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CheezItMan
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Overall nice work, some small bits could use some work. Take a look at my comments and let me know if you have questions.
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| string_stack = Stack.new() | ||
| length = 0 | ||
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| string.each_char do |char| | ||
| string_stack.push(Node.new(char)) | ||
| length += 1 | ||
| end | ||
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| return false if length % 2 != 0 | ||
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| length /= 2 | ||
| comparison_string_stack = Stack.new() | ||
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| length.times do | ||
| comparison_string_stack.push(string_stack.pop) | ||
| end | ||
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| key = {"[" => "]", "]" => "[", "{" => "}", "}" => "{", "(" => ")", ")" => "("} | ||
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| until string_stack.empty? | ||
| letter1 = string_stack.pop | ||
| letter2 = comparison_string_stack.pop | ||
| return false if letter1.data != key[letter2.data] | ||
| end | ||
| return true |
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This method wouldn't work for this balanced string
"{()}[()]"
Consider just using one stack, and traverse the string. When you see a start brace of some sort, you push it on the stack, when you see a close brace you pop and see if they match.
By the end the stack should be empty.
| dequeued = @store[@front] | ||
| @store[@front] = nil | ||
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| self.size == 0 ? @front = @rear = -1 : @front = (@front + 1) % QUEUE_SIZE |
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I would say if self.size == 1 (because you'd have 1 element).
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| def size | ||
| raise NotImplementedError, "Not yet implemented" | ||
| store = @store.select { |item| item != nil } |
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You could also do this in O(1) with some math.
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Stacks and Queues
Thanks for doing some brain yoga. You are now submitting this assignment!
Comprehension Questions
OPTIONAL JobSimulation