Write a function that counts the number of times a given int occurs in a Linked List

Given a singly linked list and a key, count the number of occurrences of the given key in the linked list. For example, if the given linked list is 1->2->1->2->1->3->1 and the given key is 1, then the output should be 4.

Method 1- Without Recursion

Algorithm: 

Step 1: Start
Step 2: Create A Function Of A Linked List, Pass A Number 
        As Arguments And Provide The Count Of The Number To The Function.
Step 3: Initialize Count Equal To 0.
Step 4: Traverse In Linked List Until Equal Number Found.
Step 5: If Found A Number Equal To Update Count By 1.
Step 6: After Reaching The End Of The Linked List Return Count.
Step 7: Call The Function.
Step 8: Prints The Number Of Int Occurrences.
Step 9: Stop.

Implementation:

C++

// C++ program to count occurrences in a linked list
#include <bits/stdc++.h>
using namespace std;

/* Link list node */
class Node {
public:
    int data;
    Node* next;
};

/* Given a reference (pointer to pointer) to the head
of a list and an int, push a new node on the front
of the list. */
void 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;
}

/* Counts the no. of occurrences of a node
(search_for) in a linked list (head)*/
int count(Node* head, int search_for)
{
    Node* current = head;
    int count = 0;
    while (current != NULL) {
        if (current->data == search_for)
            count++;
        current = current->next;
    }
    return count;
}

/* Driver program to test count function*/
int main()
{
    /* Start with the empty list */
    Node* head = NULL;

    /* Use push() to construct below list
    1->2->1->3->1 */
    push(&head, 1);
    push(&head, 3);
    push(&head, 1);
    push(&head, 2);
    push(&head, 1);

    /* Check the count function */
    cout << "count of 1 is " << count(head, 1);
    return 0;
}

// This is code is contributed by rathbhupendra

C

// C program to count occurrences in a linked list
#include <stdio.h>
#include <stdlib.h>

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

/* Given a reference (pointer to pointer) to the head
  of a list and an int, push a new node on the front
  of the 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;
}

/* Counts the no. of occurrences of a node
   (search_for) in a linked list (head)*/
int count(struct Node* head, int search_for)
{
    struct Node* current = head;
    int count = 0;
    while (current != NULL) {
        if (current->data == search_for)
            count++;
        current = current->next;
    }
    return count;
}

/* Driver program to test count function*/
int main()
{
    /* Start with the empty list */
    struct Node* head = NULL;

    /* Use push() to construct below list
     1->2->1->3->1  */
    push(&head, 1);
    push(&head, 3);
    push(&head, 1);
    push(&head, 2);
    push(&head, 1);

    /* Check the count function */
    printf("count of 1 is %d", count(head, 1));
    return 0;
}

Java

// Java program to count occurrences in a linked list
import java.io.*;
class LinkedList {
    Node head; // head of list

    /* Linked list Node*/
    class Node {
        int data;
        Node next;
        Node(int d)
        {
            data = d;
            next = null;
        }
    }

    /* Inserts a new Node at front of the list. */
    public void push(int new_data)
    {
        /* 1 & 2: Allocate the Node &
                  Put in the data*/
        Node new_node = new Node(new_data);

        /* 3. Make next of new Node as head */
        new_node.next = head;

        /* 4. Move the head to point to new Node */
        head = new_node;
    }

    /* Counts the no. of occurrences of a node
    (search_for) in a linked list (head)*/
    int count(int search_for)
    {
        Node current = head;
        int count = 0;
        while (current != null) {
            if (current.data == search_for)
                count++;
            current = current.next;
        }
        return count;
    }

    /* Driver function to test the above methods */
    public static void main(String args[])
    {
        LinkedList llist = new LinkedList();

        /* Use push() to construct below list
          1->2->1->3->1  */
        llist.push(1);
        llist.push(2);
        llist.push(1);
        llist.push(3);
        llist.push(1);

        /*Checking count function*/
        System.out.println("Count of 1 is "
                           + llist.count(1));
    }
}
// This code is contributed by Rajat Mishra

Python3

# Python program to count the number of time a given
# int occurs in a linked list

# Node class
class Node:

    # Constructor to initialize the node object
    def __init__(self, data):
        self.data = data
        self.next = None

class LinkedList:

