Answer:
Yes it is possible for the following cases:-
- When the queue is full.
- When the queue is empty.
Explanation:
When the queue is full the the front and the rear references in the circular array implementation are equal because after inserting an element in the queue we increase the rear pointer.So when inserting the last element the rear pointer will be increased and it will become equal to front pointer.
When the queue is empty the front and rear pointer are equal.We remove an element from queue by deleting the element at front pointer decreasing the front pointer when there is only one element and we are deleting that element front and rear pointer will become equal after deleting that element.
Hi there! Hopefully this helps!
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1. Barbara Liskov.
2. Carl Sassenrath.
3. Larry Page.
This seems to be a ROT2 Ceasar cipher. Replace each letter with a letter two places futher in the alphabet. Do the same with DOG and get <span>FQI.</span>
<h3>OPPOSITE WORDS OF REASSEMBLING</h3>
Answer:
phrase[2:5]
Explanation:
Given:
The above code segment
Required
Which instruction returns 'u a'
First, we need to get the index of u in the phrase:
u is at the third position but in programming, index starts at 0.
So, u is at index 2
Next, we need to get the index of a in the phrase:
a is at index 4
One of the ways to return a sub string from a string in python is ![string[start:stop+1]](https://tex.z-dn.net/?f=string%5Bstart%3Astop%2B1%5D)
Where:
----- index of u
----- index of a
phrase ---- The string variable
So, the instruction that returns 'u a' is: ![phrase[2:5]](https://tex.z-dn.net/?f=phrase%5B2%3A5%5D)
<em>Where 5 = 4 + 1</em>
