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Find the Second Largest Element in a Linked List

Last Updated : 27 Jan, 2023
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Given a Linked list of integer data. The task is to write a program that efficiently finds the second largest element present in the Linked List.

Examples

Input : List = 12 -> 35 -> 1 -> 10 -> 34 -> 1
Output : The second largest element is 34.

Input : List = 10 -> 5 -> 10
Output : The second largest element is 5.

A Simple Solution will be to first sort the linked list in descending order and then print the second element from the sorted linked list. The time complexity of this solution is O(nlogn).

A Better Solution is to traverse the Linked list twice. In the first traversal find the maximum element. In the second traversal find the greatest element less than the element obtained in first traversal. The time complexity of this solution is O(n).

A more Efficient Solution can be to find the second largest element in a single traversal. 

Below is the complete algorithm for doing this: 

1) Initialize two variables first and second to INT_MIN as,
   first = second = INT_MIN
2) Start traversing the Linked List,
   a) If the current element in Linked List say list[i] is greater
      than first. Then update first and second as,
      second = first
      first = list[i]
   b) If the current element is in between first and second,
      then update second to store the value of current variable as
      second = list[i]
3) Return the value stored in second node.

Below is the implementation of the above approach: 

C++
// C++ program to print second largest
// value in a linked list
#include <climits>
#include <iostream>

using namespace std;

// A linked list node
struct Node {
    int data;
    struct Node* next;
};

// Function to add a node at the
// beginning of Linked List
void push(struct Node** head_ref, int new_data)
{
    /* allocate node */
    struct Node* new_node = (struct Node*)malloc(sizeof(struct Node));

    /* put in the data */
    new_node->data = new_data;

    /* link the old list of the new node */
    new_node->next = (*head_ref);

    /* move the head to point to the new node */
    (*head_ref) = new_node;
}

// Function to count size of list
int listSize(struct Node* node)
{
    int count = 0;

    while (node != NULL) {
        count++;

        node = node->next;
    }

    return count;
}

// Function to print the second
// largest element
void print2largest(struct Node* head)
{
    int i, first, second;

    int list_size = listSize(head);

    /* There should be atleast two elements */
    if (list_size < 2) {
        cout << "Invalid Input";
        return;
    }

    first = second = INT_MIN;

    struct Node* temp = head;

    while (temp != NULL) {
        if (temp->data > first) {
            second = first;
            first = temp->data;
        }

        // If current node's data is in between
        // first and second then update second
        else if (temp->data > second && temp->data != first)
            second = temp->data;

        temp = temp->next;
    }

    if (second == INT_MIN)
        cout << "There is no second largest element\n";
    else
        cout << "The second largest element is " << second;
}

// Driver program to test above function
int main()
{
    struct Node* start = NULL;

    /* The constructed linked list is: 
     12 -> 35 -> 1 -> 10 -> 34 -> 1 */
    push(&start, 1);
    push(&start, 34);
    push(&start, 10);
    push(&start, 1);
    push(&start, 35);
    push(&start, 12);

    print2largest(start);

    return 0;
}
Java
// Java program to print second largest 
// value in a linked list 
class GFG
{
    
// A linked list node 
static class Node 
{ 
    int data; 
    Node next; 
}; 

// Function to add a node at the 
// beginning of Linked List 
static Node push( Node head_ref, int new_data) 
{ 
    // allocate node /
    Node new_node = new Node(); 

    // put in the data /
    new_node.data = new_data; 

    // link the old list of the new node /
    new_node.next = (head_ref); 

    // move the head to point to the new node /
    (head_ref) = new_node; 
    return head_ref;
} 

// Function to count size of list 
static int listSize( Node node) 
{ 
    int count = 0; 

    while (node != null)
    { 
        count++; 

        node = node.next; 
    } 

    return count; 
} 

// Function to print the second 
// largest element 
static void print2largest( Node head) 
{ 
    int i, first, second; 

    int list_size = listSize(head); 

    // There should be atleast two elements /
    if (list_size < 2) 
    { 
        System.out.print("Invalid Input"); 
        return; 
    } 

    first = second = Integer.MIN_VALUE; 

    Node temp = head; 

    while (temp != null)
    { 
        if (temp.data > first) 
        { 
            second = first; 
            first = temp.data; 
        } 

        // If current node's data is in between 
        // first and second then update second 
        else if (temp.data > second && temp.data != first) 
            second = temp.data; 

        temp = temp.next; 
    } 

    if (second == Integer.MIN_VALUE) 
        System.out.print( "There is no second largest element\n"); 
    else
        System.out.print ("The second largest element is " + second); 
} 

// Driver program to test above function 
public static void main(String args[])
{ 
    Node start = null; 

    // The constructed linked list is: 
    //12 . 35 . 1 . 10 . 34 . 1 
    start=push(start, 1); 
    start=push(start, 34); 
    start=push(start, 10); 
    start=push(start, 1); 
    start=push(start, 35); 
    start=push(start, 12); 

    print2largest(start); 
}
} 

// This code is contributed by Arnab Kundu
Python3
# Python3 program to print second largest 
# value in a linked list 

