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 :



title-img




                        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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