Replace even nodes of a doubly linked list with the elements of array Last Updated : 17 Mar, 2023 Comments Improve Suggest changes Like Article Like Report Given a doubly linked list and an array with only odd values. Both are of equal size N. The task is replace all node which have even value with the Array elements from left to right. Examples: Input : List = 6 9 8 7 4 Arr[] = {3, 5, 23, 17, 1} Output : List = 3 9 5 7 23Input : List = 9 14 7 12 8 13 Arr[] = {5, 1, 17, 21, 11, 7} Output : List = 9 5 7 1 17 13 Approach: The idea is to traverse the nodes of the doubly linked list one by one and get the pointer of the nodes having even data then replace by the value of the array and increment the index of the array and move to the next node in the linked list. Algorithm: Create the function push to add a node to the doubly linked list's beginning. Establish a new node and assign memory to it. Incorporate data into the fresh node. Because the new node is being inserted first, set prior to null. Assign the old list's next pointer to the list's head in order to connect it to the new node. If the head is not null, change the head node's previous pointer to the new node. Adjust the head so that it faces the new node. Give the new head node back. Establish a function Make each doubly linked list node odd by using the makeOddNode function. Move the cursor to the list's top. Create a variable right after to hold the following node.should be declared as a variable to store the array's index. Follow the linked list all the way to the end node. Determine whether the data for each node is even or odd. If the node data is even, swap it out for array As i-th element and increase i. The pointer is set to the following node. Retrieve the modified doubly linked list's head. Below is the implementation of the above approach: C++ // C++ implementation to create // odd doubly linked list #include <bits/stdc++.h> using namespace std; // Node of the doubly linked list struct Node { int data; Node *prev, *next; }; // function to insert a node at the beginning // of the Doubly Linked List void push(Node** head_ref, int new_data) { // allocate node Node* new_node = (Node*)malloc(sizeof(struct Node)); // put in the data new_node->data = new_data; // since we are adding at the beginning, // prev is always NULL new_node->prev = NULL; // link the old list of the new node new_node->next = (*head_ref); // change prev of head node to new node if ((*head_ref) != NULL) (*head_ref)->prev = new_node; // move the head to point to the new node (*head_ref) = new_node; } // function to make all node is odd void makeOddNode(Node** head_ref, int A[], int n) { Node* ptr = *head_ref; Node* next; int i = 0; // traverse list till last node while (ptr != NULL) { next = ptr->next; // check if node is even then // replace it and increment in i if (ptr->data % 2 == 0) { ptr->data = A[i]; i++; } ptr = next; } // return sum of nodes which is divided by K } // function to print nodes in a // given doubly linked list void printList(Node* head) { while (head != NULL) { cout << head->data << " "; head = head->next; } } // Driver program to test above int main() { // start with the empty list Node* head = NULL; // create the doubly linked list // 6 <=> 9 <=> 8 <=> 7 <=> 4 int Arr[] = { 3, 5, 23, 17, 1 }; push(&head, 4); push(&head, 7); push(&head, 8); push(&head, 9); push(&head, 6); int n = sizeof(Arr) / sizeof(Arr[0]); cout << "Original List: "; printList(head); cout << endl; makeOddNode(&head, Arr, n); cout << "New odd List: "; printList(head); } Java // Java implementation to create // odd doubly linked list class GFG { // Node of the doubly linked list static class Node { int data; Node prev, next; }; // function to insert a node at the beginning // of the Doubly Linked List static Node push(Node head_ref, int new_data) { // allocate node Node new_node = new Node(); // put in the data new_node.data = new_data; // since we are adding at the beginning, // prev is always null new_node.prev = null; // link the old list of the new node new_node.next = (head_ref); // change prev of head node to new node if ((head_ref) != null) (head_ref).prev = new_node; // move the head to point to the new node (head_ref) = new_node; return head_ref; } // function to make all node is odd static Node makeOddNode(Node head_ref, int A[], int n) { Node ptr = head_ref; Node next; int i = 0; // traverse list till last node while (ptr != null) { next = ptr.next; // check if node is even then // replace it and increment in i if (ptr.data % 2 == 0) { ptr.data = A[i]; i++; } ptr = next; } // return sum of nodes which is divided by K return head_ref; } // function to print nodes in a // given doubly linked list static void printList(Node head) { while (head != null) { System.out.print( head.data + " "); head = head.next; } } // Driver code public static void main(String args[]) { // start with the empty list Node head = null; // create the doubly linked list // 6 <=> 9 <=> 8 <=> 7 <=> 4 int Arr[] = { 3, 5, 23, 17, 1 }; head = push(head, 4); head = push(head, 7); head = push(head, 8); head = push(head, 9); head = push(head, 6); int n = Arr.length; System.out.print( "Original List: "); printList(head); System.out.println(); head = makeOddNode(head, Arr, n); System.out.print("New odd List: "); printList(head); } } // This code is contributed by Arnab Kundu Python3 # Python3 implementation to # create odd doubly linked list # Node of the doubly linked list class Node: def __init__(self, data): self.data = data