Leetcode: Print Zero Even Odd

Print Zero Even Odd

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• CheatSheet: Leetcode For Code Interview
• CheatSheet: Common Code Problems & Follow-ups
• Tag: #concurrency

Suppose you are given the following code:

class ZeroEvenOdd {
  public ZeroEvenOdd(int n) { ... }      // constructor
  public void zero(printNumber) { ... }  // only output 0's
  public void even(printNumber) { ... }  // only output even numbers
  public void odd(printNumber) { ... }   // only output odd numbers
}

The same instance of ZeroEvenOdd will be passed to three different threads:

1. Thread A will call zero() which should only output 0’s.
2. Thread B will call even() which should only ouput even numbers.
3. Thread C will call odd() which should only output odd numbers.

Each of the thread is given a printNumber method to output an integer. Modify the given program to output the series 010203040506… where the length of the series must be 2n.

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"

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

• Solution:

## https://code.dennyzhang.com/print-zero-even-odd
## Basic Ideas: Semaphore
## Complexity: Time O(1), Space O(1)

from threading import Semaphore

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

        # printNumber(x) outputs "x", where x is an integer.
    def zero(self, printNumber: 'Callable[[int], None]') -> None:
        for x in range(0, n):
            self.z.acquire()
            print(0)
            if x%2 == 1:
                self.o.release()
            else:
                self.e.release()

    def even(self, printNumber: 'Callable[[int], None]') -> None:
        for x in range(1, n+1):
            if x%2 == 0:
                self.e.acquire()
                print(x)
                self.z.release()

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