# A linked list node 
class Node :
    def __init__(self):
        self.data = 0
        self.next = None

# Function to add a node at the 
# beginning of Linked List 
def push( head_ref, new_data) :

    # allocate node /
    new_node = Node() 

    # put in the data /
    new_node.data = new_data 

    # link the old list of the new node /
    new_node.next = (head_ref) 

    # move the head to point to the new node /
    (head_ref) = new_node 
    return head_ref

# Function to count size of list 
def listSize( node): 

    count = 0
    while (node != None):
        count = count + 1

        node = node.next
    
    return count 

# Function to print the second 
# largest element 
def print2largest( head): 

    i = 0
    first = 0
    second = 0

    list_size = listSize(head) 

    # There should be atleast two elements /
    if (list_size < 2) :
    
        print("Invalid Input") 
        return

    first = second = -323767
    temp = head 

    while (temp != None):
    
        if (temp.data > first) :
            second = first 
            first = temp.data 
        
        # If current node's data is in between 
        # first and second then update second 
        elif (temp.data > second and temp.data != first) :
            second = temp.data 

        temp = temp.next

    if (second == -323767) :
        print( "There is no second largest element\n") 
    else:
        print ("The second largest element is " , second) 

# Driver code

start = None

# The constructed linked list is: 
# 12 . 35 . 1 . 10 . 34 . 1 
start = push(start, 1) 
start = push(start, 34) 
start = push(start, 10) 
start = push(start, 1) 
start = push(start, 35) 
start = push(start, 12) 

print2largest(start) 

# This code is contributed by Arnab Kundu
C#
// C# program to print second largest 
// value in a linked list 
using System;

class GFG
{
    
// A linked list node 
public class Node 
{ 
    public int data; 
    public Node next; 
}; 

// Function to add a node at the 
// beginning of Linked List 
static Node push( Node head_ref, int new_data) 
{ 
    // allocate node 
    Node new_node = new Node(); 

    // put in the data 
    new_node.data = new_data; 

    // link the old list of the new node 
    new_node.next = (head_ref); 

    // move the head to point to the new node 
    (head_ref) = new_node; 
    return head_ref;
} 

// Function to count size of list 
static int listSize( Node node) 
{ 
    int count = 0; 

    while (node != null)
    { 
        count++; 

        node = node.next; 
    } 

    return count; 
} 

// Function to print the second 
// largest element 
static void print2largest(Node head) 
{ 
    int first, second; 

    int list_size = listSize(head); 

    // There should be atleast two elements
    if (list_size < 2) 
    { 
        Console.Write("Invalid Input"); 
        return; 
    } 

    first = second = int.MinValue; 

    Node temp = head; 

    while (temp != null)
    { 
        if (temp.data > first) 
        { 
            second = first; 
            first = temp.data; 
        } 

        // If current node's data is in between 
        // first and second then update second 
        else if (temp.data > second && 
                 temp.data != first) 
            second = temp.data; 

        temp = temp.next; 
    } 

    if (second == int.MinValue) 
        Console.Write( "There is no second" +
                       " largest element\n"); 
    else
        Console.Write("The second largest " + 
                      "element is " + second); 
} 

// Driver Code
public static void Main(String []args)
{ 
    Node start = null; 

    // The constructed linked list is: 
    //12 . 35 . 1 . 10 . 34 . 1 
    start = push(start, 1); 
    start = push(start, 34); 
    start = push(start, 10); 
    start = push(start, 1); 
    start = push(start, 35); 
    start = push(start, 12); 

    print2largest(start); 
}
} 

// This code is contributed by 29AjayKumar
JavaScript
<script>

      // JavaScript program to print second largest
      // value in a linked list
      // A linked list node
      class Node {
        constructor() {
          this.data = 0;
          this.next = null;
        }
      }

      // Function to add a node at the
      // beginning of Linked List
      function push(head_ref, new_data) {
        // allocate node
        var new_node = new Node();

        // put in the data
        new_node.data = new_data;

        // link the old list of the new node
        new_node.next = head_ref;

        // move the head to point to the new node
        head_ref = new_node;
        return head_ref;
      }

      // Function to count size of list
      function listSize(node) {
        var count = 0;

        while (node != null) {
          count++;

          node = node.next;
        }

        return count;
      }

      // Function to print the second
      // largest element
      function print2largest(head) {
        var first, second;

        var list_size = listSize(head);

        // There should be atleast two elements
        if (list_size < 2) {
          document.write("Invalid Input");
          return;
        }

        first = second = -2147483648;

        var temp = head;

        while (temp != null) {
          if (temp.data > first) {
            second = first;
            first = temp.data;
          }

          // If current node's data is in between
          // first and second then update second
          else if (temp.data > second && temp.data != first) 
          second = temp.data;

          temp = temp.next;
        }

        if (second == -2147483648)
        document.write("There is no second" + " largest element<br>");
        else 
        document.write("The second largest " + "element is " + second);
      }

      // Driver Code
      var start = null;

      // The constructed linked list is:
      //12 . 35 . 1 . 10 . 34 . 1
      start = push(start, 1);
      start = push(start, 34);
      start = push(start, 10);
      start = push(start, 1);
      start = push(start, 35);
      start = push(start, 12);

      print2largest(start);
      
</script>

Output
The second largest element is 34

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