Given a linked list, the task is to reverse the linked list by changing the links between nodes.
Examples:
Input: head: 1 -> 2 -> 3 -> 4 -> NULL
Output: head: 4 -> 3 -> 2 -> 1 -> NULL
Explanation: Reversed Linked List:
Input: head: 1 -> 2 -> 3 -> 4 -> 5 -> NULL
Output: head: 5 -> 4 -> 3 -> 2 -> 1 -> NULL
Explanation: Reversed Linked List:
[Expected Approach] Using Iterative Method - O(n) Time and O(1) Space
The idea is to reverse the links of all nodes using three pointers:
- prev: pointer to keep track of the previous node
- curr: pointer to keep track of the current node
- next: pointer to keep track of the next node
Starting from the first node, initialize curr with the head of linked list and next with the next node of curr. Update the next pointer of curr with prev. Finally, move the three pointer by updating prev with curr and curr with next.
Follow the steps below to solve the problem:
- Initialize three pointers prev as NULL, curr as head, and next as NULL.
- Iterate through the linked list. In a loop, do the following:
- Store the next node, next = curr -> next
- Update the next pointer of curr to prev, curr -> next = prev
- Update prev as curr and curr as next, prev = curr and curr = next
C++
// Iterative C++ program to reverse a linked list
#include <iostream>
using namespace std;
class Node {
public:
int data;
Node *next;
Node(int new_data) {
data = new_data;
next = nullptr;
}
};
// Given the head of a list, reverse the list and
// return the head of reversed list
Node *reverseList(Node *head) {
// Initialize three pointers: curr, prev and next
Node *curr = head, *prev = nullptr, *next;
// Traverse all the nodes of Linked List
while (curr != nullptr) {
// Store next
next = curr->next;
// Reverse current node's next pointer
curr->next = prev;
// Move pointers one position ahead
prev = curr;
curr = next;
}
// Return the head of reversed linked list
return prev;
}
void printList(Node *node) {
while (node != nullptr) {
cout << " " << node->data;
node = node->next;
}
}
int main() {
// Create a hard-coded linked list:
// 1 -> 2 -> 3 -> 4 -> 5
Node *head = new Node(1);
head->next = new Node(2);
head->next->next = new Node(3);
head->next->next->next = new Node(4);
head->next->next->next->next = new Node(5);
cout << "Given Linked list:";
printList(head);
head = reverseList(head);
cout << "\nReversed Linked List:";
printList(head);
return 0;
}
// Iterative C program to reverse a linked list
#include <stdio.h>
struct Node {
int data;
struct Node* next;
};
// Given the head of a list, reverse the list and return the
// head of reversed list
struct Node* reverseList(struct Node* head) {
// Initialize three pointers: curr, prev and next
struct Node *curr = head, *prev = NULL, *next;
// Traverse all the nodes of Linked List
while (curr != NULL) {
// Store next
next = curr->next;
// Reverse current node's next pointer
curr->next = prev;
// Move pointers one position ahead
prev = curr;
curr = next;
}
// Return the head of reversed linked list
return prev;
}
void printList(struct Node* node) {
while (node != NULL) {
printf(" %d", node->data);
node = node->next;
}
}
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;
}
int main() {
// Create a hard-coded linked list:
// 1 -> 2 -> 3 -> 4 -> 5
struct Node* head = createNode(1);
head->next = createNode(2);
head->next->next = createNode(3);
head->next->next->next = createNode(4);
head->next->next->next->next = createNode(5);
printf("Given Linked list:");
printList(head);
head = reverseList(head);
printf("\nReversed Linked List:");
printList(head);
return 0;
}
// Iterative Java program to reverse a linked list
class Node {
int data;
Node next;
Node(int new_data) {
data = new_data;
next = null;
}
}
// Given the head of a list, reverse the list and return the
// head of reversed list
class GfG {
static Node reverseList(Node head) {
// Initialize three pointers: curr, prev and next
Node curr = head, prev = null, next;
// Traverse all the nodes of Linked List
while (curr != null) {
// Store next
next = curr.next;
// Reverse current node's next pointer
curr.next = prev;
// Move pointers one position ahead
