Python Problem:

is an element and n is the Here's the "smart" implementation of a counting queue that we saw during class. It stores pairs (2,n), where count of the number of occurrences of 2. [1] class CountingQueue (object): def __init__(self): self.queue = [] def _repr__(self): return repr(self.queue) def add(self, x, count=1): # If the element is the same as the last element, we simply # increment the count. This assumes we can test equality of # elements. if len(self.queue) > 0: XX, CC = self.queue[-1] if xx == x: self.queue [-1] = (xx, cc + count) else: self.queue.append((x, count)) else: self.queue = [(x, count)] def get(self): if len(self.queue) == 0: return None X, C = self.queue [0] if c == 1: self.queue.pop() return x else: self.queue[@] = (x, 5 - 1) return x def isempty(self): # Since the count of an element is never 0, we can just check # whether the queue is empty. return len(self.queue) == 0 Problem 1: implement _len__ for a counting queue For this exercise, you will implement the __len__ method for the CountingQueue class defined above This should require no more than 4 lines of code. def countingqueue_len(self): "*"Returns the number of elements in the queue."" # YOUR CODE HERE raise NotImplementedError() # This is a way to add a method to a class once the class # has already been defined. CountingQueue. __len__ = countingqueue_len ] ### Tests for `_len__ from nose.tools import assert_equal q = CountingQueue () for i in range(10): q.add('a') q.add('b') for i in range (3): q.add('c', count=2) assert_equal(len(9), 17)