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Find the Largest Number Using Four Nodes in a Singly Linked List

Given a singly linked list, the task is to find the largest number in the linked list using four nodes.

Examples:



Input: List: 2->5->3->4->6
Output: 6543
Explanation: Using the four nodes, we traverse the linked list and construct the largest number, which is 6543

Input: List: 1->7->6->5->4->9->2
Output: 9765
Explanation: Using the four nodes, we traverse the linked list and construct the largest number, which is 9765.



Input: List: 5->3->7->1->9->2
Output: 9753
Explanation: By using four nodes to traverse the linked list, we can construct the largest number, which is 9753.

Approach: To solve the problem follow the below idea:

Approach to find the largest number in a singly linked list using four nodes. It does this by iteratively traversing the linked list while maintaining four maximum values: max1, max2, max3, and max4. As it traverses the list, it compares the current node’s value with these maximum values and updates them accordingly. This approach ensures that it identifies the four largest digits in the list, allowing the code to concatenate them in reverse order, forming the largest number.

Steps of this approach:

Implementation of the above approach:




// C++ code for the above approach:
#include <bits/stdc++.h>
using namespace std;
 
// Define a structure for a singly linked list node
struct Node {
 
    // Data stored in the node
    int data;
 
    // Pointer to the next node
    Node* next;
 
    Node(int val)
    {
 
        // Constructor to initialize the node
        // with a value
        data = val;
 
        // Initialize the next pointer to null
        next = NULL;
    }
};
 
// Function to find the largest number using
// four nodes in the linked list
long long formLargestNumber(Node* head)
{
    // Initialize the maximum values to the smallest
    // possible integer values
    int max1 = INT_MIN, max2 = INT_MIN, max3 = INT_MIN,
        max4 = INT_MIN;
 
    // Traverse the linked list and update the maximum
    // values
    for (Node* curr = head; curr != NULL;
         curr = curr->next) {
        int val = curr->data;
 
        if (val > max1) {
            max4 = max3;
            max3 = max2;
            max2 = max1;
 
            // Update the first maximum if the
            // value is greater
            max1 = val;
        }
        else if (val > max2 && val <= max1) {
            max4 = max3;
            max3 = max2;
 
            // Update the second maximum if the
            // value is greater but not greater
            // than the first
            max2 = val;
        }
        else if (val > max3 && val <= max2) {
            max4 = max3;
 
            // Update the third maximum if the
            // value is greater but not greater
            // than the first two
            max3 = val;
        }
 
        // Update the fourth maximum if the
        // value is greater but not greater
        // than the first three
        else if (val > max4 && val <= max3) {
            max4 = val;
        }
    }
 
    // Concatenate the four maximum values to form the
    // largest number
    long long largestNum
        = max1 * 1000LL + max2 * 100LL + max3 * 10LL + max4;
 
    return largestNum;
}
 
// Drivers code
int main()
{
    // Create a linked list
    Node* head = new Node(1);
    head->next = new Node(7);
    head->next->next = new Node(6);
    head->next->next->next = new Node(5);
    head->next->next->next->next = new Node(4);
    head->next->next->next->next->next = new Node(9);
    head->next->next->next->next->next->next = new Node(2);
 
    // Find the largest number using four nodes in the
    // linked list
    long long largestNum = formLargestNumber(head);
 
    // Print the largest number
    cout << "The largest number formed using four nodes in "
            "the linked list is: "
         << largestNum << endl;
 
    return 0;
}




// Java code for the above approach:
public class LargestNumberFromLinkedList {
 
    // Define a structure for a singly linked list node
    static class Node {
 
        // Data stored in the node
        int data;
 
        // Pointer to the next node
        Node next;
 
        Node(int val) {
            // Constructor to initialize the node
            // with a value
            data = val;
 
            // Initialize the next pointer to null
            next = null;
        }
    }
 
    // Function to find the largest number using
    // four nodes in the linked list
    static long formLargestNumber(Node head) {
        // Initialize the maximum values to the smallest
        // possible integer values
        int max1 = Integer.MIN_VALUE, max2 = Integer.MIN_VALUE, max3 = Integer.MIN_VALUE,
                max4 = Integer.MIN_VALUE;
 
