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? - PrepBharat

Register Now

Lost Password

Lost your password? Please enter your email address. You will receive a link and will create a new password via email.

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: d
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.

Join The Discussion