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Allisa [31]
2 years ago
14

In a linked chain implementation of a queue, the performance of the enqueue operation

Computers and Technology
1 answer:
alexdok [17]2 years ago
4 0

Answer:

A.O(1)

Explanation:

In the implementation of queue by using linked chain the performance of  the enqueue operation is O(1).We have to  maintain  two pointers one  head and the other tailand  for  enqueue operation  we have to insert element  to the next of the tail and then  make that element  tail.Which takes O(1) time.

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The function below takes two arguments: a dictionary of strings (keys) to integers (values) a_dict and a list of strings key_lis
vitfil [10]

Answer:

  1. def getData(a_dict, key_list):
  2.    result = []
  3.    for key in key_list:
  4.        result.append(a_dict[key])
  5.    
  6.    return result  
  7. result = getData( {"puffin": 5, "corgi": 2, "three": 3} , ["three", "corgi"])
  8. print(result)

Explanation:

Let's define a function <em>getData() </em>with two parameters,<em> a_dict </em>and <em>key_list</em> as required by the question (Line 1).

Since the function is to return a list of associated values of dictionaries, a new list,<em> result</em>,  is declared and initialized with empty values (Line 2).

Next, use for-loop to traverse through every string in the input <em>key_list </em>(Line 4) and use the traversed key to address the value in the<em> a_dict </em>and add it to the <em>result</em> list (Line 5)

At last, return the <em>result </em>list as output (Line 7)

We can test the function using the test case from the question and we shall see the output as follows:

[3, 2]

8 0
2 years ago
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