# Kth Last Node of a Linked List - Google Top Interview Questions

### Problem Statement :

Given a singly linked list node, return the value of the kth last node (0-indexed).

k is guaranteed not to be larger than the size of the linked list.

This should be done in \mathcal{O}(1)O(1) space.

Constraints

n ≤ 100,000 where n is the length of node

Example 1

Input

node = [1, 2]

k = 1

Output

1

Explanation

The second last node has the val of 1

Example 2

Input

node = [0, 1, 2, 3]

k = 2

Output

1

### Solution :

                        Solution in C++ :

int solve(LLNode* node, int k) {
LLNode* tmp = node;
int i = 0;

while (tmp != NULL) {
if (i > k) {
node = node->next;
}
i++;
tmp = tmp->next;
}
return node->val;
}


                        Solution in Java :

import java.util.*;

/**
* class LLNode {
*   int val;
*   LLNode next;
* }
*/
class Solution {
public int solve(LLNode node, int k) {
int size = 0;
LLNode head = node;
while (head != null) {
size++;
head = head.next;
}
LLNode curr = node;

while (curr != null) {
size--;
if (size == k)
return curr.val;
curr = curr.next;
}
return 0;
}
}


                        Solution in Python :

class Solution:
def solve(self, node, k):
result, end = node, node
count = 0
while end:
if count > k:
result = result.next
end = end.next
count += 1

return result.val if count > k else None


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