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C++ Program For Inserting A Node After The N-th Node From The End

Last Updated : 23 Jul, 2025
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Insert a node x after the nth node from the end in the given singly linked list. It is guaranteed that the list contains the nth node from the end. Also 1 <= n.

Examples: 

Input : list: 1->3->4->5
        n = 4, x = 2
Output : 1->2->3->4->5
4th node from the end is 1 and
insertion has been done after this node.

Input : list: 10->8->3->12->5->18
        n = 2, x = 11
Output : 10->8->3->12->5->11->18

Method 1 (Using length of the list):
Find the length of the linked list, i.e, the number of nodes in the list. Let it be len. Now traverse the list from the 1st node upto the (len-n+1)th node from the beginning and insert the new node after this node. This method requires two traversals of the list. 

C++ 
// C++ implementation to insert a node after
// the n-th node from the end
#include <bits/stdc++.h>
using namespace std;

// structure of a node
struct Node {
    int data;
    Node* next;
};

// function to get a new node
Node* getNode(int data)
{
    // allocate memory for the node
    Node* newNode = (Node*)malloc(sizeof(Node));

    // put in the data
    newNode->data = data;
    newNode->next = NULL;
    return newNode;
}

// function to insert a node after the
// nth node from the end
void insertAfterNthNode(Node* head, int n, int x)
{
    // if list is empty
    if (head == NULL)
        return;

    // get a new node for the value 'x'
    Node* newNode = getNode(x);
    Node* ptr = head;
    int len = 0, i;

    // find length of the list, i.e, the
    // number of nodes in the list
    while (ptr != NULL) {
        len++;
        ptr = ptr->next;
    }

    // traverse up to the nth node from the end
    ptr = head;
    for (i = 1; i <= (len - n); i++)
        ptr = ptr->next;

    // insert the 'newNode' by making the
    // necessary adjustment in the links
    newNode->next = ptr->next;
    ptr->next = newNode;
}

// function to print the list
void printList(Node* head)
{
    while (head != NULL) {
        cout << head->data << " ";
        head = head->next;
    }
}

// Driver program to test above
int main()
{
    // Creating list 1->3->4->5
    Node* head = getNode(1);
    head->next = getNode(3);
    head->next->next = getNode(4);
    head->next->next->next = getNode(5);

    int n = 4, x = 2;

    cout << "Original Linked List: ";
    printList(head);

    insertAfterNthNode(head, n, x);

    cout << "
Linked List After Insertion: ";
    printList(head);

    return 0;
}

Output:

Original Linked List: 1 3 4 5
Linked List After Insertion: 1 2 3 4 5


Time Complexity: O(n), where n is the number of nodes in the list.

Auxiliary Space: O(1)

Method 2 (Single traversal):
This method uses two pointers, one is slow_ptr and the other is fast_ptr. First move the fast_ptr up to the nth node from the beginning. Make the slow_ptr point to the 1st node of the list. Now, simultaneously move both the pointers until fast_ptr points to the last node. At this point the slow_ptr will be pointing to the nth node from the end. Insert the new node after this node. This method requires single traversal of the list.

C++ 
// C++ implementation to insert a node after the
// nth node from the end
#include <bits/stdc++.h>

using namespace std;

// structure of a node
struct Node {
    int data;
    Node* next;
};

// function to get a new node
Node* getNode(int data)
{
    // allocate memory for the node
    Node* newNode = (Node*)malloc(sizeof(Node));

    // put in the data
    newNode->data = data;
    newNode->next = NULL;
    return newNode;
}

// function to insert a node after the
// nth node from the end
void insertAfterNthNode(Node* head, int n, int x)
{
    // if list is empty
    if (head == NULL)
        return;

    // get a new node for the value 'x'
    Node* newNode = getNode(x);

    // Initializing the slow and fast pointers
    Node* slow_ptr = head;
    Node* fast_ptr = head;

    // move 'fast_ptr' to point to the nth node
    // from the beginning
    for (int i = 1; i <= n - 1; i++)
        fast_ptr = fast_ptr->next;

    // iterate until 'fast_ptr' points to the
    // last node
    while (fast_ptr->next != NULL) {

        // move both the pointers to the
        // respective next nodes
        slow_ptr = slow_ptr->next;
        fast_ptr = fast_ptr->next;
    }

    // insert the 'newNode' by making the
    // necessary adjustment in the links
    newNode->next = slow_ptr->next;
    slow_ptr->next = newNode;
}

// function to print the list
void printList(Node* head)
{
    while (head != NULL) {
        cout << head->data << " ";
        head = head->next;
    }
}

// Driver program to test above
int main()
{
    // Creating list 1->3->4->5
    Node* head = getNode(1);
    head->next = getNode(3);
    head->next->next = getNode(4);
    head->next->next->next = getNode(5);

    int n = 4, x = 2;

    cout << "Original Linked List: ";
    printList(head);

    insertAfterNthNode(head, n, x);

    cout << "
Linked List After Insertion: ";
    printList(head);

    return 0;
}

Output: 

Original Linked List: 1 3 4 5
Linked List After Insertion: 1 2 3 4 5


Time Complexity: O(n), where n is the number of nodes in the list.

Auxiliary space: O(1) because using constant variables

Please refer complete article on Insert a node after the n-th node from the end for more details!


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