threading Archives - Huahua's Tech Road

花花酱 LeetCode 1116. Print Zero Even Odd

zxi — Wed, 24 Jul 2019 10:45:27 +0000

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:

Thread A will call zero() which should only output 0’s.
Thread B will call even() which should only ouput even numbers.
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"

Solution: Mutex

C++

// Author: Huahua, 40 ms / 9 MB
class ZeroEvenOdd {
private:
  int x;
  int n;
  mutex m_zero_;
  mutex m_odd_;
  mutex m_even_;
public:
    ZeroEvenOdd(int n) {
      this->n = n;
      this->x = 1;
      m_odd_.lock();
      m_even_.lock();
    }

    // printNumber(x) outputs "x", where x is an integer.
    void zero(function printNumber) {
      for (int i = 0; i < n; ++i) {
        m_zero_.lock();
        printNumber(0);
        if (i % 2 == 0) {
          m_odd_.unlock();
        } else {
          m_even_.unlock();
        }
      }
    }

    void even(function printNumber) {
      for (int i = 2; i <= n; i += 2) {
        m_even_.lock();
        printNumber(i);
        m_zero_.unlock();
      }
    }

    void odd(function printNumber) {
      for (int i = 1; i <= n; i += 2) {
        m_odd_.lock();
        printNumber(i);
        m_zero_.unlock();      
      }
    }
};

The post 花花酱 LeetCode 1116. Print Zero Even Odd appeared first on Huahua's Tech Road.

花花酱 LeetCode 1115. Print FooBar Alternately

zxi — Tue, 23 Jul 2019 16:22:32 +0000

Suppose you are given the following code:

class FooBar {
  public void foo() {
    for (int i = 0; i < n; i++) {
      print("foo");
    }
  }

  public void bar() {
    for (int i = 0; i < n; i++) {
      print("bar");
    }
  }
}

The same instance of FooBar will be passed to two different threads. Thread A will call foo() while thread B will call bar(). Modify the given program to output “foobar” n times.

Example 1:

Input: n = 1
Output: "foobar"
Explanation: There are two threads being fired asynchronously. One of them calls foo(), while the other calls bar(). "foobar" is being output 1 time.

Example 2:

Input: n = 2
Output: "foobarfoobar"
Explanation: "foobar" is being output 2 times.

Solution: Mutex

C++

class FooBar {
private:
  int n;
  std::mutex m1_;
  std::mutex m2_; 
public:
  FooBar(int n) {
    this->n = n;    
    m2_.lock();
  }

  void foo(function printFoo) {

      for (int i = 0; i < n; i++) {
        m1_.lock();
        // printFoo() outputs "foo". Do not change or remove this line.
        printFoo();
        m2_.unlock();
      }
  }

  void bar(function printBar) {

      for (int i = 0; i < n; i++) {
        m2_.lock();
        // printBar() outputs "bar". Do not change or remove this line.
        printBar();
        m1_.unlock();
      }
  }
};

The post 花花酱 LeetCode 1115. Print FooBar Alternately appeared first on Huahua's Tech Road.

花花酱 LeetCode 1114. Print in Order

zxi — Tue, 23 Jul 2019 15:58:56 +0000

Suppose we have a class:

public class Foo {
  public void first() { print("first"); }
  public void second() { print("second"); }
  public void third() { print("third"); }
}

The same instance of Foo will be passed to three different threads. Thread A will call first(), thread B will call second(), and thread C will call third(). Design a mechanism and modify the program to ensure that second() is executed after first(), and third() is executed after second().

Example 1:

Input: [1,2,3]
Output: "firstsecondthird"
Explanation: There are three threads being fired asynchronously. The input [1,2,3] means thread A calls first(), thread B calls second(), and thread C calls third(). "firstsecondthird" is the correct output.

Example 2:

Input: [1,3,2]
Output: "firstsecondthird"
Explanation: The input [1,3,2] means thread A calls first(), thread B calls third(), and thread C calls second(). "firstsecondthird" is the correct output.

Note:

We do not know how the threads will be scheduled in the operating system, even though the numbers in the input seems to imply the ordering. The input format you see is mainly to ensure our tests’ comprehensiveness.

Solution 1: Mutex

C++

// Author: Huahua, 128 ms / 8.2 MB
class Foo {
public:
    Foo() {
      first_.lock();
      second_.lock();
    }

    void first(function printFirst) {        
      // printFirst() outputs "first". Do not change or remove this line.
      printFirst();  
      first_.unlock();
    }

    void second(function printSecond) {            
      first_.lock();
      // printSecond() outputs "second". Do not change or remove this line.
      printSecond();
      first_.unlock();
      second_.unlock();
    }

    void third(function printThird) {
      second_.lock();
      // printThird() outputs "third". Do not change or remove this line.
      printThird();      
      second_.unlock();
    }
private:
  std::mutex first_;
  std::mutex second_;
};

Solution 2: Busy waiting

C++

// Author: Huahua, 176 ms / 9.2 MB
#include 

class Foo {
public:
    Foo() {}

    void first(function printFirst) {        
      // printFirst() outputs "first". Do not change or remove this line.
      printFirst();  
      state_ = 1;
    }

    void second(function printSecond) {            
      while (state_ != 1) usleep(1);
      // printSecond() outputs "second". Do not change or remove this line.
      printSecond();
      state_ = 2;
    }

    void third(function printThird) {
      while (state_ != 2) usleep(1);
      // printThird() outputs "third". Do not change or remove this line.
      printThird();      
      state_ = 3;
    }
private:
  int state_ = 0;
};

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