public class FoodProducer implements Runnable { private BlockingQueue<String> producerQueue = null; public FoodProducer(BlockingQueue<String> queue) { producerQueue = queue; } @Override public void run() { try { producerQueue.put("Drinks"); Thread.sleep(2000); producerQueue.put("Chocolates"); Thread.sleep(2000); producerQueue.put("Fruits"); Thread.sleep(2000); } catch (InterruptedException e) { Thread.currentThread().interrupt(); e.printStackTrace(); } } }

 

FoodConsumer.java

 

package com.techbyteslearn.lab.concurrent;

import java.util.concurrent.BlockingQueue;

public class FoodConsumer implements Runnable {

    private BlockingQueue<String> consumerQueue = null;

    public FoodConsumer(BlockingQueue<String> consumerQueue) {
        this.consumerQueue = consumerQueue;
    }

    @Override
    public void run() {
        try {
            System.out.println(consumerQueue.take());
            System.out.println(consumerQueue.take());
            System.out.println(consumerQueue.take());

            Thread.sleep(2000);

        } catch (InterruptedException e) {
            Thread.currentThread().interrupt();
            e.printStackTrace();
        }
    }
}

 

MainFoodProcess.java

 

package com.techbyteslearn.lab.concurrent; import java.util.concurrent.ArrayBlockingQueue; import java.util.concurrent.BlockingQueue; public class MainFoodProcess { public static void main(String[] args) throws InterruptedException { final BlockingQueue<String> queue = new ArrayBlockingQueue<>(2); FoodProducer producer = new FoodProducer(queue); FoodConsumer consumer = new FoodConsumer(queue); new Thread(producer).start(); new Thread(consumer).start(); Thread.sleep(3000); } }

Output:

Drinks 
Chocolates 
Fruits
The important point in this example is that the ArrayBlockingQueue has a capacity of 2, while the producer adds three items. The put() method blocks when the queue is full until the consumer takes an item from the queue.

The output here is that, every time the producer insert element into the Queue the consumer will take that element out of the queue. 

Here we have used the below 2 important methods take() and put(). There are few many method provided by the BlockingQueue implementation. Find more methods on BlockingQueue.

take() - Retrieves and removes the head of this queue, waiting if necessary until an element becomes available.
put() - Inserts the specified element into this queue, waiting if necessary for space to become available.


Happy Learning.