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CheezItMan
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Diana, nice work. You hit the main learning goals. Well done.
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| # Time Complexity: O(n), will always have to traverse through the string | ||
| # Space Complexity: O(n), because in the worst case all char in string | ||
| # are opening brackets, best case would be O(1) where we push and pop every item | ||
| def balanced(string) |
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👍 , However you can dry up the case statement by using a hash instead. The has would have the open braces as the keys and the closing braces as the values.
return false if brace_hash[current_char] != char| @back = 0 | ||
| end | ||
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| def enqueue(element) |
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| def front | ||
| raise NotImplementedError, "Not yet implemented" | ||
| @front |
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Suggested change
| @front | |
| @store[@front] |
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| def size | ||
| raise NotImplementedError, "Not yet implemented" | ||
| @size | ||
| end |
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Size should be the number of elements in the queue. You can calculate it using the position of front and rear.
| @@ -1,19 +1,20 @@ | |||
| class Stack | |||
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Stacks and Queues
Thanks for doing some brain yoga. You are now submitting this assignment!
Comprehension Questions
-pop: remove element
-is_empty: check if stack is empty
-initialize: create LinkedList base for stack
-to_s: allow for easy to read output of stack
-dequeue: removes element from front of rear
-is_empty?: checks if queue is empty
-initialize: sets up initial queue
-size: gives size of queue
Use: able to utilize without knowing the logic–application of the something
OPTIONAL JobSimulation