[02 Feb 2020] How to print even and odd numbers using threads in java



How to print even and odd numbers using threads in java

You are given two threads. You need to print odd numbers using one thread and even numbers using another thread.You need to print in natural order up to MAX.
For example:
If MAX is 10, you need to print:

1 2 3 4 5 6 7 8 9 10
So 1 3 5 7 9 will be printed by odd thread
2 4 6 8 10 will be printed by even thread.

Solution 1
We will use wait and notify to solve how to print even and odd numbers using threads in java.

Use a variable called boolean odd. If you want to print odd number, it’s value should be true and vice versa for even number.
Create two methods(printOdd and printEven), one will print odd numbers and other will print even numbers.
Create two threads, one for odd and one for even.
t1 will call  printEven method and t2 will call printOdd method simultaneously.
If boolean odd is true in printEven method, t1 will wait.
If boolean odd is false in printOdd method, t2 will wait.
Print even and odd numbers using threads in java


public class OddEvenPrintMain {
boolean odd;
int count = 1;
int MAX = 20;
public void printOdd() {
synchronized (this) {
while (count < MAX) {
System.out.println("Checking odd loop");

while (!odd) {
try {
System.out.println("Odd waiting : " + count);
wait();
System.out.println("Notified odd :" + count);
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
System.out.println("Odd Thread :" + count);
count++;
odd = false;
notify();
}
}
}

public void printEven() {

try {
Thread.sleep(1000);
} catch (InterruptedException e1) {
e1.printStackTrace();
}
synchronized (this) {
while (count < MAX) {
System.out.println("Checking even loop");

while (odd) {
try {
System.out.println("Even waiting: " + count);
wait();
System.out.println("Notified even:" + count);
} catch (InterruptedException e) {
e.printStackTrace();
}
}
System.out.println("Even thread :" + count);
count++;
odd = true;
notify();

}
}
}

public static void main(String[] args) {

OddEvenPrintMain oep = new OddEvenPrintMain();
oep.odd = true;
Thread t1 = new Thread(new Runnable() {

@Override
public void run() {
oep.printEven();

}
});
Thread t2 = new Thread(new Runnable() {

@Override
public void run() {
oep.printOdd();

}
});

t1.start();
t2.start();

try {
t1.join();
t2.join();
} catch (InterruptedException e) {
e.printStackTrace();
}

}
}

When you run above program, you will get below output:

Checking odd loop
Odd Thread :1
Checking odd loop
Odd waiting : 2
Checking even loop
Even thread :2
Checking even loop
Even waiting: 3
Notified odd :3
Odd Thread :3
Checking odd loop
Odd waiting : 4
Notified even:4
Even thread :4
Checking even loop
Even waiting: 5
Notified odd :5
Odd Thread :5
Checking odd loop
Odd waiting : 6
Notified even:6
Even thread :6
Checking even loop
Even waiting: 7
Notified odd :7
Odd Thread :7
Checking odd loop
Odd waiting : 8
Notified even:8
Even thread :8
Checking even loop
Even waiting: 9
Notified odd :9
Odd Thread :9
Checking odd loop
Odd waiting : 10
Notified even:10
Even thread :10
Checking even loop
Even waiting: 11
Notified odd :11
Odd Thread :11
Checking odd loop
Odd waiting : 12
Notified even:12
Even thread :12
Checking even loop
Even waiting: 13
Notified odd :13
Odd Thread :13
Checking odd loop
Odd waiting : 14
Notified even:14
Even thread :14
Checking even loop
Even waiting: 15
Notified odd :15
Odd Thread :15
Checking odd loop
Odd waiting : 16
Notified even:16
Even thread :16
Checking even loop
Even waiting: 17
Notified odd :17
Odd Thread :17
Checking odd loop
Odd waiting : 18
Notified even:18
Even thread :18
Checking even loop
Even waiting: 19
Notified odd :19
Odd Thread :19
Notified even:20
Even thread :20


Solution 2: Using remainder
You can use concept of remainder here.

