nth node from end

Author: sachuverma

DSA Crack Sheet/README.md

@@ -157,6 +157,7 @@
 - [Multiply two linked lists](https://practice.geeksforgeeks.org/problems/multiply-two-linked-lists/1# "view question") - [Cpp Solution](./solutions/Multiply%20two%20linked%20lists.cpp)
 - [Delete nodes having greater value on right](https://practice.geeksforgeeks.org/problems/delete-nodes-having-greater-value-on-right/1# "view question") - [Cpp Solution](./solutions/Delete%20nodes%20having%20greater%20value%20on%20right.cpp)
 - [Segregate even and odd nodes in a Link List](https://practice.geeksforgeeks.org/problems/segregate-even-and-odd-nodes-in-a-linked-list5035/1 "view question") - [Cpp Solution](./solutions/Segregate%20even%20and%20odd%20nodes%20in%20a%20Link%20List.cpp)
+- [Nth node from end of linked list](https://practice.geeksforgeeks.org/problems/nth-node-from-end-of-linked-list/1# "view question") - [Cpp Solution](./solutions/Nth%20node%20from%20end%20of%20linked%20list.cpp)
 - []( "view question") - [Cpp Solution](./solutions/.cpp)
 
 ### Binary Tree

DSA Crack Sheet/solutions/Nth node from end of linked list.cpp

@@ -0,0 +1,74 @@
+/*
+Nth node from end of linked list
+================================
+
+Given a linked list consisting of L nodes and given a number N. The task is to find the Nth node from the end of the linked list.
+
+Example 1:
+Input:
+N = 2
+LinkedList: 1->2->3->4->5->6->7->8->9
+Output: 8
+Explanation: In the first example, there
+are 9 nodes in linked list and we need
+to find 2nd node from end. 2nd node
+from end os 8.  
+
+Example 2:
+Input:
+N = 5
+LinkedList: 10->5->100->5
+Output: -1
+Explanation: In the second example, there
+are 4 nodes in the linked list and we
+need to find 5th from the end. Since 'n'
+is more than the number of nodes in the
+linked list, the output is -1.
+Your Task:
+The task is to complete the function getNthFromLast() which takes two arguments: reference to head and N and you need to return Nth from the end or -1 in case node doesn't exist..
+
+Note:
+Try to solve in single traversal.
+
+Expected Time Complexity: O(N).
+Expected Auxiliary Space: O(1).
+
+Constraints:
+1 <= L <= 103
+1 <= N <= 103
+*/
+
+int getNthFromLast(Node *head, int n)
+{
+  Node *main_ptr = head;
+  Node *ref_ptr = head;
+
+  int count = 0;
+  if (head != NULL)
+  {
+    while (count < n)
+    {
+      if (ref_ptr == NULL)
+        return -1;
+      ref_ptr = ref_ptr->next;
+      count++;
+    }
+
+    if (ref_ptr == NULL)
+    {
+      head = head->next;
+      if (head != NULL)
+        return (main_ptr->data);
+    }
+    else
+    {
+      while (ref_ptr != NULL)
+      {
+        main_ptr = main_ptr->next;
+        ref_ptr = ref_ptr->next;
+      }
+      return (main_ptr->data);
+    }
+  }
+  return -1;
+}