prev = curr;
curr = next;
}
// Return the head of reversed linked list
return prev;
}
// This function prints the contents
// of the linked list starting from the head
static void printList(Node node) {
while (node != null) {
System.out.print(" " + node.data);
node = node.next;
}
}
public static void main(String[] args) {
// Create a hard-coded linked list:
// 1 -> 2 -> 3 -> 4 -> 5
Node head = new Node(1);
head.next = new Node(2);
head.next.next = new Node(3);
head.next.next.next = new Node(4);
head.next.next.next.next = new Node(5);
System.out.print("Given Linked list:");
printList(head);
head = reverseList(head);
System.out.print("\nReversed Linked List:");
printList(head);
}
}
# Iterative Python program to reverse a linked list
class Node:
def __init__(self, newData):
self.data = newData
self.next = None
# Given the head of a list, reverse the list and return the
# head of reversed list
def reverseList(head):
# Initialize three pointers: curr, prev and next
curr = head
prev = None
# Traverse all the nodes of Linked List
while curr is not None:
# Store next
nextNode = curr.next
# Reverse current node's next pointer
curr.next = prev
# Move pointers one position ahead
prev = curr
curr = nextNode
# Return the head of reversed linked list
return prev
def printList(node):
while node is not None:
print(f" {node.data}", end="")
node = node.next
print()
if __name__ == "__main__":
# Create a hard-coded linked list:
# 1 -> 2 -> 3 -> 4 -> 5
head = Node(1)
head.next = Node(2)
head.next.next = Node(3)
head.next.next.next = Node(4)
head.next.next.next.next = Node(5)
print("Given Linked list:", end="")
printList(head)
head = reverseList(head)
print("Reversed Linked List:", end="")
printList(head)
// Iterative C# program to reverse a linked list
using System;
class Node {
public int Data;
public Node Next;
public Node(int newData) {
Data = newData;
Next = null;
}
}
// Given the head of a list, reverse the list and return the
// head of reversed list
class GfG {
static Node ReverseList(Node head) {
// Initialize three pointers: curr, prev and next
Node curr = head;
Node prev = null;
Node next;
// Traverse all the nodes of Linked List
while (curr != null) {
// Store next
next = curr.Next;
// Reverse current node's next pointer
curr.Next = prev;
// Move pointers one position ahead
prev = curr;
curr = next;
}
// Return the head of reversed linked list
return prev;
}
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:
// 1 -> 2 -> 3 -> 4 -> 5
Node head = new Node(1);
head.Next = new Node(2);
head.Next.Next = new Node(3);
head.Next.Next.Next = new Node(4);
head.Next.Next.Next.Next = new Node(5);
Console.Write("Given Linked list:");
PrintList(head);
head = ReverseList(head);
Console.Write("Reversed Linked List:");
PrintList(head);
}
}
// Iterative JavaScript program to reverse a linked list
class Node {
constructor(newData) {
this.data = newData;
this.next = null;
}
}
// Given the head of a list, reverse the list and return the
// head of reversed list
function reverseList(head) {
// Initialize three pointers: curr, prev and next
let curr = head;
let prev = null;
let next;
// Traverse all the nodes of Linked List
while (curr !== null) {
// Store next
next = curr.next;
// Reverse current node's next pointer
curr.next = prev;
// Move pointers one position ahead
prev = curr;
curr = next;
}
// Return the head of reversed linked list
return prev;
}
function printList(node) {
while (node !== null) {
console.log(" " + node.data);
node = node.next;
}
console.log();
}
// Driver Code
// Create a hard-coded linked list:
// 1 -> 2 -> 3 -> 4 -> 5
let head = new Node(1);
head.next = new Node(2);
head.next.next = new Node(3);
head.next.next.next = new Node(4);
head.next.next.next.next = new Node(5);
console.log("Given Linked list:");
printList(head);
head = reverseList(head);
console.log("Reversed Linked List:");
printList(head);
OutputGiven Linked list: 1 2 3 4 5
Reversed Linked List: 5 4 3 2 1
[Alternate Approach - 1] Using Recursion - O(n) Time and O(n) Space
The idea is to reach the last node of the linked list using recursion then start reversing the linked list from the last node.
