# Count of Nodes in a LinkedList whose value is equal to their frequency

• Last Updated : 09 Jul, 2021

Given a Singly linked list, the task is to count the number of nodes whose data value is equal to its frequency.

Examples:

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Input: Linked list = 2 -> 3 -> 3 -> 3 -> 4 -> 2
Output:
Frequency of element 2 is 2
Frequency of element 3 is 3
Frequency of element 4 is 1
So, 2 and 3 are elements which have same frequency as itâ€™s value

Input: Linked list = 1 -> 2 -> 3 -> 4 -> 5 -> 6
Output: 1

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:
Approach to solve this problem is as following

• Iterate over the linked list and store the frequency of every element of the array using a map
• Iterate over the map and count the number of elements whose frequency is equal to their value

Below is the implementation of the above approach:

## C++

 `/* Link list node */``#include ``using` `namespace` `std;` `class` `Node {``public``:``    ``int` `data;``    ``Node* next;``};` `// Function to add a node at the``// beginning of List``void` `push(Node** head_ref, ``int` `data) ``{ ``    ``/* allocate node */``    ``Node* new_node =``new` `Node();``  ` `    ``/* put in the data */``    ``new_node->data = data; ``  ` `    ``// link the old list off 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 occurences of a``// node in a linked list``int` `countValuesWithSameFreq(Node* start)``{``    ``map<``int``, ``int``> mpp;``    ``Node* current = start;``    ``int` `count = 0;``    ``while` `(current != NULL) {``        ``mpp[current->data] += 1;``        ``current = current->next;``    ``}``    ``int` `ans = 0;``    ``for` `(``auto` `x : mpp) {``        ``int` `value = x.first;``        ``int` `freq = x.second;` `        ``// Check if value equals to frequency``        ``// and increment the count``        ``if` `(value == freq) {``            ``ans++;``        ``}``    ``}``    ``return` `ans;``}` `// main program``int` `main()``{``    ``Node* head = NULL;``    ``push(&head, 3);``    ``push(&head, 4);``    ``push(&head, 3);``    ``push(&head, 2);``    ``push(&head, 2);``    ``push(&head, 3);` `    ``cout << countValuesWithSameFreq(head);``    ``return` `0;``}`

## Java

 `/* Link list node */``import` `java.util.*;` `class` `GFG{` `public` `static` `class` `Node``{``    ``int` `data;``    ``Node next;``};` `// Function to add a node at the``// beginning of List``static` `Node push(Node head_ref, ``int` `data)``{``    ` `    ``// Allocate node``    ``Node new_node = ``new` `Node();` `    ``// Put in the data``    ``new_node.data = data;` `    ``// Link the old list off the new``    ``// node``    ``new_node.next = head_ref;` `    ``// Move the head to point to the``    ``// new node``    ``head_ref = new_node;``    ``return` `head_ref;``}` `// Counts the no. of occurences of a``// node in a linked list``static` `int` `countValuesWithSameFreq(Node start)``{``    ``HashMap mpp = ``new` `HashMap<>();``    ` `    ``Node current = start;``    ``while` `(current != ``null``)``    ``{``        ``if` `(mpp.containsKey(current.data))``        ``{``            ``mpp.put(current.data,``                     ``mpp.get(current.data) + ``1``);``        ``}``        ``else``        ``{``            ``mpp.put(current.data, ``1``);``        ``}``        ``current = current.next;``    ``}``    ``int` `ans = ``0``;``    ``for``(Map.Entry x : mpp.entrySet())``    ``{``        ``int` `value = x.getKey();``        ``int` `freq = x.getValue();` `        ``// Check if value equals to frequency``        ``// and increment the count``        ``if` `(value == freq)``        ``{``            ``ans++;``        ``}``    ``}``    ``return` `ans;``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``Node head = ``null``;``    ``head = push(head, ``3``);``    ``head = push(head, ``4``);``    ``head = push(head, ``3``);``    ``head = push(head, ``2``);``    ``head = push(head, ``2``);``    ``head = push(head, ``3``);` `    ``System.out.print(countValuesWithSameFreq(head));``}``}` `// This code is contributed by amal kumar choubey`

