# Find the string formed by joining k consecutive nodes of linked list

Given an integer K and a linked list in which each node stores a single character. The task is to join every K consecutive nodes of the linked list to form a single word. Finally, print the string obtained by joining these words (space separated).

Examples:

Input: List = ‘a’ -> ‘b’ -> ‘c’ ->’d’ -> ‘e’ -> NULL, k = 3
Output: abc de
The first three nodes form the first word “abc”
and next two nodes form the second word “de”.

Input: List = ‘a’ -> ‘b’ -> ‘c’ -> ‘d’ -> ‘e’ -> ‘f’ -> NULL, k = 2
Output: ab cd ef

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

Approach: The idea is to traverse the linked list and keep adding character present at each node to the word formed so far. Keep track of the number of nodes traversed and when the count becomes equal to k, add the word formed so far to the resultant string, reset word to an empty string and reset the count to zero. Repeat this until the entire linked list is not traversed.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Structure of a node ` `struct` `Node { ` `    ``char` `data; ` `    ``Node* next; ` `}; ` ` `  `// Function to get a new node ` `Node* getNode(``char` `data) ` `{ ` `    ``// Allocate space ` `    ``Node* newNode = ``new` `Node; ` ` `  `    ``// Put in data ` `    ``newNode->data = data; ` `    ``newNode->next = NULL; ` `    ``return` `newNode; ` `} ` ` `  `// Function to find string formed by joining words ` `// obtained by joining k consecutive nodes of ` `// linked list. ` `string findKWordString(Node* head, ``int` `k) ` `{ ` `    ``// Stores the final string ` `    ``string ans = ``""``; ` ` `  `    ``// Keep track of the number of ` `    ``// nodes traversed ` `    ``int` `cnt = 0; ` ` `  `    ``// Stores the word formed by k consecutive ` `    ``// nodes of the linked list ` `    ``string word = ``""``; ` ` `  `    ``while` `(head) { ` ` `  `        ``// Check if k nodes are traversed ` `        ``// If yes then add the word obtained ` `        ``// to the result string ` `        ``if` `(cnt == k) { ` `            ``if` `(ans != ``""``) { ` `                ``ans = ans + ``" "``; ` `            ``} ` ` `  `            ``ans = ans + word; ` `            ``word = ``""``; ` `            ``cnt = 0; ` `        ``} ` ` `  `        ``// Add the current character to the word ` `        ``// formed so far and increase the count ` `        ``word = word + string(1, head->data); ` `        ``cnt++; ` `        ``head = head->next; ` `    ``} ` ` `  `    ``// Add the final word to the result ` `    ``// Length of the final word can be less than k ` `    ``if` `(ans != ``" "``) { ` `        ``ans = ans + ``" "``; ` `    ``} ` `    ``ans = ans + word; ` ` `  `    ``return` `ans; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` ` `  `    ``// Create list: a -> b -> c -> d -> e ` `    ``Node* head = getNode(``'a'``); ` `    ``head->next = getNode(``'b'``); ` `    ``head->next->next = getNode(``'c'``); ` `    ``head->next->next->next = getNode(``'d'``); ` `    ``head->next->next->next->next = getNode(``'e'``); ` ` `  `    ``int` `k = 3; ` ` `  `    ``cout << findKWordString(head, k); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `class` `GFG{ ` ` `  `// Class of a node ` `static` `class` `Node ` `{ ` `    ``char` `data; ` `    ``Node next; ` `}; ` ` `  `// Function to get a new node ` `static` `Node getNode(``char` `data) ` `{ ` `     `  `    ``// Allocate space ` `    ``Node newNode = ``new` `Node(); ` ` `  `    ``// Put in data ` `    ``newNode.data = data; ` `    ``newNode.next = ``null``; ` `    ``return` `newNode; ` `} ` ` `  `// Function to find string formed by  ` `// joining words obtained by joining  ` `// k consecutive nodes of linked list. ` `static` `String findKWordString(Node head, ``int` `k) ` `{ ` `     `  `    ``// Stores the final string ` `    ``String ans = ``""``; ` ` `  `    ``// Keep track of the number of ` `    ``// nodes traversed ` `    ``int` `cnt = ``0``; ` ` `  `    ``// Stores the word formed by k consecutive ` `    ``// nodes of the linked list ` `    ``String word = ``""``; ` ` `  `    ``while` `(head != ``null``) ` `    ``{ ` ` `  `        ``// Check if k nodes are traversed ` `        ``// if yes then add the word obtained ` `        ``// to the result String ` `        ``if` `(cnt == k) ` `        ``{ ` `            ``if` `(ans != ``""``)  ` `            ``{ ` `                ``ans = (ans + ``" "``); ` `            ``} ` ` `  `            ``ans = ans + word; ` `            ``word = ``""``; ` `            ``cnt = ``0``; ` `        ``} ` ` `  `        ``// Add the current character to the word ` `        ``// formed so far and increase the count ` `        ``word = word + head.data; ` `        ``cnt++; ` `        ``head = head.next; ` `    ``} ` ` `  `    ``// Add the final word to the result ` `    ``// Length of the final word can be ` `    ``// less than k ` `    ``if` `(ans != ``" "``) ` `    ``{ ` `        ``ans = (ans + ``" "``); ` `    ``} ` `    ``ans = ans + word; ` `    ``return` `ans; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` ` `  `    ``// Create list: a.b.c.d.e ` `    ``Node head = getNode(``'a'``); ` `    ``head.next = getNode(``'b'``); ` `    ``head.next.next = getNode(``'c'``); ` `    ``head.next.next.next = getNode(``'d'``); ` `    ``head.next.next.next.next = getNode(``'e'``); ` ` `  `    ``int` `k = ``3``; ` ` `  `    ``System.out.print(findKWordString(head, k)); ` `} ` `} ` ` `  `// This code is contributed by GauravRajput1 `