    # Function to initialize head
    def __init__(self):
        self.head = None

    # Counts the no . of occurrences of a node
    # (search_for) in a linked list (head)
    def count(self, search_for):
        current = self.head
        count = 0
        while(current is not None):
            if current.data == search_for:
                count += 1
            current = current.next
        return count

    # Function to insert a new node at the beginning
    def push(self, new_data):
        new_node = Node(new_data)
        new_node.next = self.head
        self.head = new_node

    # Utility function to print the LinkedList
    def printList(self):
        temp = self.head
        while(temp):
            print (temp.data)
            temp = temp.next


# Driver program
llist = LinkedList()
llist.push(1)
llist.push(3)
llist.push(1)
llist.push(2)
llist.push(1)

# Check for the count function
print ("count of 1 is % d" %(llist.count(1)))

# This code is contributed by Nikhil Kumar Singh(nickzuck_007)

C#

// C# program to count occurrences in a linked list
using System;
class LinkedList {
    Node head; // head of list

    /* Linked list Node*/
    public class Node {
        public int data;
        public Node next;
        public Node(int d)
        {
            data = d;
            next = null;
        }
    }

    /* Inserts a new Node at front of the list. */
    public void push(int new_data)
    {
        /* 1 & 2: Allocate the Node &
                Put in the data*/
        Node new_node = new Node(new_data);

        /* 3. Make next of new Node as head */
        new_node.next = head;

        /* 4. Move the head to point to new Node */
        head = new_node;
    }

    /* Counts the no. of occurrences of a node
    (search_for) in a linked list (head)*/
    int count(int search_for)
    {
        Node current = head;
        int count = 0;
        while (current != null) {
            if (current.data == search_for)
                count++;
            current = current.next;
        }
        return count;
    }

    /* Driver code */
    public static void Main(String[] args)
    {
        LinkedList llist = new LinkedList();

        /* Use push() to construct below list
        1->2->1->3->1 */
        llist.push(1);
        llist.push(2);
        llist.push(1);
        llist.push(3);
        llist.push(1);

        /*Checking count function*/
        Console.WriteLine("Count of 1 is " + llist.count(1));
    }
}

// This code is contributed by Arnab Kundu

Javascript

<script>
// javascript program to count occurrences in a linked list
var head; // head of list

    /* Linked list Node */

     class Node {
            constructor(val) {
                this.data = val;
                this.next = null;
            }
        }

    /* Inserts a new Node at front of the list. */
    function push(new_data) {
        /*
         * 1 & 2: Allocate the Node & Put in the data
         */
         new_node = new Node(new_data);

        /* 3. Make next of new Node as head */
        new_node.next = head;

        /* 4. Move the head to point to new Node */
        head = new_node;
    }

    /*
     * Counts the no. of occurrences of a node (search_for) in a linked list (head)
     */
    function count(search_for) {
         current = head;
        var count = 0;
        while (current != null) {
            if (current.data == search_for)
                count++;
            current = current.next;
        }
        return count;
    }

    /* Driver function to test the above methods */



        /*
         * Use push() to construct below list 1->2->1->3->1
         */
        push(1);
        push(2);
        push(1);
        push(3);
        push(1);

        /* Checking count function */
        document.write("Count of 1 is " + count(1));

// This code contributed by gauravrajput1
</script>

Output

count of 1 is 3

Time Complexity: O(n) 
Auxiliary Space: O(1)

Method 2- With Recursion

Algorithm: 

Algorithm
count(head, key);
if head is NULL
return frequency
if(head->data==key)
increase frequency by 1
count(head->next, key)

Implementation:

C++

// C++ program to count occurrences in a linked list using
// recursion
#include <bits/stdc++.h>
using namespace std;

/* Link list node */
struct Node {
    int data;
    struct Node* next;
};
// global variable for counting frequency of
// given element k
int frequency = 0;

/* Given a reference (pointer to pointer) to the head
of a list and an int, push a new node on the front
of the 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;
}

/* Counts the no. of occurrences of a node
(search_for) in a linked list (head)*/
int count(struct Node* head, int key)
{
    if (head == NULL)
        return frequency;
    if (head->data == key)
        frequency++;
    return count(head->next, key);
}

/* Driver program to test count function*/
int main()
{
    /* Start with the empty list */
    struct Node* head = NULL;

    /* Use push() to construct below list
     1->2->1->3->1  */
    push(&head, 1);
    push(&head, 3);
    push(&head, 1);
    push(&head, 2);
    push(&head, 1);

