Given a singly linked list and an integer k, the task is to rotate the linked list to the left by k places.
Examples:
Input: linked list = 10 -> 20 -> 30 -> 40 -> 50, k = 4
Output: 50 -> 10 -> 20 -> 30 -> 40
Explanation: After rotating the linked list to the left by 4 places, the 5th node, i.e node 50 becomes the head of the linked list and next pointer of node 50 points to node 10.
Input: linked list = 10 -> 20 -> 30 -> 40, k = 6
Output: 30 -> 40 -> 10 -> 20
Explanation: After rotating the linked list to the left by 6 places (same as rotating by 2 places as (k % len) = (6 % 4 = 2)) , the 3rd node, i.e node 30 becomes the head of the linked list and next pointer of node 40 points to node 10.
[Naive Approach] Shifting head node to the end k times - O(n * k) Time and O(1) Space
To rotate a linked list to the left k places, we can repeatedly move the head node to the end of the linked list k times.
C++
// C++ program to rotate a linked list by
// changing moving first node to end k times
#include <iostream>
using namespace std;
class Node {
public:
int data;
Node *next;
Node(int new_data) {
data = new_data;
next = nullptr;
}
};
// Function to rotate the linked list
// left by k nodes
Node *rotate(Node *head, int k) {
if (k == 0 || head == nullptr)
return head;
// Rotate the list by k nodes
for (int i = 0; i < k; ++i) {
Node *curr = head;
while (curr->next != nullptr)
curr = curr->next;
// Move the first node to the last
curr->next = head;
curr = curr->next;
head = head->next;
curr->next = nullptr;
}
return head;
}
void printList(Node *node) {
while (node != nullptr) {
cout << node->data << " ";
node = node->next;
}
cout << endl;
}
int main() {
// Create a hard-coded linked list:
// 10 -> 20 -> 30 -> 40
Node *head = new Node(10);
head->next = new Node(20);
head->next->next = new Node(30);
head->next->next->next = new Node(40);
head = rotate(head, 6);
printList(head);
return 0;
}
// C program to rotate a linked list
// by changing moving first node to end k times.
#include <stdio.h>
#include <stdlib.h>
// A linked list node
struct Node {
int data;
struct Node* next;
};
// Function to create a new node
struct Node* createNode(int new_data) {
struct Node* new_node =
(struct Node*)malloc(sizeof(struct Node));
new_node->data = new_data;
new_node->next = NULL;
return new_node;
}
// Function to rotate the linked list
// left by k nodes
struct Node* rotate(struct Node *head, int k) {
if (k == 0 || head == NULL)
return head;
// Rotate the list by k nodes
for (int i = 0; i < k; ++i) {
struct Node *curr = head;
while (curr->next != NULL)
curr = curr->next;
// Move the first node to the last
curr->next = head;
curr = curr->next;
head = head->next;
curr->next = NULL;
}
return head;
}
void printList(struct Node *node) {
while (node != NULL) {
printf("%d ", node->data);
node = node->next;
}
printf("\n");
}
int main() {
// Create a hard-coded linked list:
// 10 -> 20 -> 30 -> 40
struct Node *head = createNode(10);
head->next = createNode(20);
head->next->next = createNode(30);
head->next->next->next = createNode(40);
head = rotate(head, 6);
printList(head);
return 0;
}
// Java program to rotate a linked list
// by changing moving first node to end k times.
class Node {
int data;
Node next;
Node(int new_data) {
data = new_data;
next = null;
}
}
class GfG {
// Function to rotate the linked list
// left by k nodes
static Node rotate(Node head, int k) {
if (k == 0 || head == null)
return head;
// Rotate the list by k nodes
for (int i = 0; i < k; ++i) {
Node curr = head;
while (curr.next != null)
curr = curr.next;
// Move the first node to the last
curr.next = head;
curr = curr.next;
head = head.next;
curr.next = null;
}
return head;
}
static void printList(Node node) {
while (node != null) {
System.out.print(node.data + " ");
node = node.next;
}
System.out.println();
}
public static void main(String[] args) {
// Create a hard-coded linked list:
// 10 -> 20 -> 30 -> 40
Node head = new Node(10);
head.next = new Node(20);
head.next.next = new Node(30);
head.next.next.next = new Node(40);
head = rotate(head, 6);
printList(head);
