1116. Print Zero Even Odd

Description

You have a function printNumber that can be called with an integer parameter and prints it to the console.

  • For example, calling printNumber(7) prints 7 to the console.

You are given an instance of the class ZeroEvenOdd that has three functions: zero, even, and odd. The same instance of ZeroEvenOdd will be passed to three different threads:

  • Thread A: calls zero() that should only output 0's.
  • Thread B: calls even() that should only output even numbers.
  • Thread C: calls odd() that should only output odd numbers.

Modify the given class to output the series "010203040506..." where the length of the series must be 2n.

Implement the ZeroEvenOdd class:

  • ZeroEvenOdd(int n) Initializes the object with the number n that represents the numbers that should be printed.
  • void zero(printNumber) Calls printNumber to output one zero.
  • void even(printNumber) Calls printNumber to output one even number.
  • void odd(printNumber) Calls printNumber to output one odd number.

 

Example 1:

Input: n = 2
Output: "0102"
Explanation: There are three threads being fired asynchronously.
One of them calls zero(), the other calls even(), and the last one calls odd().
"0102" is the correct output.

Example 2:

Input: n = 5
Output: "0102030405"

 

Constraints:

  • 1 <= n <= 1000

Solutions

Solution 1: Multithreading + Semaphore

We use three semaphores $z$, $e$, and $o$ to control the execution order of the three threads, where $z$ is initially set to $1$, and $e$ and $o$ are set to $0$.

  • Semaphore $z$ controls the execution of the zero function. When the value of semaphore $z$ is $1$, the zero function can be executed. After execution, the value of semaphore $z$ is set to $0$, and the value of semaphore $e$ or $o$ is set to $1$, depending on whether the even function or the odd function needs to be executed next.
  • Semaphore $e$ controls the execution of the even function. When the value of semaphore $e$ is $1$, the even function can be executed. After execution, the value of semaphore $z$ is set to $1$, and the value of semaphore $e$ is set to $0$.
  • Semaphore $o$ controls the execution of the odd function. When the value of semaphore $o$ is $1$, the odd function can be executed. After execution, the value of semaphore $z$ is set to $1$, and the value of semaphore $o$ is set to $0$.

The time complexity is $O(n)$, and the space complexity is $O(1)$.

Python Code
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from threading import Semaphore


class ZeroEvenOdd:
    def __init__(self, n):
        self.n = n
        self.z = Semaphore(1)
        self.e = Semaphore(0)
        self.o = Semaphore(0)

    # printNumber(x) outputs "x", where x is an integer.
    def zero(self, printNumber: 'Callable[[int], None]') -> None:
        for i in range(self.n):
            self.z.acquire()
            printNumber(0)
            if i % 2 == 0:
                self.o.release()
            else:
                self.e.release()

    def even(self, printNumber: 'Callable[[int], None]') -> None:
        for i in range(2, self.n + 1, 2):
            self.e.acquire()
            printNumber(i)
            self.z.release()

    def odd(self, printNumber: 'Callable[[int], None]') -> None:
        for i in range(1, self.n + 1, 2):
            self.o.acquire()
            printNumber(i)
            self.z.release()

Java Code
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class ZeroEvenOdd {
    private int n;
    private Semaphore z = new Semaphore(1);
    private Semaphore e = new Semaphore(0);
    private Semaphore o = new Semaphore(0);

    public ZeroEvenOdd(int n) {
        this.n = n;
    }

    // printNumber.accept(x) outputs "x", where x is an integer.
    public void zero(IntConsumer printNumber) throws InterruptedException {
        for (int i = 0; i < n; ++i) {
            z.acquire(1);
            printNumber.accept(0);
            if (i % 2 == 0) {
                o.release(1);
            } else {
                e.release(1);
            }
        }
    }

    public void even(IntConsumer printNumber) throws InterruptedException {
        for (int i = 2; i <= n; i += 2) {
            e.acquire(1);
            printNumber.accept(i);
            z.release(1);
        }
    }

    public void odd(IntConsumer printNumber) throws InterruptedException {
        for (int i = 1; i <= n; i += 2) {
            o.acquire(1);
            printNumber.accept(i);
            z.release(1);
        }
    }
}

C++ Code
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#include <semaphore.h>

class ZeroEvenOdd {
private:
    int n;
    sem_t z, e, o;

public:
    ZeroEvenOdd(int n) {
        this->n = n;
        sem_init(&z, 0, 1);
        sem_init(&e, 0, 0);
        sem_init(&o, 0, 0);
    }

    // printNumber(x) outputs "x", where x is an integer.
    void zero(function<void(int)> printNumber) {
        for (int i = 0; i < n; ++i) {
            sem_wait(&z);
            printNumber(0);
            if (i % 2 == 0) {
                sem_post(&o);
            } else {
                sem_post(&e);
            }
        }
    }

    void even(function<void(int)> printNumber) {
        for (int i = 2; i <= n; i += 2) {
            sem_wait(&e);
            printNumber(i);
            sem_post(&z);
        }
    }

    void odd(function<void(int)> printNumber) {
        for (int i = 1; i <= n; i += 2) {
            sem_wait(&o);
            printNumber(i);
            sem_post(&z);
        }
    }
};