Is this correct one for this Question?

#1


Is this correct one for this Question?

#2

@cbcao263 hey,yes the logic is looking fine according to the question.
(By making enQueue operation costly) This method makes sure that oldest entered element is always at the top of stack 1, so that deQueue operation just pops from stack1. To put the element at top of stack1, stack2 is used.

enQueue(q, x):

  • While stack1 is not empty, push everything from stack1 to stack2.
  • Push x to stack1 (assuming size of stacks is unlimited).
  • Push everything back to stack1.

Here time complexity will be O(n)

deQueue(q):

  • If stack1 is empty then error
  • Pop an item from stack1 and return it

Here time complexity will be O(1)

#3

Sir, I have printed that one but still it is not giving correct output. Please tell why so and how to do it properly.

#4

@cbcao263 hey ,I have corrected this code ,please check:

#5

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