}
}
# Python program to rotate a linked list
# by changing moving first node to end k times.
class Node:
def __init__(self, newData):
self.data = newData
self.next = None
# Function to rotate the linked list
# left by k nodes
def rotate(head, k):
if k == 0 or head is None:
return head
# Rotate the list by k nodes
for _ in range(k):
curr = head
while curr.next is not None:
curr = curr.next
# Move the first node to the last
curr.next = head
curr = curr.next
head = head.next
curr.next = None
return head
def printList(node):
while node is not None:
print(node.data, end=" ")
node = node.next
print()
if __name__ == "__main__":
# Create a hard-coded linked list:
# 10 -> 20 -> 30 -> 40
head = Node(10)
head.next = Node(20)
head.next.next = Node(30)
head.next.next.next = Node(40)
head = rotate(head, 6)
printList(head)
// C# program to rotate a linked list
// by changing moving first node to end k times
using System;
class Node {
public int data;
public Node next;
public Node(int new_data) {
data = new_data;
next = null;
}
}
// Function to rotate the linked list
// left by k nodes
class GfG {
static Node Rotate(Node head, int k) {
if (k == 0 || head == null)
return head;
// Rotate the list by k nodes
for (int i = 0; i < k; ++i) {
Node curr = head;
while (curr.next != null)
curr = curr.next;
// Move the first node to the last
curr.next = head;
curr = curr.next;
head = head.next;
curr.next = null;
}
return head;
}
static void PrintList(Node node) {
while (node != null) {
Console.Write(node.data + " ");
node = node.next;
}
Console.WriteLine();
}
static void Main() {
// Create a hard-coded linked list:
// 10 -> 20 -> 30 -> 40
Node head = new Node(10);
head.next = new Node(20);
head.next.next = new Node(30);
head.next.next.next = new Node(40);
head = Rotate(head, 6);
PrintList(head);
}
}
// Javascript program to rotate a linked list
// by changing moving first node to end k times.
class Node {
constructor(new_data) {
this.data = new_data;
this.next = null;
}
}
// Function to rotate the linked list
// left by k nodes
function rotate(head, k) {
if (k === 0 || head === null) {
return head;
}
// Rotate the list by k nodes
for (let i = 0; i < k; ++i) {
let curr = head;
while (curr.next !== null) {
curr = curr.next;
}
// Move the first node to the last
curr.next = head;
curr = curr.next;
head = head.next;
curr.next = null;
}
return head;
}
function printList(node) {
let result = '';
while (node !== null) {
result += node.data + ' ';
node = node.next;
}
console.log(result.trim());
}
// Create a hard-coded linked list:
// 10 -> 20 -> 30 -> 40
let head = new Node(10);
head.next = new Node(20);
head.next.next = new Node(30);
head.next.next.next = new Node(40);
head = rotate(head, 6);
printList(head);
Time Complexity: O(n * k), where n is the number of nodes in Linked List and k is the number of rotations.
Auxiliary Space: O(1)
[Expected Approach] By changing pointer of kth node - O(n) Time and O(1) Space
The idea is to first convert the linked list to circular linked list by updating the next pointer of last node to the head of linked list. Then, traverse to the kth node and update the head of the linked list to the (k+1)th node. Finally, break the loop by updating the next pointer of kth node to NULL..
How to handle large values of k?
For a linked list of size n, if we rotate the linked list to the left by n places, then the linked list will remain unchanged and if we rotate the list to the left by (n + 1) places, then it is same as rotating the linked list to the left by 1 place. Similarly, if we rotate the linked list k (k >= n) places to the left, then it is same as rotating the linked list by (k % n) places. So, we can simply update k with k % n to handle large values of k.
