Given a singly Linked List, the task is to swap the first odd valued node from the beginning and the first even valued node from the end of the Linked List. If the list contains node values of a single parity, then no modifications are required.
Examples:
Input: 4 -> 3 -> 5 -> 2 -> 3 -> NULL
Output: 4 -> 2 -> 5 -> 3 -> 3 -> NULL
Explanation: 4 -> 3 -> 5 -> 2 -> 3 -> NULL ===> 4 -> 2 -> 5 -> 3 -> 3 -> NULL The first odd value in any node from the beginning is 3. The first even value in any node from the end is 2. After swapping the above two node values, the linked list modifies to 4 -> 2 -> 5 -> 3 -> 3 -> NULL.Input: LL: 2 -> 6 -> 8 -> 2 -> NULL
Output: 2 -> 6 -> 8 -> 2 -> NULL
Approach: The given problem can be solved by keeping track of the first and the last occurrences of odd and even valued nodes respectively and swapping them. Follow the steps below to solve the problem:
- Initialize two variables, say firstOdd and firstEven, to store the first node having odd and even values from the beginning and the end respectively.
- Initialize two variables, say firstOdd and firstEven (initially NULL).
- Traverse the linked list and perform the following steps:
- If the value of the current node is odd and firstOdd is NULL, then update the firstOdd node to the current node.
- Otherwise, update the firstEven as the current node.
- After completing the above steps, if firstOdd and firstEven is not NULL, then swap the values at both the pointers.
- Print the modified Linked List as the resultant Linked List.
Below is the implementation of the above approach:
C++
#include <iostream>using namespace std;class Node {public: int val; Node* next; Node(int x) { val = x; next = NULL; }};void printLL(Node* head) { while (head != NULL) { cout << head->val << " "; head = head->next; } cout << endl;}Node* swapNodes(Node* head, Node* even, Node* odd) { Node* prevEven = NULL; Node* currEven = head; while (currEven != NULL && currEven != even) { prevEven = currEven; currEven = currEven->next; } Node* prevOdd = NULL; Node* currOdd = head; while (currOdd != NULL && currOdd != odd) { prevOdd = currOdd; currOdd = currOdd->next; } if (currEven == NULL || currOdd == NULL) { return head; } if (prevEven != NULL) { prevEven->next = currOdd; } else { head = currOdd; } if (prevOdd != NULL) { prevOdd->next = currEven; } else { head = currEven; } Node* temp = currEven->next; currEven->next = currOdd->next; currOdd->next = temp; return head;}void swapOddAndEvenNodes(Node* head) { Node* even = NULL; Node* curr = head; while (curr != NULL) { if (curr->val % 2 == 0) { even = curr; } curr = curr->next; } Node* odd = NULL; curr = head; while (curr != NULL) { if (curr->val % 2 != 0) { odd = curr; break; } curr = curr->next; } if (odd != NULL && even != NULL) { head = swapNodes(head, even, odd); } printLL(head);}Node* linkedList(int arr[], int n) { Node* head = NULL; Node* ptr = NULL; for (int i = 0; i < n; i++) { if (head == NULL) { head = new Node(arr[i]); ptr = head; } else { Node* newNode = new Node(arr[i]); ptr->next = newNode; ptr = newNode; } } return head;}int main() { int arr[] = { 4, 3, 5, 2, 3 }; int n = sizeof(arr) / sizeof(arr[0]); Node* head = linkedList(arr, n); swapOddAndEvenNodes(head); return 0;} |
Java
class Node { int val; Node next; Node(int x) { val = x; }}public class Main { // Function to display the Linked List public static void printLL(Node head) { // Traverse until end of list is reached while (head != null) { // Print the value stored in the head System.out.print(head.val + " "); // Move to the next of head head = head.next; } System.out.println(); } // Function to swap the nodes public static Node swapNodes(Node head, Node even, Node odd) { // Keeps the track of prevEven and CurrEven Node prevEven = null; Node currEven = head; while (currEven != null && currEven != even) { prevEven = currEven; currEven = currEven.next; } // Keeps the track of prevOdd and currOdd Node prevOdd = null; Node currOdd = head; while (currOdd != null && currOdd != odd) { prevOdd = currOdd; currOdd = currOdd.next; } // If list contains nodes of a single parity if (currEven == null || currOdd == null) { return head; } // If head of the linked list does not contain even // value if (prevEven != null) { prevEven.next = currOdd; } // Make odd node the new head else { head = currOdd; } // If head of the linked list does not contain odd // value if (prevOdd != null) { prevOdd.next = currEven; } // Make