Check if One String is a Rotation of Another

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ProblemExamplesVisualizationSolutionAlgorithmCodeTimeSpace

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.