A stack is implemented using an array and the top is always pointing to the first position in the array. Assuming the stack is not empty, the time complexities of push() and pop() operations are:Group of answer choicesO(n) and O(1)O(1) and O(1)O(1) and O(n)O(n) and O(n)
Question
A stack is implemented using an array and the top is always pointing to the first position in the array. Assuming the stack is not empty, the time complexities of push() and pop() operations are:Group of answer choicesO(n) and O(1)O(1) and O(1)O(1) and O(n)O(n) and O(n)
Solution
The time complexities of push() and pop() operations in this case are O(n) and O(1) respectively.
Here's why:
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Push Operation: Since the top is always pointing to the first position in the array, every time we push an element into the stack, we need to shift all existing elements to the right to make room for the new element at the first position. This operation takes linear time, hence the time complexity is O(n), where n is the number of elements in the stack.
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Pop Operation: To pop an element from the stack, we simply remove the element at the first position of the array, which is a constant time operation, hence the time complexity is O(1).
Similar Questions
Select the correct answerA stack is implemented with an array of 'A [0..N - 1]' and a varlable 'pos'. The push and pop operations are defined by the following code.push(x) A[pos] ← X pos ← pos - 1end pushpop( ) pos ← pos + 1 return A[pos]end popWhich of the following will initialize an empty stack with capacity N for the above Implementation?Optionspos ← −1pos ← 1pos ← N−1pos
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Which of the following operations doesn’t take constant time O(1) in Stack ? Answer choicesSelect only one optionREVISITRemoving the elementSearching the elementInserting the elementFinding the topmost element
The time complexity of both push and pop operations on a stack implemented with an array is O(1).a.TRUEb.FALSE
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