Question

Consider yourself to be in a planet where the computational power of chips to be slow. You have an array of size 10.You want to perform enqueue some element into this array. But you can perform only push and pop operations .Push and pop operation both take 1 sec respectively. The total time required to perform enQueue operation is?

(a) 20

(b) 40

(c) 42

(d) 43

Answer with explanation needed.

Answer 1

Correct answer is (d) 43

Easy explanation - First you have to empty all the elements of the current stack into the temporary stack, push the required element and empty the elements of the temporary stack into the original stack. Therfore taking 10+10+1+11+11= 43 seconds.