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
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 <bits/stdc++.h> 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 |
Python3
# Python3 implementation of the approach # Structure of a node class Node: def __init__( self , d): self .data = d self . next = None # Function to find formed by joining words # obtained by joining k consecutive nodes of # linked list. def findKWordString(head,k): # Stores the final ans = "" # Keep track of the number of # nodes traversed cnt = 0 # Stores the word formed by k consecutive # nodes of the linked list word = "" while (head): # Check if k nodes are traversed # If yes then add the word obtained # to the result 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 + = 1 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 if __name__ = = '__main__' : #Create list: a . b . c . d . e head = Node( 'a' ) head. next = Node( 'b' ) head. next . next = Node( 'c' ) head. next . next . next = Node( 'd' ) head. next . next . next . next = Node( 'e' ) k = 3 print (findKWordString(head, k)) # This code is contributed by mohit kumar 29 |
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 |
Javascript
<script> // JavaScript implementation of the approach // Class of a node class Node { constructor(val) { this .data = val; this .next = null ; } } // Function to get a new node function getNode( data) { // Allocate space var 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. function findKWordString(head , k) { // Stores the final string var ans = "" ; // Keep track of the number of // nodes traversed var cnt = 0; // Stores the word formed by k consecutive // nodes of the linked list var 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 // Create list: a.b.c.d.e var 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' ); var k = 3; document.write(findKWordString(head, k)); // This code contributed by aashish1995 </script> |
abc de
Time Complexity: O(N)
Auxiliary Space: O(1)
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