Binary TreesBinary Trees36
  1. 1Preorder Traversal of a Binary Tree using Recursion
  2. 2Preorder Traversal of a Binary Tree using Iteration
  3. 3Inorder Traversal of a Binary Tree using Recursion
  4. 4Inorder Traversal of a Binary Tree using Iteration
  5. 5Postorder Traversal of a Binary Tree Using Recursion
  6. 6Postorder Traversal of a Binary Tree using Iteration
  7. 7Level Order Traversal of a Binary Tree using Recursion
  8. 8Level Order Traversal of a Binary Tree using Iteration
  9. 9Reverse Level Order Traversal of a Binary Tree using Iteration
  10. 10Reverse Level Order Traversal of a Binary Tree using Recursion
  11. 11Find Height of a Binary Tree
  12. 12Find Diameter of a Binary Tree
  13. 13Find Mirror of a Binary Tree
  14. 14Left View of a Binary Tree
  15. 15Right View of a Binary Tree
  16. 16Top View of a Binary Tree
  17. 17Bottom View of a Binary Tree
  18. 18Zigzag Traversal of a Binary Tree
  19. 19Check if a Binary Tree is Balanced
  20. 20Diagonal Traversal of a Binary Tree
  21. 21Boundary Traversal of a Binary Tree
  22. 22Construct a Binary Tree from a String with Bracket Representation
  23. 23Convert a Binary Tree into a Doubly Linked List
  24. 24Convert a Binary Tree into a Sum Tree
  25. 25Find Minimum Swaps Required to Convert a Binary Tree into a BST
  26. 26Check if a Binary Tree is a Sum Tree
  27. 27Check if All Leaf Nodes are at the Same Level in a Binary Tree
  28. 28Lowest Common Ancestor (LCA) in a Binary Tree
  29. 29Solve the Tree Isomorphism Problem
  30. 30Check if a Binary Tree Contains Duplicate Subtrees of Size 2 or More
  31. 31Check if Two Binary Trees are Mirror Images
  32. 32Calculate the Sum of Nodes on the Longest Path from Root to Leaf in a Binary Tree
  33. 33Print All Paths in a Binary Tree with a Given Sum
  34. 34Find the Distance Between Two Nodes in a Binary Tree
  35. 35Find the kth Ancestor of a Node in a Binary Tree
  36. 36Find All Duplicate Subtrees in a Binary Tree
GraphsGraphs46
  1. 1Breadth-First Search in Graphs
  2. 2Depth-First Search in Graphs
  3. 3Number of Provinces in an Undirected Graph
  4. 4Connected Components in a Matrix
  5. 5Rotten Oranges Problem - BFS in Matrix
  6. 6Flood Fill Algorithm - Graph Based
  7. 7Detect Cycle in an Undirected Graph using DFS
  8. 8Detect Cycle in an Undirected Graph using BFS
  9. 9Distance of Nearest Cell Having 1 - Grid BFS
  10. 10Surrounded Regions in Matrix using Graph Traversal
  11. 11Number of Enclaves in Grid
  12. 12Word Ladder - Shortest Transformation using Graph
  13. 13Word Ladder II - All Shortest Transformation Sequences
  14. 14Number of Distinct Islands using DFS
  15. 15Check if a Graph is Bipartite using DFS
  16. 16Topological Sort Using DFS
  17. 17Topological Sort using Kahn's Algorithm
  18. 18Cycle Detection in Directed Graph using BFS
  19. 19Course Schedule - Task Ordering with Prerequisites
  20. 20Course Schedule 2 - Task Ordering Using Topological Sort
  21. 21Find Eventual Safe States in a Directed Graph
  22. 22Alien Dictionary Character Order
  23. 23Shortest Path in Undirected Graph with Unit Distance
  24. 24Shortest Path in DAG using Topological Sort
  25. 25Dijkstra's Algorithm Using Set - Shortest Path in Graph
  26. 26Dijkstra’s Algorithm Using Priority Queue
  27. 27Shortest Distance in a Binary Maze using BFS
  28. 28Path With Minimum Effort in Grid using Graphs
  29. 29Cheapest Flights Within K Stops - Graph Problem
  30. 30Number of Ways to Reach Destination in Shortest Time - Graph Problem
  31. 31Minimum Multiplications to Reach End - Graph BFS
  32. 32Bellman-Ford Algorithm for Shortest Paths
  33. 33Floyd Warshall Algorithm for All-Pairs Shortest Path
  34. 34Find the City With the Fewest Reachable Neighbours
  35. 35Minimum Spanning Tree in Graphs
  36. 36Prim's Algorithm for Minimum Spanning Tree
  37. 37Disjoint Set (Union-Find) with Union by Rank and Path Compression
  38. 38Kruskal's Algorithm - Minimum Spanning Tree
  39. 39Minimum Operations to Make Network Connected
  40. 40Most Stones Removed with Same Row or Column
  41. 41Accounts Merge Problem using Disjoint Set Union
  42. 42Number of Islands II - Online Queries using DSU
  43. 43Making a Large Island Using DSU
  44. 44Bridges in Graph using Tarjan's Algorithm
  45. 45Articulation Points in Graphs
  46. 46Strongly Connected Components using Kosaraju's Algorithm