self.prev = None self.next = None # Function to insert a node at the # beginning of the Doubly Linked List def push(head_ref, new_data): # allocate node new_node = Node(new_data) # link the old list of the new node new_node.next = head_ref # change prev of head node to new node if head_ref != None: head_ref.prev = new_node # move the head to point to the new node head_ref = new_node return head_ref # Function to make all node is odd def makeOddNode(head_ref, A, n): ptr = head_ref i = 0 # traverse list till last node while ptr != None: next = ptr.next # check if node is even then # replace it and increment in i if ptr.data % 2 == 0: ptr.data = A[i] i += 1 ptr = next # return sum of nodes which is divided by K # Function to print nodes in a # given doubly linked list def printList(head): while head != None: print(head.data, end = " ") head = head.next # Driver Code if __name__ == "__main__": # start with the empty list head = None # create the doubly linked list # 6 <=> 9 <=> 8 <=> 7 <=> 4 Arr = [3, 5, 23, 17, 1] head = push(head, 4) head = push(head, 7) head = push(head, 8) head = push(head, 9) head = push(head, 6) n = len(Arr) print("Original List:", end = " ") printList(head) print() makeOddNode(head, Arr, n) print("New odd List:", end = " ") printList(head) # This code is contributed by Rituraj Jain C# // C# implementation to create // odd doubly linked list using System; class GFG { // Node of the doubly linked list public class Node { public int data; public Node prev, next; }; // function to insert a node at the beginning // of the Doubly Linked List static Node push(Node head_ref, int new_data) { // allocate node Node new_node = new Node(); // put in the data new_node.data = new_data; // since we are adding at the beginning, // prev is always null new_node.prev = null; // link the old list of the new node new_node.next = (head_ref); // change prev of head node to new node if ((head_ref) != null) (head_ref).prev = new_node; // move the head to point to the new node (head_ref) = new_node; return head_ref; } // function to make all node is odd static Node makeOddNode(Node head_ref, int []A, int n) { Node ptr = head_ref; Node next; int i = 0; // traverse list till last node while (ptr != null) { next = ptr.next; // check if node is even then // replace it and increment in i if (ptr.data % 2 == 0) { ptr.data = A[i]; i++; } ptr = next; } // return sum of nodes which is divided by K return head_ref; } // function to print nodes in a // given doubly linked list static void printList(Node head) { while (head != null) { Console.Write( head.data + " "); head = head.next; } } // Driver code public static void Main(String []args) { // start with the empty list Node head = null; // create the doubly linked list // 6 <=> 9 <=> 8 <=> 7 <=> 4 int []Arr = { 3, 5, 23, 17, 1 }; head = push(head, 4); head = push(head, 7); head = push(head, 8); head = push(head, 9); head = push(head, 6); int n = Arr.Length; Console.WriteLine( "Original List: "); printList(head); Console.WriteLine(); head = makeOddNode(head, Arr, n); Console.WriteLine("New odd List: "); printList(head); } } // This code contributed by Rajput-Ji JavaScript <script> // JavaScript implementation to create // odd doubly linked list // Node of the doubly linked list class Node { constructor(val) { this.data = val; this.prev = null; this.next = null; } } // function to insert a node at the beginning // of the Doubly Linked List function push(head_ref , new_data) { // allocate node var new_node = new Node(); // put in the data new_node.data = new_data; // since we are adding at the beginning, // prev is always null new_node.prev = null; // link the old list of the new node new_node.next = (head_ref); // change prev of head node to new node if ((head_ref) != null) (head_ref).prev = new_node; // move the head to point to the new node (head_ref) = new_node; return head_ref; } // function to make all node is odd function makeOddNode(head_ref , A , n) { var ptr = head_ref; var next; var i = 0; // traverse list till last node while (ptr != null) { next = ptr.next; // check if node is even then // replace it and increment in i if (ptr.data % 2 == 0) { ptr.data = A[i]; i++; } ptr = next; } // return sum of nodes which is divided by K return head_ref; } // function to print nodes in a // given doubly linked list function printList(head) { while (head != null) { document.write(head.data + " "); head = head.next; } } // Driver code // start with the empty list var head = null; // create the doubly linked list // 6 <=> 9 <=> 8 <=> 7 <=> 4 var Arr = [ 3, 5, 23, 17, 1 ]; head = push(head, 4); head = push(head, 7); head = push(head, 8); head = push(head, 9); head = push(head, 6); var n = Arr.length; document.write("Original List: "); printList(head); document.write("<br/>"); head = makeOddNode(head, Arr, n); document.write("New odd List: "); printList(head); // This code contributed by umadevi9616 </script> Output: Original List: 6 9 8 7 4 New odd List: 3 9 5 7 23 Time Complexity: O(N), where N is the total number of nodes.Auxiliary Space: O(1) because it is using constant space Comment More infoAdvertise with us R rajput-ji Follow Improve Article Tags : Linked List Technical Scripter DSA Arrays Technical Scripter 2018 doubly linked list +2 More Practice Tags : ArraysLinked List Similar Reads Basics & PrerequisitesLogic Building ProblemsLogic building is about creating clear, step-by-step methods to solve problems using simple rules and principles. 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