Find the Highest Occurring Digit in a Linked List Nodes

Given a linked list, the task is to find the highest occurring digit in the linked list. Each node in the linked list contains a positive integer value. Analyze the digits present in these values and determine the digit that occurs the most frequently throughout the linked list.

Examples:

Input: 2 -> 11 -> 3 -> 2 -> 1
Output: 1
Explanation: In this example, the linked list is: 2 -> 11 -> 3 -> 2 -> 1. The digit 2 occurs twice, the digit 1 occurs thrice, and the digit 3 also occurs once. The digit 2 occurs in two nodes: the first node (2) and the fourth node (2). The digit 1 occurs in two nodes: the second node contains two times (11) and the fifth node (1). The digit 3 occurs in one node: the third node (3). Among these digits, the digit 1 has the highest occurrence with a count of 3. Therefore, the output is 1.

Input: 5 -> 5 -> 3 -> 2 -> 12
Output: 5
Explanation: In this example, the linked list is: 5 -> 5 -> 3 -> 2 -> 12. The digit 5 occurs twice, the digit 3 occurs once, and the digit 2 also occurs twice. The digit 5 occurs in two nodes: the first node (5) and the second node (5). The digit 3 occurs in one node : the third node (3). The digit 2 occurs in two node: the fourth node (2) and in the fifth node (2). Among these digits, the digit 5 and 2 has the highest occurrence with a count of 2 . Therefore, the output is 5 because 5 digit occurs before digit 2.

Approach: To solve the problem follow the below idea:

The intuition behind the approach is to iterate through the linked list and examine each node's value. By extracting individual digits from the value, we can keep track of their occurrences using an unordered map. The algorithm maintains a count for each digit encountered, updating the count whenever a digit is encountered again. Throughout the traversal, it identifies the digit with the highest count. Finally, it returns the digit that occurs the most frequently in the linked list. This approach efficiently determines the highest occurring digit by analyzing the digits within the linked list nodes and selecting the one with the maximum count.

Below are the steps of the approach:

• Initialize variables: maxCount and digitWithMaxCount.
• Start traversing the linked list from the head.
• For each node, extract its value.
• Count how many times each digit appears in the value.
• Update maxCount and digitWithMaxCount if a digit appears more times.
• Move to the next node and repeat steps 3-5.
• After traversing all nodes, return the digit with the highest count.

Below is the implementation of the above approach:

// C++ code for the above approach:
#include <bits/stdc++.h>
using namespace std;

// Define a structure for
// the linked list node
struct Node {

    // Data value of the node
    int data;

    // Pointer to the next node
    // in the list
    struct Node* next;
};

// Function to find the highest occurring
// digit in the linked list
int highestOccurringDigit(Node* head)
{

    // Initialize a map to store
    // digit counts
    unordered_map<int, int> digitCount;

    // Initialize maximum count
    int maxCount = 0;

    // Initialize digit with the
    // maximum count
    int digitWithMaxCount = 0;

    // Start traversing the
    // linked list
    Node* current = head;
    while (current != NULL) {

        // Get the value stored in
        // the current node
        int num = current->data;

        // Extract individual digits from
        // the number and count them
        while (num != 0) {

            // Extract the last digit
            int digit = num % 10;

            // Increment the count of
            // the current digit
            digitCount[digit]++;

            // Update maxCount and
            // digitWithMaxCount if needed
            if (digitCount[digit] > maxCount
                || (digitCount[digit] == maxCount
                    && digit > digitWithMaxCount)) {
                maxCount = digitCount[digit];
                digitWithMaxCount = digit;
            }

            // Remove the last digit
            num /= 10;
        }

        // Move to the next node
        /// in the linked list
        current = current->next;
    }

    // Return the digit with
    // the highest occurrence
    return digitWithMaxCount;
}

// Function to create a new node with
// the given data value
Node* createNode(int data)
{
    Node* newNode = new Node();
    newNode->data = data;
    newNode->next = NULL;
    return newNode;
}