## Python3

 `# Link list node``class` `Node:``    ` `    ``def` `__init(``self``, ``next``):``        ` `        ``self``.data ``=` `0``        ``self``.``next` `=` `None`` ` `# Function to add a node at the``# beginning of List``def` `push(head_ref, data):``    ` `    ``# Allocate node``    ``new_node ``=` `Node()``    ` `    ``# Put in the data``    ``new_node.data ``=` `data``   ` `    ``# Link the old list off the new``    ``# node``    ``new_node.``next` `=` `(head_ref) ``   ` `    ``# Move the head to point to the``    ``# new node``    ``(head_ref) ``=` `new_node``    ` `    ``return` `head_ref``    ` `# Counts the no. of occurences of a``# node in a linked list``def` `countValuesWithSameFreq(start):``    ` `    ``mpp ``=` `dict``()``    ``current ``=` `start``    ``count ``=` `0``    ` `    ``while` `(current !``=` `None``):``        ``if` `current.data ``not` `in` `mpp:``            ``mpp[current.data] ``=` `0``            ` `        ``mpp[current.data] ``+``=` `1``        ``current ``=` `current.``next``    ` `    ``ans ``=` `0``    ` `    ``for` `x ``in` `mpp.keys():``        ``value ``=` `x``        ``freq ``=` `mpp[x]``        ` `        ``# Check if value equals to frequency``        ``# and increment the count``        ``if` `(value ``=``=` `freq):``            ``ans ``+``=` `1``            ` `    ``return` `ans`` ` `# Driver code``if` `__name__``=``=``'__main__'``:``    ` `    ``head ``=` `None``    ``head ``=` `push(head, ``3``)``    ``head ``=` `push(head, ``4``)``    ``head ``=` `push(head, ``3``)``    ``head ``=` `push(head, ``2``)``    ``head ``=` `push(head, ``2``)``    ``head ``=` `push(head, ``3``)`` ` `    ``print``(countValuesWithSameFreq(head))``    ` `# This code is contributed by rutvik_56`

## C#

 `/* Link list node */``using` `System;``using` `System.Collections.Generic;` `class` `GFG{` `public` `class` `Node``{``    ``public` `int` `data;``    ``public` `Node next;``};` `// Function to add a node at the``// beginning of List``static` `Node push(Node head_ref, ``int` `data)``{``    ` `    ``// Allocate node``    ``Node new_node = ``new` `Node();` `    ``// Put in the data``    ``new_node.data = data;` `    ``// Link the old list off the new``    ``// node``    ``new_node.next = head_ref;` `    ``// Move the head to point to the``    ``// new node``    ``head_ref = new_node;``    ``return` `head_ref;``}` `// Counts the no. of occurences of a``// node in a linked list``static` `int` `countValuesWithSameFreq(Node start)``{``    ``Dictionary<``int``,``                 ``int``> mpp = ``new` `Dictionary<``int``,``                                           ``int``>();``    ` `    ``Node current = start;``    ``while` `(current != ``null``)``    ``{``        ``if` `(mpp.ContainsKey(current.data))``        ``{``            ``mpp[current.data] = mpp[current.data] + 1;``        ``}``        ``else``        ``{``            ``mpp.Add(current.data, 1);``        ``}``        ``current = current.next;``    ``}``    ``int` `ans = 0;``    ` `    ``foreach``(KeyValuePair<``int``, ``int``> x ``in` `mpp)``    ``{``        ``int` `value = x.Key;``        ``int` `freq = x.Value;` `        ``// Check if value equals to frequency``        ``// and increment the count``        ``if` `(value == freq)``        ``{``            ``ans++;``        ``}``    ``}``    ``return` `ans;``}` `// Driver code``public` `static` `void` `Main(String[] args)``{``    ``Node head = ``null``;``    ``head = push(head, 3);``    ``head = push(head, 4);``    ``head = push(head, 3);``    ``head = push(head, 2);``    ``head = push(head, 2);``    ``head = push(head, 3);` `    ``Console.Write(countValuesWithSameFreq(head));``}``}` `// This code is contributed by Rohit_ranjan`

## Javascript

 ``

Output:

`2`

Complexity Analysis:
Time Complexity: For a given linked list of size n, we are iterating over it once. So the time complexity of this solution is O(n)
Space Complexity: For a given linked list of size n, we are using an extra map which can have maximum of n key-values, so space complexity of this solution is O(n)

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