GATE | GATE-CS-2003 | Question 64
Let S be a stack of size n ≥ 1. Starting with the empty stack, suppose we push the first n natural numbers in sequence, and then perform n pop operations. Assume that Push and pop operation take X seconds each, and Y seconds elapse between the end of one such stack operation and the start of the next operation. For m ≥ 1, define the stack-life of m as the time elapsed from the end of Push(m) to the start of the pop operation that removes m from S. The average stack-life of an element of this stack is
(A) n (X + Y)
(B) 3Y + 2X
(C) n (X + Y) – X
(D) Y + 2X
Explanation: Background required – Stack and Basic Maths
Let Tn be time span of nth element of stack. Let us first find out the sum of Tn for n = 1 to n
Stack Lifetime of last element, Tn = Y (Since it is popped as soon as it is pushed on the stack) Stack Lifetime of last element, Tn-1 = Tn + 2X + 2Y (The time needed to push and then pop nth element plus two pauses Y each). = 2X + 3Y Stack Lifetime of last element, Tn-2 = Tn-1 + 2X + 2Y (Using the Same reasoning above) = 4X + 5Y . . . Stack Lifetime of 1st element = 2(n-1)X + (2n-1)Y (Generalizing the pattern) Sum of all the time spans of all the elements = (Σ 2(n-1)X) + (Σ (2n-1)Y) for n = 1 to n = 2X(1 + 2 + . . . + n-1) + Y(1 + 3 + 5 + . . . + (2n-1))
Using 2 identities
- Sum of n natural numbers = (n*(n+1))/2 for the first summation
- Sn = (n/2)(a+l) Sum of AP series with a as first term and l being last for second summation
Above sum is,
= (2X(n-1)n)/2 + Y(n/2)*(1 + 2n-1)
Therefore Average = Sum/n = n(X+Y)-X . Hence Option (c)
This explanation has been contributed by Pranjul Ahuja.
Quiz of this Question
Attention reader! Don’t stop learning now. Practice GATE exam well before the actual exam with the subject-wise and overall quizzes available in GATE Test Series Course.
Learn all GATE CS concepts with Free Live Classes on our youtube channel.