Array rotate and delete

Question

Based on the problem in GeeksForGeeks here. I came across a solution here.

Can someone please help me understand the solution. Primarily I need help in the following block:

if(n==1) cout<<arr[0]<<endl;
else if(n%2) {
    ll ind = n-3;
    ind = floor(ind/4);
    ind = 3+ind;
    cout<<arr[ind-1]<<endl;
} else {
    ll ind = n-2;
    ind = floor(ind/4);
    ind = 2+ind;
    cout<<arr[ind-1]<<endl;
}

Answer 1

For each size of array, a particular position is the answer (i.e .independent of array elements).

For any array of size 8, 2 nd position (i.e. 3rd element) gives the answer.

Lets look at some examples:

• size=1, position=0

---

• size=2, position=1
• size=3, position=2
• size=4, position=1
• size=5, position=2

---

• size=6, position=2
• size=7, position=3
• size=8, position=2
• size=9, position=3

---

• size=10, position=3
• size=11, position=4
• size=12, position=3
• size=13, position=4

---

• size=14, position=4
• size=15, position=5
• size=16, position=4
• size=17, position=5

---

• size=18, position=5
• size=19, position=6
• size=20, position=5
• size=21, position=6

  ---

And so on.

For even size : floor( (n-3)/4 )+2 gives the position.

For odd size : floor( (n-2)/4 )+1 gives the position.

Answer 2

For even size : floor( (n-3)/4 )+2 gives the position.

For odd size : floor( (n-2)/4 )+1 gives the position.

CORRECT:

For Odd size : floor( (n-3)/4 )+2 gives the position.

For Even size : floor( (n-2)/4 )+1 gives the position.