Follow the steps below to solve the problem:
- Divide the list in two parts - first node and rest of the linked list.
- Call reverse for the rest of the linked list.
- Link the rest linked list to first.
- Fix head pointer to NULL.
C++
// Recursive C++ program to reverse a linked list
#include <bits/stdc++.h>
using namespace std;
class Node {
public:
int data;
Node *next;
Node(int new_data) {
data = new_data;
next = nullptr;
}
};
// Given the head of a list, reverse the list
// and return the head of reversed list
Node *reverseList(Node *head) {
if (head == NULL || head->next == NULL)
return head;
// reverse the rest of linked list and put
// the first element at the end
Node *rest = reverseList(head->next);
// Make the current head as last node of
// remaining linked list
head->next->next = head;
// Update next of current head to NULL
head->next = NULL;
// Return the reversed linked list
return rest;
}
void printList(Node *node) {
while (node != nullptr) {
cout << " " << node->data;
node = node->next;
}
}
int main() {
// Create a hard-coded linked list:
// 1 -> 2 -> 3 -> 4 -> 5
Node *head = new Node(1);
head->next = new Node(2);
head->next->next = new Node(3);
head->next->next->next = new Node(4);
head->next->next->next->next = new Node(5);
cout << "Given Linked List:";
printList(head);
head = reverseList(head);
cout << "\nReversed Linked List:";
printList(head);
return 0;
}
// Recursive C program to reverse a linked list
#include <stdio.h>
struct Node {
int data;
struct Node* next;
};
// Given the head of a list, reverse the list and
// return the head of reversed list
struct Node* reverseList(struct Node* head) {
if (head == NULL || head->next == NULL)
return head;
// reverse the rest of linked list and put
// the first element at the end
struct Node* rest = reverseList(head->next);
// Make the current head as last node of
// remaining linked list
head->next->next = head;
// Update next of current head to NULL
head->next = NULL;
// Return the reversed linked list
return rest;
}
// This function prints the contents
// of the linked list starting from the head
void printList(struct Node* node) {
while (node != NULL) {
printf(" %d", node->data);
node = node->next;
}
printf("\n");
}
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;
}
int main() {
// Create a hard-coded linked list:
// 1 -> 2 -> 3 -> 4 -> 5
struct Node* head = createNode(1);
head->next = createNode(2);
head->next->next = createNode(3);
head->next->next->next = createNode(4);
head->next->next->next->next = createNode(5);
printf("Given Linked List:");
printList(head);
head = reverseList(head);
printf("Reversed Linked List:");
printList(head);
return 0;
}
// Recursive Java program to reverse a linked list
class Node {
int data;
Node next;
Node(int new_data) {
data = new_data;
next = null;
}
}
class GfG {
// Given the head of a list, reverse the list
// and return the head of reversed list
static Node reverseList(Node head) {
// If we have reached last node or linked
// list is empty, return head of linked list
if (head == null || head.next == null)
return head;
// reverse the rest of linked list and put
// the first element at the end
Node rest = reverseList(head.next);
// Make the current head as last node of
// remaining linked list
head.next.next = head;
// Update next of current head to NULL
head.next = null;
// Return the reversed linked list
return rest;
}
// This function prints the contents
// of the linked list starting from the head
static void printList(Node node) {
while (node != null) {
System.out.print(" " + node.data);
node = node.next;
}
}
public static void main(String[] args)
{
// Create a hard-coded linked list:
// 1 -> 2 -> 3 -> 4 -> 5
Node head = new Node(1);
head.next = new Node(2);
head.next.next = new Node(3);
head.next.next.next = new Node(4);
head.next.next.next.next = new Node(5);
System.out.print("Given Linked List:");
printList(head);
head = reverseList(head);
System.out.print("\nReversed Linked List:");
printList(head);
}
}
