Question

Consider the following stack of characters implemented as an array of 4 elements:

Stack = ["A", "J","P","N"]

Display the Stack as the following operations take place:(a) Stack.pop()

(b) Stack.append("K")

(c) Stack.append("S")

(d) Stack.pop()

(e) Stack.append("G")

(f) Stack.pop()​

Answer 1

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['A', 'J', 'P']

['A', 'J', 'P', 'K']

['A', 'J', 'P', 'K', 'S']

['A', 'J', 'P', 'K']

['A', 'J', 'P', 'K', 'G']

['A', 'J', 'P', 'K']

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Answer 2

Answer:

Operations that would take place are as follows:

Explanation:

• (a) Stack.pop(): [‘A’, ‘J’, ‘P’] the last element would be deleted since stack works on the technology of "Last-In-First-Out", and the element is deleted from the top of the stack.
• (b) Stack.append("K"): [‘A’, ‘J’, ‘P’, ‘K’] again using the same logic of LIFO, element 'K' would be added on top of the stack.
• (c) Stack.append("S"): [‘A’, ‘J’, ‘P’, ‘K’, ‘S’]
• (d) Stack.pop(): [‘A’, ‘J’, ‘P’, ‘K’]
• (e) Stack.append("G"): [‘A’, ‘J’, ‘P’, ‘K’, ‘G’]
• (f) Stack.pop()​: [‘A’, ‘J’, ‘P’, ‘K’]

Hence, in this way, these operations would be performed on the stack of the characters.

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