DSA interview question

Rotate String

Easy · Algorithms · Topics: string

The problem

A marquee sign scrolls text by moving its first character to the end, over and over. Given the current text s and a target goal, can the sign ever show goal? Trying every rotation is O(n²). The elegant observation: every rotation of s appears as a substring of s + s. For example, "abcdeabcde" contains "cdeab". So the answer is "same length, and goal occurs in s + s", which is a single substring search.

Example: ("abcde", "cdeab") → true; ("abcde", "abced") → false.

Function signature

bool rotateString(const string& s, const string& goal);

Test cases

Input `(s, goal)`	Output	Explanation
`("abcde", "cdeab")`	true	Rotate by 2
`("abcde", "abced")`	false	Not a rotation
`("aa", "a")`	false	Different lengths
`("a", "a")`	true	Rotation by 0
`("abab", "baba")`	true	Rotate by 1

(Confirm: a rotation moves characters from the front to the back; zero rotations count.)

How interviewers phrase it

"Return true if s can become goal after some number of shifts" (LeetCode 796).
"Check whether one string is a rotation of another, using only one call to a substring function" (a classic).

Clarifying questions to ask

Do the strings have to be the same length? (If not, false.)
Does rotating by 0 count?
Is the character set limited?

Follow-ups to expect

"Why does s + s contain every rotation?"
"Why check the lengths first?"
"How fast is the substring search?"
"What's the brute force?"
"Complexity?"

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