# Recursive Python program to reverse a linked list
class Node:
def __init__(self, newData):
self.data = newData
self.next = None
# Given the head of a list, reverse the list and
# return the head of reversed list
def reverseList(head):
if head is None or head.next is None:
return head
# reverse the rest of linked list and put the
# first element at the end
rest = reverseList(head.next)
# Make the current head as last node of
# remaining linked list
head.next.next = head
# Update next of current head to NULL
head.next = None
# Return the reversed linked list
return rest
def printList(node):
while node is not None:
print(f" {node.data}", end='')
node = node.next
print()
if __name__ == "__main__":
# Create a hard-coded linked list:
# 1 -> 2 -> 3 -> 4 -> 5
head = Node(1)
head.next = Node(2)
head.next.next = Node(3)
head.next.next.next = Node(4)
head.next.next.next.next = Node(5)
print("Given Linked List:", end='')
printList(head)
head = reverseList(head)
print("\nReversed Linked List:", end='')
printList(head)
// Recursive C# program to reverse a linked list
using System;
class Node {
public int Data;
public Node Next;
public Node(int newData) {
Data = newData;
Next = null;
}
}
class GfG {
// Given the head of a list, reverse the list
// and return the head of reversed list
static Node ReverseList(Node head) {
if (head == null || head.Next == null)
return head;
// reverse the rest of linked list and
// put the first element at the end
Node rest = ReverseList(head.Next);
// Make the current head as last node
// of remaining linked list
head.Next.Next = head;
// Update next of current head to NULL
head.Next = null;
// Return the reversed linked list
return rest;
}
// This function prints the contents
// of the linked list starting from the 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:
// 1 -> 2 -> 3 -> 4 -> 5
Node head = new Node(1);
head.Next = new Node(2);
head.Next.Next = new Node(3);
head.Next.Next.Next = new Node(4);
head.Next.Next.Next.Next = new Node(5);
Console.Write("Given Linked List:");
PrintList(head);
head = ReverseList(head);
Console.Write("\nReversed Linked List:");
PrintList(head);
}
}
// Recursive javascript program to reverse a linked list
class Node {
constructor(new_data) {
this.data = new_data;
this.next = null;
}
}
// Given the head of a list, reverse the list
// and return the head of reversed list
function reverseList(head) {
if (head === null || head.next === null)
return head;
// reverse the rest of linked list and
// put the first element at the end
let rest = reverseList(head.next);
// Make the current head as last node of
// remaining linked list
head.next.next = head;
// Update next of current head to NULL
head.next = null;
// Return the reversed linked list
return rest;
}
function printList(node) {
while (node !== null) {
console.log(` ${node.data}`);
node = node.next;
}
console.log();
}
// Driver Code
// Create a hard-coded linked list:
// 1 -> 2 -> 3 -> 4 -> 5
let head = new Node(1);
head.next = new Node(2);
head.next.next = new Node(3);
head.next.next.next = new Node(4);
head.next.next.next.next = new Node(5);
console.log("Given Linked List:");
printList(head);
head = reverseList(head);
console.log("\nReversed Linked List:");
printList(head);
OutputGiven Linked List: 1 2 3 4 5
Reversed Linked List: 5 4 3 2 1
[Alternate Approach - 2] Using Stack - O(n) Time and O(n) Space
The idea is to traverse the linked list and push all nodes except the last node into the stack. Make the last node as the new head of the reversed linked list. Now, start popping the element and append each node to the reversed Linked List. Finally, return the head of the reversed linked list.
Follow the steps below to solve the problem:
- Push all the nodes(values and address) except the last node in the stack.
- Once the nodes are pushed, update the Head pointer to the last node.
- Start popping the nodes and push them at the end of the linked list in the same order until the stack is empty.
- Update the next pointer of last node in the stack by NULL.
