## algorithm to find nth last element of a single linked list

We can do this in two ways

• Recursive Approach
• Iterative Approach

Problem Statement: Lets say the Single Linked List (SLL) is 0-1-2-3-4-5-6-7-8-9, and we want to find last 3rd element (say ‘pos=3’) in this SLL

Recursive Approach:
Take a instance/global variable to track the post recursive call and when it is same as ‘pos’, print the value.
Order of complexity: O(n) for pre recursive calls and O(n) for post recursive calls, which is = O(2n)

```	public void getElementRecursive(Node node, int pos) {
if(node != null) {
getElementRecursive(node.next, pos);
ctr++;
if(pos==ctr) {
System.out.println("Middle element of the list is: " + node.value);
}
}
}
```

Iterative Approach:
Take two pointers ‘n1’ and ‘n2’, as first step – move pointer n1 to ‘n-1’ positions and as second step – move n1 and n2 simultaneously till n1 is not equal to null.
Order of complexity: O(pos) for first step and O(n-pos) for second step, which is = O(n)

```	public void getElementIterative(Node node, int pos) {

Node n1 = node;
Node n2 = node;

for(int i=0; i<pos-1; i++) {
n1 = n1.next;
}

while(n1 != null && n1.next != null) {
n1 = n1.next;
n2 = n2.next;
}
System.out.println("Middle element of the list is: " + n2.value);
}
```

Note: Out of these two approaches, recursive first comes to mind, but looking at the complexities iterative approach is better to implement this algorithm.

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