Question

What is the output of the StackQueueMystery algorithm (below) when we have enqueued 1, 2, 5, 6, 3, 7, 4, 10, 8, 9 on q and pushed 8, 6, 9, 7, 10, 5, 4, 3, 2, 1 on stk (in that order) before calling this algorithm. Show your work.

StackQueueMystery:

Input: stk: stack with integers 1-n

Input: q: queue with integers 1-n

Input: n: size and range of stk and q

Pseudocode:

count = 0

while stk and q are not empty:

{

    x = x0 = stk.pop()

    y = y0 = q.dequeue()

    while x != y0 and stk is not empty

        x = stk.pop()

    while y != x0 and q is not empty

        y = q.dequeue()

    count = count + 1

}

return count

Answer 1

Solution:-

The output of the StackQueueMystery algorithm will be 5 as the value of the count will be 5 at the end.

In this question we are given with a stack and a queue. The sequence of push and enqueue that is given can be shown as

as per the pseudo code, the value of count initially is 0 and changes as the while loop is executed

for 1st iteration

• x = x0 = stk.pop // x = x0 = 1
• y = y0 = q.dequee() // 1
• both inner while loop = false.
• count = count + 1 // 1

for 2nd iteration

• x = x0 = stk.pop // x = x0 = 2
• y = y0 = q.dequee() // 2
• both inner while loop = false.
• count = count + 1 // 2

For the 3rd iteration

• x = x0 = stk.pop // x = x0 = 3
• y = y0 = q.dequee() // 5

while x! = y0 and stk is not empty // true .
it pops elements until its found 5 or stk becomes empty.
5 was found at 5th pos.

while y!=x= and q is not empty // true.
it continues until q becomes empty or its finds 3 .
3 was found at 5th pos.

• count = count + 1 // 3

for the 4th iteration

• x = x0 = stk.pop // x = x0 = 10
• y = y0 = q.dequee() // 3

while x! = y0 and stk is not empty // true .
it pops elements until its found 7 or stk becomes empty.
7 was found at 7th pos.

while y!=x= and q is not empty // true.
it continues until q becomes empty or its finds 10 .
10 was found at 8th pos.

• count = count + 1 // 4

for the 5th iteration

• x = x0 = stk.pop // x = x0 = 8
• y = y0 = q.dequee() // 9

while x! = y0 and stk is not empty // true .
it pops elements until its found 8 or stk becomes empty.
8 was found at 10th pos.

while y!=x= and q is not empty // true.
it continues until q becomes empty or its finds 9 .
9 was found at 10th pos.

• count = count + 1 // 5 ( result)

here both the stk and q got empted and the value of count becomes 5.

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