// Drivers code
int main()
{

    // Create a linked list with
    // sample data
    Node* head = createNode(2);
    head->next = createNode(11);
    head->next->next = createNode(3);
    head->next->next->next = createNode(2);
    head->next->next->next->next = createNode(1);

    // Find and print the highest occurring
    // digit in the linked list
    int highestDigit = highestOccurringDigit(head);
    cout << "Highest Occurring Digit: " << highestDigit
         << endl;

    return 0;
}

import java.util.HashMap;
import java.util.Map;

// Define a class for the linked list node
class Node {
    // Data value of the node
    int data;

    // Pointer to the next node in the list
    Node next;

    Node(int data) {
        this.data = data;
        this.next = null;
    }
}

public class Main {
    // Function to find the highest occurring digit in the linked list
    static int highestOccurringDigit(Node head) {
        // Initialize a map to store digit counts
        Map<Integer, Integer> digitCount = new HashMap<>();

        // Initialize maximum count
        int maxCount = 0;

        // Initialize digit with the maximum count
        int digitWithMaxCount = 0;

        // Start traversing the linked list
        Node current = head;
        while (current != null) {
            // Get the value stored in the current node
            int num = current.data;

            // Extract individual digits from the number and count them
            while (num != 0) {
                // Extract the last digit
                int digit = num % 10;

                // Increment the count of the current digit
                digitCount.put(digit, digitCount.getOrDefault(digit, 0) + 1);

                // Update maxCount and digitWithMaxCount if needed
                if (digitCount.get(digit) > maxCount ||
                        (digitCount.get(digit) == maxCount && digit > digitWithMaxCount)) {
                    maxCount = digitCount.get(digit);
                    digitWithMaxCount = digit;
                }

                // Remove the last digit
                num /= 10;
            }

            // Move to the next node in the linked list
            current = current.next;
        }

        // Return the digit with the highest occurrence
        return digitWithMaxCount;
    }

    // Function to create a new node with the given data value
    static Node createNode(int data) {
        return new Node(data);
    }

    public static void main(String[] args) {
        // Create a linked list with sample data
        Node head = createNode(2);
        head.next = createNode(11);
        head.next.next = createNode(3);
        head.next.next.next = createNode(2);
        head.next.next.next.next = createNode(1);

        // Find and print the highest occurring digit in the linked list
        int highestDigit = highestOccurringDigit(head);
        System.out.println("Highest Occurring Digit: " + highestDigit);
    }
}
// This code is contributed by Vikram_Shirsat

# Python code for the above approach

# Define a class for the linked list node
class Node:
    # Constructor to initialize the node
    def __init__(self, data):
        self.data = data
        self.next = None

# Function to find the highest occurring
# digit in the linked list
def highest_occuring_digit(head):
    # Initialize a dictionary to store digit counts
    digit_count = {}

    # Initialize maximum count
    max_count = 0

    # Initialize digit with the maximum count
    digit_with_max_count = 0

    # Start traversing the linked list
    current = head
    while current is not None:
        # Get the value stored in the current node
        num = current.data

        # Extract individual digits from the number and count them
        while num != 0:
            # Extract the last digit
            digit = num % 10

            # Increment the count of the current digit
            digit_count[digit] = digit_count.get(digit, 0) + 1

            # Update max_count and digit_with_max_count if needed
            if digit_count[digit] > max_count or (digit_count[digit] == max_count and digit > digit_with_max_count):
                max_count = digit_count[digit]
                digit_with_max_count = digit

            # Remove the last digit
            num //= 10

        # Move to the next node in the linked list
        current = current.next

    # Return the digit with the highest occurrence
    return digit_with_max_count

# Function to create a new node with the given data value
def create_node(data):
    return Node(data)

# Driver code
if __name__ == "__main__":
    # Create a linked list with sample data
    head = create_node(2)
    head.next = create_node(11)
    head.next.next = create_node(3)
    head.next.next.next = create_node(2)
    head.next.next.next.next = create_node(1)

    # Find and print the highest occurring digit in the linked list
    highest_digit = highest_occuring_digit(head)
    print(f"Highest Occurring Digit: {highest_digit}")

// C# Code

using System;
using System.Collections.Generic;