If number%2==1 then Odd will print the number and increment it else will go in the wait state.
If number%2==0 then Even will print the number and increment it else will go in the wait state.
Let’s check with the help of example.
Create a class named “OddEvenRunnable” and implement Runnable interface.
Create a class named "OddEvenRunnable". It will implement Runnable interface.

public class OddEvenRunnable implements Runnable{

public int PRINT_NUMBERS_UPTO=10;
static int  number=1;
int remainder;
static Object lock=new Object();

OddEvenRunnable(int remainder)
{
this.remainder=remainder;
}

@Override
public void run() {
while (number < PRINT_NUMBERS_UPTO) {
synchronized (lock) {
while (number % 2 != remainder) { // wait for numbers other than remainder
try {
lock.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
System.out.println(Thread.currentThread().getName() + " " + number);
number++;
lock.notifyAll();
}
}
}
}
public class PrintOddEvenMain {
public static void main(String[] args) {

OddEvenRunnable oddRunnable=new OddEvenRunnable(1);
OddEvenRunnable evenRunnable=new OddEvenRunnable(0);

Thread t1=new Thread(oddRunnable,"Odd");
Thread t2=new Thread(evenRunnable,"Even");

t1.start();
t2.start();
}
}

When you run above program, you will get below output

Odd 1
Even 2
Odd 3
Even 4
Odd 5
Even 6
Odd 7
Even 8
Odd 9
Even 10
This is all about printing even and odd numbers using threads in java. Please comment if the explanation is not very clear.

Comments

Post a Comment

Popular posts from this blog

Very Basic Java Programs

[1] Count ways to reach a score using 1 and 2 with no consecutive 2s A cricket player has to score N runs, with condition he can take either 1 or 2 runs only and consecutive runs should not be 2. Find all the possible combination he can take. Examples: Input : N = 4 Output : 4 1+1+1+1, 1+2+1, 2+1+1, 1+1+2 Input : N = 5 Output : 6 This problem is a variation of count number of ways to reach given score in a game and can be solved in O(n) time and O(n) auxiliary space. Below is the recursive solution of above problem. // A simple recursive implementation for // counting ways to reach a score using 1 and 2 with // consecutive 2 allowed   import java.io.*;   class GFG {    static  int CountWays(int n, boolean flag) {     if (n == 0) // base case         return 1;       int sum = 0;       // 2 is not scored last time so we can score either 2 or 1     if (flag == fals...
Read more

[08 Feb 2020] Minimum number of platforms required for a railway station

Image
Problem: You are given arrival and departure time of trains reaching to a particular station. You need to find minimum number of platforms required to accommodate the trains at any point of time. For example: arrival[] = {1:00, 1:40, 1:50, 2:00, 2:15, 4:00}  departure[] = {1:10, 3:00, 2:20, 2:30, 3:15, 6:00} No. of platforms required in above scenario = 4 Please note that arrival time is in chronological order. Solution : If you notice we need to find maximum number of trains that can be on the station with the help of arrival and departure time. Solution 1: You can iterate over all intervals and check how many other intervals are overlapping with it but that will require o(N^2) time complexity.  Solution 2: We will use logic very much similar to  merge sort . Sort both arrival(arr) and departure(dep) arrays. Compare current element in arrival and departure array and pick smaller one among both. If element is pick up from arrival array ...
Read more

Interview Questions asked in Goldman sach interview

Introduction, Project introduction, He noted highlighted points from intro. In CV, you have written that you improved slow loading of application. how did you do that? How will you improve it for a mobile application? How to improve API response time / optimise the APIs ? How to make database queries faster or optimise the queries? How do you choose sharding key ? What are the factors considered for sharding the table into multiple shards ? Difference between Horizontal Sharding vs Horizontal Partitioning? What is horizontal Partitioning ? Difference between partitioning and sharding? How will you design Amazon’s database architecture in terms of sharding and partitioning ? How will you make sure locking while writing in a heavily read extensive environment ? — I explained Master slave, master master slave with multiple shards but He wasn’t satisfied. Asked for which version of java are you comfortable with ? I replied java 8. What is lambda ? How does it improve/help in...
Read more