Delete middle element from stack
You are having a stack "ARR" of size 'N+1', your task is to delete the middlemost element so that the size of resulting stack is 'N'.

A stack is a linear data structure where both insertion and deletion of elements take place at the top. It follows FILO (First In Last Out) or LIFO (Last In First Out) approaches. Books piled on top of each other is an example of a stack, where you can only remove a single book at a time, which is at the top of the stack. Likewise, you can only add a single book at a time, on the top of the stack only.

Example :-

INPUT : ARR [ ] = [ 1 , 2 , 3 , 4 , 5 ] , N = 4 OUTPUT: ARR [ ] = [ 1 , 2 , 4, 5 ] The above example contains an odd number of elements, hence the middle element is clearly the (N+1) / 2th element, which is removed from the stack in the output. INPUT : ARR [ ] = [ 5, 6, 7, 8 ] , N = 3 OUTPUT: ARR [ ] = [ 5, 7, 8 ] The above example contains an even number of elements, so out of the two middle elements, we consider the one which occurs first. Hence, the middle element would be ((N+1) / 2 - 1) element, which is 6 and is removed from the stack in the output.

Input Format

The first line of input contains an integer 'T' representing the number of the test case. Then the test case follows. The first line of each test case contains an integer 'N', where 'N+1' denotes the number of elements in the stack initially. The second line of each test case contains 'N+1' space-separated integers, denoting the elements of the stack.

Output Format:

For every test case, print 'N' space-separated integer, denoting the elements in the stack after removing the middle element from the input stack. The output of every test case will be printed in a separate line.

Note:

You don’t have to print anything, it has already been taken care of. Just implement the given function.

Constraints:

1

Question

Delete middle element from stack

You are having a stack "ARR" of size 'N+1', your task is to delete the middlemost element so that the size of resulting stack is 'N'.

A stack is a linear data structure where both insertion and deletion of elements take place at the top. It follows FILO (First In Last Out) or LIFO (Last In First Out) approaches. Books piled on top of each other is an example of a stack, where you can only remove a single book at a time, which is at the top of the stack. Likewise, you can only add a single book at a time, on the top of the stack only.

Example :-

INPUT : ARR [ ] = [ 1 , 2 , 3 , 4 , 5 ] , N = 4
OUTPUT: ARR [ ] = [ 1 , 2 , 4,  5 ]

The above example contains an odd number of elements, hence the middle element is clearly the (N+1) / 2th element, which is removed from the stack in the output.

INPUT : ARR [ ] = [ 5, 6, 7, 8 ] , N = 3
OUTPUT: ARR [ ] = [ 5, 7, 8 ]

The above example contains an even number of elements, so out of the two middle elements, we consider the one which occurs first. Hence, the middle element would be ((N+1) / 2 - 1) element, which is 6 and is removed from the stack in the output.

Input Format

The first line of input contains an integer 'T' representing the number of the test case. Then the test case follows.

The first line of each test case contains an integer 'N', where 'N+1' denotes the number of elements in the stack initially.

The second line of each test case contains 'N+1' space-separated integers, denoting the elements of the stack.

Output Format:

For every test case, print 'N' space-separated integer, denoting the elements in the stack after removing the middle element from the input stack. 
The output of every test case will be printed in a separate line.

Note:

You don’t have to print anything, it has already been taken care of. Just implement the given function.

Constraints:

1 <= T <= 100    
1 <= N+1 <= 3000
0 <= data <= 10^9

Where ‘T’ is the number of test cases, ‘N+1’ is the number of elements in the input Stack. ‘data’ is the value of each element in the stack.

Time limit: 1 second

Accepted Answer

Recursive approach

The idea is to use recursive calls. We first remove all items one by one, then we recur. After recursive calls, we push all items back except for the middle item.

We have an input stack as “INPUTSTACK”, ‘N’ denotes the number of elements in the stack.
We define a function "DELETEMIDDLE" that accepts, a stack of integers "STACK" and an integer “COUNT” (initially 0) as input parameters.
Now we define the base condition. If the stack is empty or, if all the elements are traversed (COUNT == N), we return.
Now we create a new integer variable, say “TOP”, and we store the top of the "STACK" in "TOP", simultaneously pop out the top element of the stack.
Call the recursive function “DELETEMIDDLE” by incrementing the value of "COUNT" by 1.
Add "TOP" back to “STACK”, if COUNT != N / 2.
At the end of recursive calls, our stack will contain “N - 1” elements, after removing the middle element of the input stack.

Algorithm:

Initialise “DELETEMIDDLE"( STACK, N, COUNT = 0)
IF "STACK" is empty OR “COUNT” is equal to ‘N’:
- Return
If “TOP” is equal to the top element of "STACK"
Pop the top element of "STACK"
Recursively call “DELETEMIDDLE”( STACK, N, COUNT+1 )
STACK.push(TOP) if COUNT != N / 2

Space Complexity: O(n)Explanation:

O(N), where 'N' is the number of elements in the input stack.

We are storing a total of N number of elements in the recursion stack. Hence the space complexity is O(N).

Time Complexity: O(n)Explanation:

O(N) where N is the number of elements in the input stack.

We are doing recursive calls at most N times and every call is doing a linear amount of work. Hence the worst-case time complexity is O(N).

Accepted Answer

Iterative Approach

We have the given input stack, with the name "INPUTSTACK" with 'N' integers.
Maintain a new temporary stack, say “TEMPSTACK” and a counter variable “COUNT”, and initialize it with 0. "COUNT" variable will help us to find the middle element of the stack. When "COUNT" is equal to “N / 2”, it means we have arrived at the middle element of the stack.
Now while the value of “COUNT” is less than “N / 2”, push the top of "INPUTSTACK" into the “TEMPSTACK”. Simultaneously pop the top element out of the "INPUTSTACK".
Before the end of every iteration, increment the “COUNT” variable by 1.
When the "COUNT" is equal to "N / 2", and we are out of the loop, pop the top element out from the “INPUTSTACK”. This element is the middle element of the stack.
Now we have to add the stored elements back into the “INPUTSTACK”.
While “TEMPSTACK” is not empty, push the top of "TEMPSTACK" into INPUTSTACK. Simultaneously pop out the top of “TEMPSTACK”.
Now our “INPUTSTACK” contains "N - 1" integers, and the middle element is removed.

Algorithm :

Initialise a variable COUNT = 0
Until "COUNT" is less than N / 2:
- push top element of “INPUTSTACK” in “TEMPSTACK”.
- pop() the top element of "INPUTSTACK"
- Increment "COUNT" by 1.
pop() the top element of “INPUTSTACK” to remove the middle element.
Until “TEMPSTACK” is not empty:
- push top element of “TEMPSTACK” into “INPUTSTACK”.
- pop() the top element of "TEMPSTACK".

Space Complexity: O(n)Explanation:

O(N), where ‘N’ is the number of elements in the input stack.

We are storing N / 2 numbers of elements in the temporary stack. Hence our worst-case space complexity is of linear order i.e O(N).