Binary TreesBinary Trees36

  1. 1Preorder Traversal of a Binary Tree using Recursion
  2. 2Preorder Traversal of a Binary Tree using Iteration
  3. 3Postorder Traversal of a Binary Tree Using Recursion
  4. 4Postorder Traversal of a Binary Tree using Iteration
  5. 5Level Order Traversal of a Binary Tree using Recursion
  6. 6Level Order Traversal of a Binary Tree using Iteration
  7. 7Reverse Level Order Traversal of a Binary Tree using Iteration
  8. 8Reverse Level Order Traversal of a Binary Tree using Recursion
  9. 9Find Height of a Binary Tree
  10. 10Find Diameter of a Binary Tree
  11. 11Find Mirror of a Binary Tree - Todo
  12. 12Inorder Traversal of a Binary Tree using Recursion
  13. 13Inorder Traversal of a Binary Tree using Iteration
  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

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.

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

Examples

s1s2OutputDescription
waterbottleerbottlewattrues2 is a rotation of s1
helloloheltrueCharacters rotated from front to back
abcdcdabtrueValid rotation ("ab" moved to the end)
abcdacbdfalseCharacters are shuffled, not rotated
abcabcdfalseDifferent lengths, cannot be rotations
abcdabcefalseSame length, but different characters
''''trueBoth strings are empty — considered valid rotation
''afalseEmpty string cannot rotate to form a non-empty string
a''falseNon-empty string cannot match an empty one

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.

Visualization

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.
Average 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.



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