    /* Check the count function */
    cout << "count of 1 is " << count(head, 1);
    return 0;
}

// This code is contributed by Aditya Kumar (adityakumar129)

C

// C program to count occurrences in a linked list using
// recursion
#include <stdio.h>
#include <stdlib.h>

/* Link list node */
typedef struct Node {
    int data;
    struct Node* next;
} Node;
// global variable for counting frequency of
// given element k
int frequency = 0;

/* Given a reference (pointer to pointer) to the head
of a list and an int, push a new node on the front
of the list. */
void push(Node** head_ref, int new_data)
{
    /* allocate node */
    Node* new_node = (Node*)malloc(sizeof(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;
}

/* Counts the no. of occurrences of a node
(search_for) in a linked list (head)*/
int count(Node* head, int key)
{
    if (head == NULL)
        return frequency;
    if (head->data == key)
        frequency++;
    return count(head->next, key);
}

/* Driver program to test count function*/
int main()
{
    /* Start with the empty list */
    Node* head = NULL;

    /* Use push() to construct below list
     1->2->1->3->1  */
    push(&head, 1);
    push(&head, 3);
    push(&head, 1);
    push(&head, 2);
    push(&head, 1);

    /* Check the count function */
    printf("count of 1 is %d", count(head, 1));
    return 0;
}

// This code is contributed by Aditya Kumar (adityakumar129)

Java

// Java program to count occurrences in
// a linked list using recursion
import java.io.*;
import java.util.*;

// Represents node of a linkedlist
class Node {
    int data;
    Node next;
    Node(int val)
    {
        data = val;
        next = null;
    }
}

class GFG {

    // global variable for counting frequency of
    // given element k
    static int frequency = 0;

    /* Given a reference (pointer to pointer) to the head
    of a list and an int, push a new node on the front
    of the list. */

    static Node push(Node head, int new_data)
    {
        // allocate node
        Node new_node = new Node(new_data);

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

        // move the head to point to the new node
        head = new_node;

        return head;
    }

    /* Counts the no. of occurrences of a node
    (search_for) in a linked list (head)*/
    static int count(Node head, int key)
    {
        if (head == null)
            return frequency;
        if (head.data == key)
            frequency++;
        return count(head.next, key);
    }

    // Driver Code
    public static void main(String args[])
    {
        // Start with the empty list
        Node head = null;

        /* Use push() to construct below list
        1->2->1->3->1 */
        head = push(head, 1);
        head = push(head, 3);
        head = push(head, 1);
        head = push(head, 2);
        head = push(head, 1);

        /* Check the count function */
        System.out.print("count of 1 is " + count(head, 1));
    }
}

// This code is contributed by rachana soma

Python3

# Python program to count the number of
# time a given int occurs in a linked list
# Node class
class Node:

    # Constructor to initialize the node object
    def __init__(self, data):
        self.data = data
        self.next = None

class LinkedList:

    # Function to initialize head
    def __init__(self):
        self.head = None
        self.counter = 0

    # Counts the no . of occurrences of a node
    # (seach_for) in a linked list (head)
    def count(self, li, key):

        # Base case
        if(not li):
            return self.counter

        # If key is present in
        # current node, return true
        if(li.data == key):
            self.counter = self.counter + 1

        # Recur for remaining list
        return self.count(li.next, key)

    # Function to insert a new node
    # at the beginning
    def push(self, new_data):
        new_node = Node(new_data)
        new_node.next = self.head
        self.head = new_node

    # Utility function to print the
    # linked LinkedList
    def printList(self):
        temp = self.head
        while(temp):
            print (temp.data)
            temp = temp.next

# Driver Code
llist = LinkedList()
llist.push(1)
llist.push(3)
llist.push(1)
llist.push(2)
llist.push(1)

# Check for the count function
print ("count of 1 is", llist.count(llist.head, 1))

# This code is contributed by
# Gaurav Kumar Raghav

C#

// C# program to count occurrences in
// a linked list using recursion
using System;

// Represents node of a linkedlist
public class Node {
    public int data;
    public Node next;
    public Node(int val)
    {
        data = val;
        next = null;
    }
}

class GFG {

    // global variable for counting frequency of
    // given element k
    static int frequency = 0;

    /* Given a reference (pointer to pointer) to the head
    of a list and an int, push a new node on the front
    of the list. */

    static Node push(Node head, int new_data)
    {
        // allocate node
        Node new_node = new Node(new_data);

