1115 Print FooBar Alternately

Author: idf

1115 Print FooBar Alternately.py

@@ -0,0 +1,60 @@
+#!/usr/bin/python3
+"""
+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.
+"""
+from threading import Lock
+from typing import Callable
+
+
+class FooBar:
+    def __init__(self, n):
+        self.n = n
+        self.locks = [Lock(), Lock()]
+        self.locks[1].acquire()
+
+
+    def foo(self, printFoo: Callable[[], None]) -> None:
+        for i in range(self.n):
+            self.locks[0].acquire()
+            # printFoo() outputs "foo". Do not change or remove this line.
+            printFoo()
+            self.locks[1].release()
+
+
+    def bar(self, printBar: Callable[[], None]) -> None:
+        for i in range(self.n):
+            self.locks[1].acquire()
+            # printBar() outputs "bar". Do not change or remove this line.
+            printBar()
+            self.locks[0].release()