# Delete middle element from stack

Posted: 24 Dec, 2020
Difficulty: Easy

## PROBLEM STATEMENT

#### 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 = 5
OUTPUT: ARR [ ] = [ 1 , 2 , 4,  5 ]

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

INPUT : ARR [ ] = [ 5, 6, 7, 8 ] , N = 4
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 / 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, denoting the number of elements in the stack.

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

#### Output Format:

``````For every test case, print N - 1 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 <= 3000
0 <= data <= 10^9

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

Time limit: 1 second
`````` Approach 1

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.

1. We have an input stack as “INPUTSTACK”, ‘N’ denotes the number of elements in the stack.
2. We define a function "DELETEMIDDLE" that accepts, a stack of integers "STACK" and an integer “COUNT” (initially 0) as input parameters.
3. Now we define the base condition. If the stack is empty or, if all the elements are traversed (COUNT == N), we return.
4. 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.
5. Call the recursive function “DELETEMIDDLE”  by incrementing the value of "COUNT" by 1.
6. Add "TOP" back to “STACK”, if COUNT != N / 2.
7. 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