C++
// C++ program to reverse linked list using Stack
#include <bits/stdc++.h>
using namespace std;
class Node {
public:
int data;
Node* next;
Node(int new_data) {
data = new_data;
next = nullptr;
}
};
// Function to reverse the linked list
Node* reverseList(Node* head) {
// Create a stack to store the nodes
stack<Node*> s;
Node* temp = head;
// Push all nodes except the last node into stack
while (temp->next != NULL) {
s.push(temp);
temp = temp->next;
}
// Make the last node as new head of the linked list
head = temp;
// Pop all the nodes and append to the linked list
while (!s.empty()) {
// append the top value of stack in list
temp->next = s.top();
// Pop the value from stack
s.pop();
// move to the next node in the list
temp = temp->next;
}
// Update the next pointer of last node of stack to NULL
temp->next = NULL;
return head;
}
void printList(Node* node) {
while (node != nullptr) {
cout << " " << node->data;
node = node->next;
}
}
int main() {
// Create a hard-coded linked list:
// 1 -> 2 -> 3 -> 4 -> 5
Node* head = new Node(1);
head->next = new Node(2);
head->next->next = new Node(3);
head->next->next->next = new Node(4);
head->next->next->next->next = new Node(5);
cout << "Given Linked List:";
printList(head);
head = reverseList(head);
cout << "\nReversed Linked List:";
printList(head);
return 0;
}
// C program to reverse linked list using Stack
#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 reverse the linked list
struct Node* reverseList(struct Node* head) {
// Create a stack to store the nodes
struct Node* stack[100000];
int top = -1;
struct Node* temp = head;
// Push all nodes except the last node into stack
while (temp != NULL) {
stack[++top] = temp;
temp = temp->next;
}
// Make the last node as new head of the linked list
if (top >= 0) {
head = stack[top];
temp = head;
// Pop all the nodes and append to the linked list
while (top > 0) {
// append the top value of stack in list and
// pop the top value by decrementing top by 1
temp->next = stack[--top];
// move to the next node in the list
temp = temp->next;
}
// Update the next pointer of last node of stack to NULL
temp->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:
// 1 -> 2 -> 3 -> 4 -> 5
struct Node* head = createNode(1);
head->next = createNode(2);
head->next->next = createNode(3);
head->next->next->next = createNode(4);
head->next->next->next->next = createNode(5);
printf("Given Linked List:");
printList(head);
head = reverseList(head);
printf("\nReversed Linked List:");
printList(head);
return 0;
}
// Java program to reverse linked list using Stack
import java.util.Stack;
class Node {
int data;
Node next;
Node(int new_data) {
data = new_data;
next = null;
}
}
class GfG {
// Function to reverse the linked list
static Node reverseList(Node head) {
// Create a stack to store the nodes
Stack<Node> stack = new Stack<>();
Node temp = head;
// Push all nodes except the last node into stack
while (temp != null) {
stack.push(temp);
temp = temp.next;
}
// Make the last node as new head of the linked list
if (!stack.isEmpty()) {
head = stack.pop();
temp = head;
// Pop all the nodes and append to the linked list
while (!stack.isEmpty()) {
// append the top value of stack in list
temp.next = stack.pop();
// move to the next node in the list
temp = temp.next;
}
// Update the next pointer of last node
// of stack to NULL
temp.next = null;
}
return head;
}
// This function prints the contents
// of the linked list starting from the 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:
// 1 -> 2 -> 3 -> 4 -> 5
Node head = new Node(1);
head.next = new Node(2);
head.next.next = new Node(3);
head.next.next.next = new Node(4);
head.next.next.next.next = new Node(5);
System.out.print("Given Linked List:");
printList(head);
head = reverseList(head);
System.out.print("\nReversed Linked List:");
printList(head);
}
}
# Python program to reverse linked list using Stack
class Node:
def __init__(self, new_data):
self.data = new_data
self.next = None
def reverseList(head):
# Create a stack to store the nodes
stack = []
temp = head
# Push all nodes except the last node into stack
while temp.next is not None:
stack.append(temp)
temp = temp.next
# Make the last node as new head of the linked list
head = temp
# Pop all the nodes and append to the linked list
while stack:
# append the top value of stack in list
temp.next = stack.pop()
# move to the next node in the list
temp = temp.next
# Update the next pointer of last node
# of stack to None
temp.next = None
return head
def printList(node):
while node is not None:
print(f" {node.data}", end="")
node = node.next
print()
# Create a hard-coded linked list:
# 1 -> 2 -> 3 -> 4 -> 5
head = Node(1)