// Define a class for the linked list node
class Node
{
    // Data value of the node
    public int Data { get; set; }

    // Pointer to the next node in the list
    public Node Next { get; set; }

    public Node(int data)
    {
        Data = data;
        Next = null;
    }
}

public class MainClass
{
    // Function to find the highest occurring digit in the linked list
    static int HighestOccurringDigit(Node head)
    {
        // Initialize a dictionary to store digit counts
        Dictionary<int, int> digitCount = new Dictionary<int, int>();

        // Initialize maximum count
        int maxCount = 0;

        // Initialize digit with the maximum count
        int digitWithMaxCount = 0;

        // Start traversing the linked list
        Node current = head;
        while (current != null)
        {
            // Get the value stored in the current node
            int num = current.Data;

            // Extract individual digits from the number and count them
            while (num != 0)
            {
                // Extract the last digit
                int digit = num % 10;

                // Increment the count of the current digit
                digitCount[digit] = digitCount.TryGetValue(digit, out int count) ? count + 1 : 1;

                // Update maxCount and digitWithMaxCount if needed
                if (digitCount[digit] > maxCount ||
                    (digitCount[digit] == maxCount && digit > digitWithMaxCount))
                {
                    maxCount = digitCount[digit];
                    digitWithMaxCount = digit;
                }

                // Remove the last digit
                num /= 10;
            }

            // Move to the next node in the linked list
            current = current.Next;
        }

        // Return the digit with the highest occurrence
        return digitWithMaxCount;
    }

    // Function to create a new node with the given data value
    static Node CreateNode(int data)
    {
        return new Node(data);
    }

    public static void Main(string[] args)
    {
        // Create a linked list with sample data
        Node head = CreateNode(2);
        head.Next = CreateNode(11);
        head.Next.Next = CreateNode(3);
        head.Next.Next.Next = CreateNode(2);
        head.Next.Next.Next.Next = CreateNode(1);

        // Find and print the highest occurring digit in the linked list
        int highestDigit = HighestOccurringDigit(head);
        Console.WriteLine("Highest Occurring Digit: " + highestDigit);
    }
}

// This code is contributed by guptapratik

// Define a class for the linked list node
class Node {
    constructor(data) {
        this.data = data; // Data value of the node
        this.next = null; // Pointer to the next node in the list
    }
}

// Function to find the highest occurring digit in the linked list
function highestOccurringDigit(head) {
    // Initialize a map to store digit counts
    const digitCount = new Map();

    // Initialize maximum count
    let maxCount = 0;

    // Initialize digit with the maximum count
    let digitWithMaxCount = 0;

    // Start traversing the linked list
    let current = head;
    while (current !== null) {
        // Get the value stored in the current node
        let num = current.data;

        // Extract individual digits from the number and count them
        while (num !== 0) {
            // Extract the last digit
            const digit = num % 10;

            // Increment the count of the current digit
            digitCount.set(digit, (digitCount.get(digit) || 0) + 1);

            // Update maxCount and digitWithMaxCount if needed
            if (digitCount.get(digit) > maxCount || (digitCount.get(digit) === maxCount && digit > digitWithMaxCount)) {
                maxCount = digitCount.get(digit);
                digitWithMaxCount = digit;
            }

            // Remove the last digit
            num = Math.floor(num / 10);
        }

        // Move to the next node in the linked list
        current = current.next;
    }

    // Return the digit with the highest occurrence
    return digitWithMaxCount;
}

// Function to create a new node with the given data value
function createNode(data) {
    return new Node(data);
}

// Create a linked list with sample data
const head = createNode(2);
head.next = createNode(11);
head.next.next = createNode(3);
head.next.next.next = createNode(2);
head.next.next.next.next = createNode(1);

// Find and print the highest occurring digit in the linked list
const highestDigit = highestOccurringDigit(head);
console.log("Highest Occurring Digit: " + highestDigit);

Highest Occurring Digit: 1

Time Complexity: O(N), where N is the number of nodes in the linked list.
Axillary Space: O(1) (constant) because the maximum number of entries in the unordered map remains constant regardless of the size of the linked list.

vishaldhaygude01

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