        // Traverse the linked list and update the maximum
        // values
        for (Node curr = head; curr != null; curr = curr.next) {
            int val = curr.data;
 
            if (val > max1) {
                max4 = max3;
                max3 = max2;
                max2 = max1;
 
                // Update the first maximum if the
                // value is greater
                max1 = val;
            } else if (val > max2 && val <= max1) {
                max4 = max3;
                max3 = max2;
 
                // Update the second maximum if the
                // value is greater but not greater
                // than the first
                max2 = val;
            } else if (val > max3 && val <= max2) {
                max4 = max3;
 
                // Update the third maximum if the
                // value is greater but not greater
                // than the first two
                max3 = val;
            }
 
            // Update the fourth maximum if the
            // value is greater but not greater
            // than the first three
            else if (val > max4 && val <= max3) {
                max4 = val;
            }
        }
 
        // Concatenate the four maximum values to form the
        // largest number
        long largestNum = max1 * 1000L + max2 * 100L + max3 * 10L + max4;
 
        return largestNum;
    }
 
    // Drivers code
    public static void main(String[] args) {
        // Create a linked list
        Node head = new Node(1);
        head.next = new Node(7);
        head.next.next = new Node(6);
        head.next.next.next = new Node(5);
        head.next.next.next.next = new Node(4);
        head.next.next.next.next.next = new Node(9);
        head.next.next.next.next.next.next = new Node(2);
 
        // Find the largest number using four nodes in the
        // linked list
        long largestNum = formLargestNumber(head);
 
        // Print the largest number
        System.out.println("The largest number formed using four nodes in "
                + "the linked list is: " + largestNum);
    }
}




# Define a class for a singly linked list node
class Node:
    def __init__(self, val):
        # Constructor to initialize the node with a value
        self.data = val
        # Initialize the next pointer to None
        self.next = None
 
# Function to find the largest number using four nodes in the linked list
def form_largest_number(head):
    # Initialize the maximum values to the smallest possible integer values
    max1 = float('-inf')
    max2 = float('-inf')
    max3 = float('-inf')
    max4 = float('-inf')
 
    # Traverse the linked list and update the maximum values
    curr = head
    while curr is not None:
        val = curr.data
 
        if val > max1:
            max4 = max3
            max3 = max2
            max2 = max1
 
            # Update the first maximum if the value is greater
            max1 = val
        elif val > max2 and val <= max1:
            max4 = max3
            max3 = max2
 
            # Update the second maximum if the value is greater but not greater than the first
            max2 = val
        elif val > max3 and val <= max2:
            max4 = max3
 
            # Update the third maximum if the value is greater but not greater than the first two
            max3 = val
        elif val > max4 and val <= max3:
            # Update the fourth maximum if the value is greater but not greater than the first three
            max4 = val
 
        curr = curr.next
 
    # Concatenate the four maximum values to form the largest number
    largest_num = max1 * 1000 + max2 * 100 + max3 * 10 + max4
 
    return largest_num
 
# Driver code
if __name__ == "__main__":
    # Create a linked list
    head = Node(1)
    head.next = Node(7)
    head.next.next = Node(6)
    head.next.next.next = Node(5)
    head.next.next.next.next = Node(4)
    head.next.next.next.next.next = Node(9)
    head.next.next.next.next.next.next = Node(2)
 
    # Find the largest number using four nodes in the linked list
    largest_num = form_largest_number(head)
 
    # Print the largest number
    print("The largest number formed using four nodes in the linked list is:", largest_num)




using System;
 
// Define a structure for a singly linked list node
class Node {
    // Data stored in the node
    public int data;
 
    // Pointer to the next node
    public Node next;
 
    public Node(int val)
    {
        // Constructor to initialize the node
        // with a value
        data = val;
 
        // Initialize the next pointer to null
        next = null;
    }
}
 
public class GFG {
    // Function to find the largest number using
    // four nodes in the linked list
    static long FormLargestNumber(Node head)
    {
        // Initialize the maximum values to the smallest
        // possible integer values
        int max1 = int.MinValue, max2 = int.MinValue,
            max3 = int.MinValue, max4 = int.MinValue;
 