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

        // move the head to point to the new node
        head = new_node;

        return head;
    }

    /* Counts the no. of occurrences of a node
    (search_for) in a linked list (head)*/
    static int count(Node head, int key)
    {
        if (head == null)
            return frequency;
        if (head.data == key)
            frequency++;
        return count(head.next, key);
    }

    // Driver Code
    public static void Main(String[] args)
    {
        // Start with the empty list
        Node head = null;

        /* Use push() to construct below list
        1->2->1->3->1 */
        head = push(head, 1);
        head = push(head, 3);
        head = push(head, 1);
        head = push(head, 2);
        head = push(head, 1);

        /* Check the count function */
        Console.Write("count of 1 is " + count(head, 1));
    }
}

/* This code contributed by PrinciRaj1992 */

Javascript

<script>

// Javascript program to count occurrences in
// a linked list using recursion

// Represents node of a linkedlist
class Node
{
    constructor(val)
    {
        this.data = val;
        this.next = null;
    }
}

// Global variable for counting
// frequency of given element k
let frequency = 0;

/* Given a reference (pointer to pointer)
to the head of a list and an int, push a
new node on the front of the list. */
function push(head, new_data)
{

    // Allocate node
    let new_node = new Node(new_data);

    // Link the old list of the new node
    new_node.next = head;

    // Move the head to point to the new node
    head = new_node;

    return head;
}

/* Counts the no. of occurrences of a node
(search_for) in a linked list (head)*/
function count(head, key)
{
    if (head == null)
        return frequency;
    if (head.data == key)
        frequency++;

    return count(head.next, key);
}

// Driver code

// Start with the empty list
let head = null;

/* Use push() to construct below list
      1->2->1->3->1 */
head = push(head, 1);
head = push(head, 3);
head = push(head, 1);
head = push(head, 2);
head = push(head, 1);

/* Check the count function */
document.write("count of 1 is " +
               count(head, 1));

// This code is contributed by divyeshrabadiya07

</script>

Output

count of 1 is 3

Time complexity: O(n) where n is size of linked list
Auxiliary Space: O(n) for call stack since using recursion 

Below method can be used to avoid Global variable ‘frequency'(counter in case of Python 3 Code). 

C++

// method can be used to avoid
// Global variable 'frequency'

/* Counts the no. of occurrences of a node
(search_for) in a linked list (head)*/
int count(struct Node* head, int key)
{
    if (head == NULL)
        return 0;
    if (head->data == key)
        return 1 + count(head->next, key);
    return count(head->next, key);
}

Java

// method can be used to avoid
// Global variable 'frequency'

/* Counts the no. of occurrences of a node
(search_for) in a linked list (head)*/
import java.io.*;
int count(Node head, int key)
{
    if (head == null)
        return 0;
    if (head.data == key)
        return 1 + count(head.next, key);
    return count(head.next, key);
}

// This code is contributed by rachana soma

C#

// method can be used to avoid
// Global variable 'frequency'
using System;

/* Counts the no. of occurrences of a node
(search_for) in a linked list (head)*/
static int count(Node head, int key)
{
    if (head == null)
        return 0;
    if (head.data == key)
        return 1 + count(head.next, key);
    return count(head.next, key);
}

// This code is contributed by SHUBHAMSINGH10

Python3

def count(self, temp, key):

    # during the initial call, temp
    # has the value of head

    # Base case
    if temp is None:
        return 0

    # if a match is found, we
    # increment the counter
    if temp.data == key:
        return 1 + count(temp.next, key)
    return count(temp.next, key)

# to call count, use
# linked_list_name.count(head, key)

Javascript

<script>

// method can be used to afunction
// Global variable 'frequency'

/* Counts the no. of occurrences of a node
(search_for) in a linked list (head)*/
function count(head , key)
{
    if (head == null)
        return 0;
    if (head.data == key)
        return 1 + count(head.next, key);
    return count(head.next, key);
}

// This code is contributed by todaysgaurav

</script>

The above method implements head recursion. Below given is the tail recursive implementation for the same. Thanks to Puneet Jain for suggesting this approach: 

int count(struct Node* head, int key)
{
    if(head == NULL)
        return 0;
        
   int frequency = count(head->next, key);
   if(head->data == key)
     return 1 + frequency;
    
    // else 
    return frequency;
}

Time Complexity : O(n)
Auxiliary Space : O(n)

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