head.next = Node(2)
head.next.next = Node(3)
head.next.next.next = Node(4)
head.next.next.next.next = Node(5)
print("Given Linked List:", end="")
printList(head)
head = reverseList(head)
print("Reversed Linked List:", end="")
printList(head)
// C# program to reverse linked list using stack
using System;
using System.Collections.Generic;
class Node {
public int Data;
public Node Next;
public Node(int newData) {
Data = newData;
Next = null;
}
}
class GfG {
// Function to reverse the linked list
static Node ReverseList(Node head) {
// Create a stack to store the nodes
Stack<Node> stack = new Stack<Node>();
Node temp = head;
// Push all nodes except the last node into stack
while (temp.Next != null) {
stack.Push(temp);
temp = temp.Next;
}
// Make the last node as new head of the linked list
head = temp;
// Pop all the nodes and append to the linked list
while (stack.Count > 0) {
// append the top value of stack in list
temp.Next = stack.Pop();
// move to the next node in the list
temp = temp.Next;
}
// Update the next pointer of last node of stack to null
temp.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:
// 1 -> 2 -> 3 -> 4 -> 5
Node head = new Node(1);
head.Next = new Node(2);
head.Next.Next = new Node(3);
head.Next.Next.Next = new Node(4);
head.Next.Next.Next.Next = new Node(5);
Console.Write("Given Linked List:");
PrintList(head);
head = ReverseList(head);
Console.Write("\nReversed Linked List:");
PrintList(head);
}
}
// JavaScript program to reverse linked list using Stack
class Node {
constructor(newData) {
this.data = newData;
this.next = null;
}
}
// Function to reverse the linked list
function reverseList(head) {
// Create a stack to store the nodes
let stack = [];
let temp = head;
// Push all nodes except the last node into stack
while (temp.next !== null) {
stack.push(temp);
temp = temp.next;
}
// Make the last node as new head of the Linked List
head = temp;
// Pop all the nodes and append to the linked list
while (stack.length > 0) {
// append the top value of stack in list
temp.next = stack.pop();
// move to the next node in the list
temp = temp.next;
}
// Update the next pointer of last node of stack to null
temp.next = null;
return head;
}
function printList(node) {
while (node !== null) {
console.log(" " + node.data);
node = node.next;
}
console.log();
}
// Create a hard-coded linked list:
// 1 -> 2 -> 3 -> 4 -> 5
let head = new Node(1);
head.next = new Node(2);
head.next.next = new Node(3);
head.next.next.next = new Node(4);
head.next.next.next.next = new Node(5);
console.log("Given Linked List:");
printList(head);
// Function call to return the reversed list
head = reverseList(head);
console.log("\nReversed Linked List:");
printList(head);
OutputGiven Linked List: 1 2 3 4 5
Reversed Linked List: 5 4 3 2 1
Reverse a Linked List
Reverse a linked list iterative
Recursive Reverse A Linked List (Part 1)
Reverse a Linked List | DSA Problem
Similar Reads
Basics & Prerequisites
Data Structures
Array Data StructureIn this article, we introduce array, implementation in different popular languages, its basic operations and commonly seen problems / interview questions. An array stores items (in case of C/C++ and Java Primitive Arrays) or their references (in case of Python, JS, Java Non-Primitive) at contiguous
3 min read
String in Data StructureA string is a sequence of characters. The following facts make string an interesting data structure.Small set of elements. Unlike normal array, strings typically have smaller set of items. For example, lowercase English alphabet has only 26 characters. ASCII has only 256 characters.Strings are immut
2 min read
Hashing in Data StructureHashing is a technique used in data structures that efficiently stores and retrieves data in a way that allows for quick access. Hashing involves mapping data to a specific index in a hash table (an array of items) using a hash function. It enables fast retrieval of information based on its key. The
2 min read
Linked List Data StructureA linked list is a fundamental data structure in computer science. It mainly allows efficient insertion and deletion operations compared to arrays. Like arrays, it is also used to implement other data structures like stack, queue and deque. Here’s the comparison of Linked List vs Arrays Linked List:
2 min read
Stack Data StructureA Stack is a linear data structure that follows a particular order in which the operations are performed. The order may be LIFO(Last In First Out) or FILO(First In Last Out). LIFO implies that the element that is inserted last, comes out first and FILO implies that the element that is inserted first
2 min read