        // Traverse the linked list and
        // update the maximum values
        for (Node curr = head; curr != null;
             curr = curr.next) {
            int val = curr.data;
 
            if (val > max1) {
                max4 = max3;
                max3 = max2;
                max2 = max1;
 
                // Update the first maximum if the
                // value is greater
                max1 = val;
            }
            else if (val > max2 && val <= max1) {
                max4 = max3;
                max3 = max2;
 
                // Update the second maximum if the
                // value is greater but not greater
                // than the first
                max2 = val;
            }
            else if (val > max3 && val <= max2) {
                max4 = max3;
 
                // Update the third maximum if the
                // value is greater but not greater
                // than the first two
                max3 = val;
            }
            else if (val > max4 && val <= max3) {
                // Update the fourth maximum if the
                // value is greater but not greater
                // than the first three
                max4 = val;
            }
        }
 
        // Concatenate the four maximum values to form the
        // largest number
        long largestNum = max1 * 1000L + max2 * 100L
                          + max3 * 10L + max4;
 
        return largestNum;
    }
 
    // Driver code
    static public void Main()
    {
        // Create a linked list
        Node head = new Node(1);
        head.next = new Node(7);
        head.next.next = new Node(6);
        head.next.next.next = new Node(5);
        head.next.next.next.next = new Node(4);
        head.next.next.next.next.next = new Node(9);
        head.next.next.next.next.next.next = new Node(2);
 
        // Find the largest number using four nodes in the
        // linked list
        long largestNum = FormLargestNumber(head);
 
        // Print the largest number
        Console.WriteLine(
            "The largest number formed using four nodes in "
            + "the linked list is: " + largestNum);
    }
}




// javaScript code for the above approach
 
// Define a class for a singly linked list node
class Node {
    // Constructor to initialize the
    // node with a value
    constructor(val) {
        this.data = val;
        // Initialize the next pointer to null
        this.next = null;
    }
}
 
// Function to find the largest number using
// four nodes in the linked list
function formLargestNumber(head) {
    // Initialize the maximum values to the
    // smallest possible integer values
    let max1 = -Infinity,
    max2 = -Infinity,
    max3 = -Infinity,
    max4 = -Infinity;
 
    // Traverse the linked list and
    // update the maximum values
    let curr = head;
    while (curr !== null) {
        const val = curr.data;
 
        if (val > max1) {
            max4 = max3;
            max3 = max2;
            max2 = max1;
            // Update the first maximum
            // if the value is greater
            max1 = val;
        }
        else if (val > max2 && val <= max1) {
            max4 = max3;
            max3 = max2;
            // Update the second maximum if the value is
            // greater but not greater than the first
            max2 = val;
        }
        else if (val > max3 && val <= max2) {
            max4 = max3;
            // Update the third maximum if the value is
            // greater but not greater than the first two
            max3 = val;
        }
        else if (val > max4 && val <= max3) {
            // Update the fourth maximum if the value is
            // greater but not greater than the first three
            max4 = val;
        }
        curr = curr.next;
    }
 
    // Concatenate the four maximum values
    // to form the largest number
    const largestNum = max1 * 1000 + max2 * 100 + max3 * 10 + max4;
 
    return largestNum;
}
 
// Create a linked list
const head = new Node(1);
head.next = new Node(7);
head.next.next = new Node(6);
head.next.next.next = new Node(5);
head.next.next.next.next = new Node(4);
head.next.next.next.next.next = new Node(9);
head.next.next.next.next.next.next = new Node(2);
 
// Find the largest number using
// four nodes in the linked list
const largestNum = formLargestNumber(head);
 
// Print the largest number
console.log("The largest number formed using four nodes in the linked list is:", largestNum);

Output
The largest number formed using four nodes in the linked list is: 9765









Time Complexity: O(n) where ‘n’ is the number of nodes in the linked list.
Auxiliary Space: O(1) as we are not using any extra space.


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