## C#

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG{ ` ` `  `// Class of a node ` `class` `Node ` `{ ` `    ``public` `char` `data; ` `    ``public` `Node next; ` `}; ` ` `  `// Function to get a new node ` `static` `Node getNode(``char` `data) ` `{ ` `     `  `    ``// Allocate space ` `    ``Node newNode = ``new` `Node(); ` ` `  `    ``// Put in data ` `    ``newNode.data = data; ` `    ``newNode.next = ``null``; ` `    ``return` `newNode; ` `} ` ` `  `// Function to find string formed by  ` `// joining words obtained by joining  ` `// k consecutive nodes of linked list. ` `static` `String findKWordString(Node head, ``int` `k) ` `{ ` `     `  `    ``// Stores the final string ` `    ``String ans = ``""``; ` ` `  `    ``// Keep track of the number  ` `    ``// of nodes traversed ` `    ``int` `cnt = 0; ` ` `  `    ``// Stores the word formed by k  ` `    ``// consecutive nodes of the  ` `    ``// linked list ` `    ``String word = ``""``; ` ` `  `    ``while` `(head != ``null``) ` `    ``{ ` ` `  `        ``// Check if k nodes are traversed ` `        ``// if yes then add the word obtained ` `        ``// to the result String ` `        ``if` `(cnt == k) ` `        ``{ ` `            ``if` `(ans != ``""``)  ` `            ``{ ` `                ``ans = (ans + ``" "``); ` `            ``} ` ` `  `            ``ans = ans + word; ` `            ``word = ``""``; ` `            ``cnt = 0; ` `        ``} ` ` `  `        ``// Add the current character to the word ` `        ``// formed so far and increase the count ` `        ``word = word + head.data; ` `        ``cnt++; ` `        ``head = head.next; ` `    ``} ` ` `  `    ``// Add the readonly word to the result ` `    ``// Length of the readonly word can be ` `    ``// less than k ` `    ``if` `(ans != ``" "``) ` `    ``{ ` `        ``ans = (ans + ``" "``); ` `    ``} ` `    ``ans = ans + word; ` `    ``return` `ans; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(String[] args) ` `{ ` ` `  `    ``// Create list: a.b.c.d.e ` `    ``Node head = getNode(``'a'``); ` `    ``head.next = getNode(``'b'``); ` `    ``head.next.next = getNode(``'c'``); ` `    ``head.next.next.next = getNode(``'d'``); ` `    ``head.next.next.next.next = getNode(``'e'``); ` ` `  `    ``int` `k = 3; ` ` `  `    ``Console.Write(findKWordString(head, k)); ` `} ` `} ` ` `  `// This code is contributed by gauravrajput1 `

Output:

```abc de
```

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

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Improved By : GauravRajput1