Working:
Below is the implementation of the above algorithm:
C++
// C++ program to rotate a linked list
// by changing pointer of kth node
#include <iostream>
using namespace std;
class Node {
public:
int data;
Node *next;
Node(int new_data) {
data = new_data;
next = nullptr;
}
};
// Function to rotate the linked list
// to the left by k places
Node *rotate(Node *head, int k) {
// If the linked list is empty or no rotations are
// needed, then return the original linked list
if (k == 0 || head == nullptr)
return head;
Node *curr = head;
int len = 1;
// Find the length of linked list
while (curr->next != nullptr) {
curr = curr->next;
len += 1;
}
// Modulo k with length of linked list to handle
// large values of k
k %= len;
if (k == 0)
return head;
// Make the linked list circular
curr->next = head;
// Traverse the linked list to find the kth node
curr = head;
for (int i = 1; i < k; i++)
curr = curr->next;
// Update the (k + 1)th node as the new head
head = curr->next;
// Break the loop by updating next pointer of kth node
curr->next = nullptr;
return head;
}
void printList(Node *node) {
while (node != nullptr) {
cout << node->data << " ";
node = node->next;
}
cout << endl;
}
int main() {
// Create a hard-coded linked list:
// 10 -> 20 -> 30 -> 40
Node *head = new Node(10);
head->next = new Node(20);
head->next->next = new Node(30);
head->next->next->next = new Node(40);
head = rotate(head, 6);
printList(head);
return 0;
}
// C program to rotate a linked list
// by changing pointer of kth node
#include <stdio.h>
struct Node {
int data;
struct Node *next;
};
// Function to create a new node
struct Node* createNode(int new_data) {
struct Node *new_node =
(struct Node*)malloc(sizeof(struct Node));
new_node->data = new_data;
new_node->next = NULL;
return new_node;
}
// Function to rotate the linked list
// to the left by k places
struct Node* rotate(struct Node *head, int k) {
// If the linked list is empty or no rotations are
// needed, then return the original linked list
if (k == 0 || head == NULL)
return head;
struct Node *curr = head;
int len = 1;
// Find the length of linked list
while (curr->next != NULL) {
curr = curr->next;
len += 1;
}
// Modulo k with length of linked list to handle
// large values of k
k %= len;
if (k == 0)
return head;
// Make the linked list circular
curr->next = head;
// Traverse the linked list to find the kth node
curr = head;
for (int i = 1; i < k; i++)
curr = curr->next;
// Update the (k + 1)th node as the new head
head = curr->next;
// Break the loop by updating next pointer of kth node
curr->next = NULL;
return head;
}
void printList(struct Node *node) {
while (node != NULL) {
printf("%d ", node->data);
node = node->next;
}
printf("\n");
}
int main() {
// Create a hard-coded linked list:
// 10 -> 20 -> 30 -> 40
struct Node *head = createNode(10);
head->next = createNode(20);
head->next->next = createNode(30);
head->next->next->next = createNode(40);
head = rotate(head, 6);
printList(head);
return 0;
}
// Java program to rotate a linked list
// by changing pointer of kth node
class Node {
int data;
Node next;
Node(int new_data) {
data = new_data;
next = null;
}
}
// Function to rotate the linked list
// to the left by k places
class GfG {
static Node rotate(Node head, int k) {
// If the linked list is empty or no rotations are
// needed, then return the original linked list
if (k == 0 || head == null)
return head;
Node curr = head;
int len = 1;
// Find the length of linked list
while (curr.next != null) {
curr = curr.next;
len += 1;
}
// Modulo k with length of linked list to handle
// large values of k
k %= len;
if (k == 0)
return head;
// Make the linked list circular
curr.next = head;
// Traverse the linked list to find the kth node
curr = head;
for (int i = 1; i < k; i++)
curr = curr.next;
// Update the (k + 1)th node as the new head
head = curr.next;
// Break the loop by updating next pointer of kth node
curr.next = null;
return head;
}
static void printList(Node node) {
while (node != null) {
System.out.print(node.data + " ");
node = node.next;
}
System.out.println();
}
public static void main(String[] args) {
// Create a hard-coded linked list:
// 10 -> 20 -> 30 -> 40
Node head = new Node(10);
