Find the Largest Number Using Four Nodes in a Singly Linked List
Last Updated :
31 Dec, 2023
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:
- Initialize four variables max1, max2, max3, and max4 to the smallest possible integer values (INT_MIN). These variables will keep track of the four largest digits in the linked list.
- Traverse the linked list using a pointer curr. For each node, retrieve the value and store it in the variable val.
- Compare val with the current maximum values (max1, max2, max3, and max4) to determine where it fits among the four largest digits.
- If val is greater than max1, update all four maximum values in the following order: max4 takes the value of max3, max3 takes the value of max2, max2 takes the value of max1, and max1 takes the value of val.
- If val is greater than max2 but less than or equal to max1, update the maximum values for max2, max3, and max4 in a similar fashion.
- Repeat similar updates for max3 and max4 as needed while traversing the linked list.
- After the traversal, you will have the four largest digits in the linked list stored in max1, max2, max3, and max4.
- Concatenate these four maximum values in reverse order to form the largest possible number using arithmetic operations.
- Return the result as a long long integer.
Implementation of the above approach:
C++
// 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);
OutputThe 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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