Queue Data StructureA Queue Data Structure is a fundamental concept in computer science used for storing and managing data in a specific order. It follows the principle of "First in, First out" (FIFO), where the first element added to the queue is the first one to be removed. It is used as a buffer in computer systems
2 min read
Tree Data StructureTree Data Structure is a non-linear data structure in which a collection of elements known as nodes are connected to each other via edges such that there exists exactly one path between any two nodes. Types of TreeBinary Tree : Every node has at most two childrenTernary Tree : Every node has at most
4 min read
Graph Data StructureGraph Data Structure is a collection of nodes connected by edges. It's used to represent relationships between different entities. If you are looking for topic-wise list of problems on different topics like DFS, BFS, Topological Sort, Shortest Path, etc., please refer to Graph Algorithms. Basics of
3 min read
Trie Data StructureThe Trie data structure is a tree-like structure used for storing a dynamic set of strings. It allows for efficient retrieval and storage of keys, making it highly effective in handling large datasets. Trie supports operations such as insertion, search, deletion of keys, and prefix searches. In this
15+ min read
Algorithms
Searching AlgorithmsSearching algorithms are essential tools in computer science used to locate specific items within a collection of data. In this tutorial, we are mainly going to focus upon searching in an array. When we search an item in an array, there are two most common algorithms used based on the type of input
2 min read
Sorting AlgorithmsA Sorting Algorithm is used to rearrange a given array or list of elements in an order. For example, a given array [10, 20, 5, 2] becomes [2, 5, 10, 20] after sorting in increasing order and becomes [20, 10, 5, 2] after sorting in decreasing order. There exist different sorting algorithms for differ
3 min read
Introduction to RecursionThe process in which a function calls itself directly or indirectly is called recursion and the corresponding function is called a recursive function. A recursive algorithm takes one step toward solution and then recursively call itself to further move. The algorithm stops once we reach the solution
14 min read
Greedy AlgorithmsGreedy algorithms are a class of algorithms that make locally optimal choices at each step with the hope of finding a global optimum solution. At every step of the algorithm, we make a choice that looks the best at the moment. To make the choice, we sometimes sort the array so that we can always get
3 min read
Graph AlgorithmsGraph is a non-linear data structure like tree data structure. The limitation of tree is, it can only represent hierarchical data. For situations where nodes or vertices are randomly connected with each other other, we use Graph. Example situations where we use graph data structure are, a social net
3 min read
Dynamic Programming or DPDynamic Programming is an algorithmic technique with the following properties.It is mainly an optimization over plain recursion. Wherever we see a recursive solution that has repeated calls for the same inputs, we can optimize it using Dynamic Programming. The idea is to simply store the results of
3 min read
Bitwise AlgorithmsBitwise algorithms in Data Structures and Algorithms (DSA) involve manipulating individual bits of binary representations of numbers to perform operations efficiently. These algorithms utilize bitwise operators like AND, OR, XOR, NOT, Left Shift, and Right Shift.BasicsIntroduction to Bitwise Algorit
4 min read
Advanced
Segment TreeSegment Tree is a data structure that allows efficient querying and updating of intervals or segments of an array. It is particularly useful for problems involving range queries, such as finding the sum, minimum, maximum, or any other operation over a specific range of elements in an array. The tree
3 min read
Pattern SearchingPattern searching algorithms are essential tools in computer science and data processing. These algorithms are designed to efficiently find a particular pattern within a larger set of data. Patten SearchingImportant Pattern Searching Algorithms:Naive String Matching : A Simple Algorithm that works i
2 min read
GeometryGeometry is a branch of mathematics that studies the properties, measurements, and relationships of points, lines, angles, surfaces, and solids. From basic lines and angles to complex structures, it helps us understand the world around us.Geometry for Students and BeginnersThis section covers key br
2 min read
Interview Preparation
Practice Problem