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 $O(1)$.

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);
        }
    }
}

#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);
        }
    }
};

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: # 'i%2==1' will have wrong answer "0201"
                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()

1116 - Print Zero Even Odd

1116. Print Zero Even Odd

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1116. Print Zero Even Odd

Description

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All Problems

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