Check if One String is a Rotation of Another

Problem Statement

Given two strings s1 and s2, your task is to check whether s2 is a rotation of s1.

  • A string s2 is considered a rotation of s1 if you can take some characters from the start of s1 and move them to the end, in the same order, to get s2.
  • For example, "erbottlewat" is a rotation of "waterbottle" because you can move "water" to the end to get "bottlewater" and then adjust the characters to match.

Return true if s2 is a rotation of s1, else return false.

Examples

s1 s2 Output Description
"waterbottle" "erbottlewat" true s2 is a rotation of s1
"hello" "lohel" true Characters rotated from front to back
"abcd" "cdab" true Valid rotation ("ab" moved to the end)
"abcd" "acbd" false Characters are shuffled, not rotated
"abc" "abcd" false Different lengths, cannot be rotations
"abcd" "abce" false Same length, but different characters
"" "" true Both strings are empty — considered valid rotation
"" "a" false Empty string cannot rotate to form a non-empty string
"a" "" false Non-empty string cannot match an empty one

Visualization Player

Solution

To check whether one string is a rotation of another, we can use a very simple and efficient trick that takes advantage of string concatenation.

Imagine we take s1 and rotate its characters in some order. Any such rotation will always be a substring of s1 + s1 (that is, s1 repeated twice).

Example: Let’s say s1 = "waterbottle" and s2 = "erbottlewat". If we concatenate s1 with itself, we get "waterbottlewaterbottle". If s2 is truly a rotation of s1, it will appear somewhere in this doubled string. And in this case, it does!

Different Cases Explained:

  • Case 1: Same strings – If s1 and s2 are exactly the same, then s2 is a rotation of s1 (rotation by 0 characters).
  • Case 2: Valid rotation – If s2 can be formed by rotating s1, such as moving some characters from the start of s1 to the end, it will appear in s1 + s1.
  • Case 3: Different lengths – If the lengths of s1 and s2 are not equal, then s2 cannot be a rotation of s1.
  • Case 4: Same length, different characters – If the lengths are equal but the characters don’t align after any rotation, then it’s not a valid rotation.
  • Case 5: Empty strings – Two empty strings are considered valid rotations of each other because there’s nothing to rotate.
  • Case 6: One string empty – If only one of the strings is empty, then it’s never a valid rotation of the other.

This method is not only simple but also highly efficient because checking for a substring is a fast operation in modern programming languages.

Time Complexity: O(n) where n is the length of the strings (assuming efficient substring check).
Space Complexity: O(n) due to the concatenated string.

Algorithm Steps

  1. Check if lengths of s1 and s2 are equal. If not, return false.
  2. Concatenate s1 with itself to create a new string temp = s1 + s1.
  3. Check if s2 is a substring of temp.
  4. If yes, return true; else, return false.

Code

Python
Java
JavaScript
C
C++
C#
Kotlin
Swift
Go
Php
def is_rotation(s1, s2):
    if len(s1) != len(s2):
        return False  # If lengths differ, cannot be rotation
    temp = s1 + s1    # Concatenate string with itself
    return s2 in temp # Check if s2 is a substring

# Sample Input
s1 = "abcde"
s2 = "deabc"
print("Is Rotation:", is_rotation(s1, s2))

Time Complexity

CaseTime ComplexityExplanation
Best CaseO(n)We check the length and then perform a substring search which in most cases is linear time.
Average CaseO(n)Checking if one string is in another (substring search) typically takes linear time in practice.
Worst CaseO(n^2)In the worst case, depending on implementation of the `in` or `contains` function, substring search can be quadratic.

Space Complexity

O(n)

Explanation: We create a temporary string which is double the size of s1, so extra space used is proportional to n.