even node the new head else { head = currEven; } // Swap the next pointers Node temp = currEven.next; currEven.next = currOdd.next; currOdd.next = temp; // Return the modified Linked List return head; } // Function to swap the first odd node // from the beginning and the first even // node from the end of the Linked List public static void swapOddAndEvenNodes(Node head) { // Find the first even node from // the end of the Linked List Node even = null; Node curr = head; while (curr != null) { if (curr.val % 2 == 0) { even = curr; } curr = curr.next; } // Find the first odd node from // the front of the Linked List Node odd = null; curr = head; while (curr != null) { if (curr.val % 2 != 0) { odd = curr; break; } curr = curr.next; } // If required odd and even // nodes are found, then swap if (odd != null && even != null) { head = swapNodes(head, even, odd); } printLL(head); } // Function to convert given array into a Linked List public static Node linkedList(int[] arr) { Node head = null; Node ptr = null; // 4 -> 3 -> 5 -> 2 -> 3 -> NULL for (int i : arr) { if (head == null) { head = new Node(i); ptr = head; } else { Node newNode = new Node(i); ptr.next = newNode; ptr = newNode; } } return head; } // Driver Code public static void main(String[] args) { // Given Linked List int[] arr = { 4, 3, 5, 2, 3 }; // Stores head of Linked List Node head = linkedList(arr); swapOddAndEvenNodes(head); }} |
Python3
# Python3 program for the above approach# Structure of a node# in the Linked Listclass Node: def __init__(self, x): self.val = x self.next = None# Function to display the Linked Listdef printLL(head): # Traverse until end # of list is reached while(head): # Print the value # stored in the head print(head.val, end =' ') # Move to the next of head head = head.next print()# Function to swap the nodesdef swapNodes(head, even, odd): # Keeps the track of # prevEven and CurrEven prevEven = None currEven = head while currEven and currEven != even: prevEven = currEven currEven = currEven.next # Keeps the track of # prevOdd and currOdd prevOdd = None currOdd = head while currOdd and currOdd != odd: prevOdd = currOdd currOdd = currOdd.next # If list contains nodes # of a single parity if not currEven or not currOdd: return head # If head of the linked list # does not contain even value if prevEven: prevEven.next = currOdd # Make odd node the new head else: head = currOdd # If head of the linked list # does not contain odd value if prevOdd: prevOdd.next = currEven # Make even node the new head else: head = currEven # Swap the next pointers temp = currEven.next currEven.next = currOdd.next currOdd.next = temp # Return the modified Linked List return head# Function to swap the first odd node# from the beginning and the first even# node from the end of the Linked Listdef swapOddAndEvenNodes(head): # Find the first even node from # the end of the Linked List even = None curr = head while curr: if not curr.val & 1: even = curr curr = curr.next # Find the first odd node from # the front of the Linked List odd = None curr = head while curr: if curr.val & 1: odd = curr break curr = curr.next # If required odd and even # nodes are found, then swap if odd and even: head = swapNodes(head, even, odd) printLL(head)# Function to convert given# array into a Linked Listdef linkedList(arr): head = None ptr = None # 4 -> 3 -> 5 -> 2 -> 3 -> NULL for i in arr: if not head: head = Node(i) ptr = head else: newNode = Node(i) ptr.next = newNode ptr = newNode return head# Driver Code# Given Linked Listarr = [4, 3, 5, 2, 3]# Stores head of Linked Listhead = linkedList(arr)swapOddAndEvenNodes(head) |
C#
using System;public class Node { public int val; public Node next; public Node(int x) { val = x; }}public class MainClass { // Function to display the Linked List public static void printLL(Node head) { // Traverse until end of list is reached while (head != null) { // Print the value stored in the head Console.Write(head.val + " "); // Move to the next of head head = head.next; } Console.WriteLine(); } // Function to swap the nodes public static Node swapNodes(Node head, Node even, Node odd) { // Keeps the track of prevEven and CurrEven Node prevEven = null; Node currEven = head; while (currEven != null && currEven != even) { prevEven = currEven; currEven = currEven.next; } // Keeps the track of prevOdd and currOdd Node prevOdd = null; Node currOdd = head; while (currOdd != null && currOdd != odd) { prevOdd = currOdd; currOdd = currOdd.next; } // If list contains nodes of a single parity if (currEven == null || currOdd == null) { return head; } // If head of the linked list does not contain even // value if (prevEven != null) { prevEven.next = currOdd; } // Make odd node the new head else { head = currOdd; } // If head of the linked list does not contain odd // value if (prevOdd != null) { prevOdd.next = currEven; } // Make even node the new head else { head = currEven; } // Swap the next pointers Node temp = currEven.next; currEven.next = currOdd.next; currOdd.next = temp; // Return the modified Linked List return head; } // Function to swap the first odd node // from the beginning and the first even // node from the end of the Linked List public static void swapOddAndEvenNodes(Node head) { // Find the first even node from the end of the // Linked List Node even = null; Node curr = head; while (curr != null) { if (curr.val % 2 == 0) { even = curr; } curr = curr.next; } // Find the first odd node from the front of the // Linked List Node odd = null; curr = head; while (curr != null) { if (curr.val % 2 != 0) { odd = curr; break; } curr = curr.next; } // If required odd and even nodes are found, then // swap if (odd != null && even != null) { head = swapNodes(head, even, odd); } printLL(head); } // Function to convert given array into a Linked List public static Node linkedList(int[] arr) { Node head = null; Node ptr = null; // 4 -> 3 -> 5 -> 2 -> 3 -> NULL foreach(int i in arr) { if (head == null) { head = new Node(i); ptr = new Node(i); ptr = head; } else { Node newNode = new Node(i); ptr.next = newNode; ptr = newNode; } } return head; } // Driver Code public static void Main(string[] args) { // Given Linked List int[] arr = { 4, 3, 5, 2, 3 }; // Stores head of Linked List Node head = linkedList(arr); swapOddAndEvenNodes(head); }} |
Javascript
// JavaScript program for the above approach// Structure of a node// in the Linked Listclass Node { constructor(x) { this.val = x; this.next = null; }}// Function to display the Linked Listfunction printLL(head) { // Traverse until end // of list is reached while (head) { // Print the value // stored in the head process.stdout.write(head.val + ' '); // Move to the next of head head = head.next; } console.log();}// Function to swap the nodesfunction swapNodes(head, even, odd) { // Keeps the track of // prevEven and CurrEven let prevEven = null; let currEven = head; while (currEven && currEven != even) { prevEven = currEven; currEven = currEven.next; } // Keeps the track of // prevOdd and currOdd let prevOdd = null; let currOdd = head; while (currOdd && currOdd != odd) { prevOdd = currOdd; currOdd = currOdd.next; } // If list contains nodes // of a single parity if (!currEven || !currOdd) { return head; } // If head of the linked list // does not contain even value if (prevEven) { prevEven.next = currOdd; } // Make odd node the new head else { head = currOdd; } // If head of the linked list // does not contain odd value if (prevOdd) { prevOdd.next = currEven; } // Make even node the new head else { head = currEven; } // Swap the next pointers let temp = currEven.next; currEven.next = currOdd.next; currOdd.next = temp; // Return the modified Linked List return head;}// Function to swap the first odd node// from the beginning and the first even// node from the end of the Linked Listfunction swapOddAndEvenNodes(head) { // Find the first even node from // the end of the Linked List let even = null; let curr = head; while (curr) { if (!(curr.val & 1)) { even = curr; } curr = curr.next; } // Find the first odd node from // the front of the Linked List let odd = null; curr = head; while (curr) { if (curr.val & 1) { odd = curr; break; } curr = curr.next; } // If required odd and even // nodes are found, then swap if (odd && even) { head = swapNodes(head, even, odd); } printLL(head);}// Function to convert given// array into a Linked Listfunction linkedList(arr) { let head = null; let ptr = null; // 4 -> 3 -> 5 -> 2 -> 3 -> NULL for (let i of arr) { if (!head) { head = new Node(i); ptr = head; } else { let newNode = new Node(i); ptr.next = newNode; ptr = newNode; } } return head;}// Driver Code// Given Linked Listlet arr = [4, 3, 5, 2, 3]// Stores head of Linked Listlet head = linkedList(arr)swapOddAndEvenNodes(head) |
Output:
4 2 5 3 3
Time Complexity: O(N)
Auxiliary Space: O(1)
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