head.next = new Node(20);
head.next.next = new Node(30);
head.next.next.next = new Node(40);
head = rotate(head, 6);
printList(head);
}
}
# Python program to rotate a linked list
# by changing pointer of kth node
class Node:
def __init__(self, newData):
self.data = newData
self.next = None
# Function to rotate the linked list to the left by k places
def rotate(head, k):
# If the linked list is empty or no rotations are needed,
# return the original linked list
if k == 0 or head is None:
return head
curr = head
length = 1
# Find the length of the linked list
while curr.next is not None:
curr = curr.next
length += 1
# Modulo k with length of linked list to handle
# large values of k
k %= length
if k == 0:
curr.next = None
return head
# Make the linked list circular
curr.next = head
# Traverse the linked list to find the kth node
curr = head
for _ in range(1, k):
curr = curr.next
# Update the (k + 1)th node as the new head
newHead = curr.next
# Break the loop by updating the next pointer
# of kth node
curr.next = None
return newHead
def printList(node):
while node is not None:
print(node.data, end=" ")
node = node.next
print()
if __name__ == "__main__":
# Create a hard-coded linked list: 10 -> 20 -> 30 -> 40
head = Node(10)
head.next = Node(20)
head.next.next = Node(30)
head.next.next.next = Node(40)
head = rotate(head, 6)
printList(head)
// C program to rotate a linked list
// by changing pointer of kth node
using System;
public class Node {
public int data;
public Node next;
public Node(int new_data) {
data = new_data;
next = null;
}
}
// Function to rotate the linked list
// to the left by k places
class GfG {
static Node Rotate(Node head, int k) {
// If the linked list is empty or no rotations are
// needed, then return the original linked list
if (k == 0 || head == null)
return head;
Node curr = head;
int len = 1;
// Find the length of linked list
while (curr.next != null) {
curr = curr.next;
len += 1;
}
// Modulo k with length of linked list to handle
// large values of k
k %= len;
if (k == 0)
return head;
// Make the linked list circular
curr.next = head;
// Traverse the linked list to find the kth node
curr = head;
for (int i = 1; i < k; i++)
curr = curr.next;
// Update the (k + 1)th node as the new head
Node newHead = curr.next;
// Break the loop by updating next pointer of kth node
curr.next = null;
return newHead;
}
static void PrintList(Node node) {
while (node != null) {
Console.Write(node.data + " ");
node = node.next;
}
Console.WriteLine();
}
public static void Main() {
// Create a hard-coded linked list:
// 10 -> 20 -> 30 -> 40
Node head = new Node(10);
head.next = new Node(20);
head.next.next = new Node(30);
head.next.next.next = new Node(40);
head = Rotate(head, 6);
PrintList(head);
}
}
// JavaScript program to rotate a linked list
// by changing pointer of kth node
class Node {
constructor(new_data) {
this.data = new_data;
this.next = null;
}
}
// Function to rotate the linked list
// to the left by k places
function rotate(head, k) {
// If the linked list is empty or no rotations are
// needed, then return the original linked list
if (k === 0 || head === null)
return head;
let curr = head;
let len = 1;
// Find the length of linked list
while (curr.next !== null) {
curr = curr.next;
len += 1;
}
// Modulo k with length of linked list to handle
// large values of k
k %= len;
if (k === 0) {
curr.next = null;
return head;
}
// Make the linked list circular
curr.next = head;
// Traverse the linked list to find the kth node
curr = head;
for (let i = 1; i < k; i++)
curr = curr.next;
// Update the (k + 1)th node as the new head
let newHead = curr.next;
// Break the loop by updating next pointer of kth node
curr.next = null;
return newHead;
}
function printList(node) {
let output = '';
while (node !== null) {
output += node.data + ' ';
node = node.next;
}
console.log(output.trim());
}
// Create a hard-coded linked list:
// 10 -> 20 -> 30 -> 40
let head = new Node(10);
head.next = new Node(20);
head.next.next = new Node(30);
head.next.next.next = new Node(40);
head = rotate(head, 6);
printList(head);
Time Complexity: O(n), where n is the number of nodes in linked list.
Auxiliary Space: O(1)
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