Aimee 👾 - Fire 🔥 #4
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CheezItMan
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Apr 7, 2021
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| Time Complexity: O(n^2) - the stack.pop() method invoked has O(n) time complexity and it is used inside a for loop. | ||
| Space Complexity: O(n) - we keep pushing values to the stack for a string of length n. | ||
| */ | ||
| const balanced = (str) => { |
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👍 I'm pretty sure the pop method has O(1) time complexity leading your solution to O(n) time complexity overall.
| } | ||
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| enqueue(element) { | ||
| throw new Error("This method has not been implemented!"); | ||
| enqueue(value) { |
| throw new Error("Queue is Full!") | ||
| } | ||
| this.store[len] = value; | ||
| this.tail = (this.tail + 1) % Queue.maxLenth; | ||
| } | ||
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| dequeue() { |
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You should also check to see if the head and tail are equal, if so the Queue is empty and you should do something to avoid removing null entries.
| } | ||
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| push() { | ||
| throw new Error("This method has not been implemented!"); | ||
| push(value) { |
| push() { | ||
| throw new Error("This method has not been implemented!"); | ||
| push(value) { | ||
| this.store.addLast(value); | ||
| } | ||
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| pop() { |
| throw new Error("This method has not been implemented!"); | ||
| const lastValue = this.store.getLast() | ||
| this.store.delete(lastValue); | ||
| return lastValue; | ||
| } | ||
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| isEmpty() { |
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