Rotate String - Practice Coding Problems

Rotate String - Problem

Given two strings s and goal, determine if string s can be transformed into goal through a series of left rotations.

A left rotation operation moves the leftmost character of a string to the rightmost position. For example:

"abcde" → "bcdea" (after 1 rotation)
"abcde" → "cdeab" (after 2 rotations)
"abcde" → "deabc" (after 3 rotations)

Return true if s can become goal after some number of rotations, false otherwise.

Note: Both strings must have the same length for a valid transformation to be possible.

Input & Output

example_1.py — Basic Rotation

$ Input: s = "abcde", goal = "cdeab"

› Output: true

💡 Note: After 2 left rotations: "abcde" → "bcdea" → "cdeab". The goal string is achievable.

example_2.py — No Valid Rotation

$ Input: s = "abcde", goal = "abced"

› Output: false

💡 Note: No amount of rotations can transform "abcde" into "abced" because the character order is different (e and d are swapped).

example_3.py — Same String

$ Input: s = "abcde", goal = "abcde"

› Output: true

💡 Note: The goal is the same as the original string (0 rotations needed).

Visualization

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Understanding the Visualization

Original String

Start with string s arranged in a circle

Double the String

Create s + s to capture all rotations linearly

Search Pattern

Look for goal as a substring in the doubled string

Result

If found, goal is a valid rotation of s

Key Takeaway

🎯 Key Insight: Any rotation of a string appears as a substring in the string concatenated with itself (s + s)

Time & Space Complexity

Time Complexity

⏱️

O(n)

String concatenation O(n) + substring search O(n) using KMP or similar

✓ Linear Growth

Space Complexity

O(n)

Space for the concatenated string s + s

⚡ Linearithmic Space

Constraints

1 ≤ s.length, goal.length ≤ 100
s and goal consist of lowercase English letters
Both strings must have equal length for a valid transformation

Asked in

G Google 25 a Amazon 18 ⊞ Microsoft 15 f Meta 12

The optimal solution uses the key insight that any rotation of string s appears as a substring in s + s. Simply check if goal is contained in the concatenated string s + s after verifying equal lengths. This achieves O(n) time complexity with elegant simplicity.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Try All Rotations)	O(n²)	O(n)	Generate all possible rotations and check if any matches the goal
String Concatenation (Optimal)	O(n)	O(n)	Check if goal is a substring of s + s

Brute Force (Try All Rotations) — Algorithm Steps

Check if lengths are equal (early return if not)
For each rotation position i from 0 to n-1:
Create rotated string by taking substring from i to end + substring from 0 to i
Compare rotated string with goal
Return true if match found, false otherwise

Visualization

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Step-by-Step Walkthrough

Original String

Start with string s = 'abcde'

Rotation 1

Move first char to end: 'bcdea'

Rotation 2

Continue rotating: 'cdeab'

Compare

Check if current rotation matches goal

Code -

solution.c — C

#include <stdio.h>
#include <string.h>
#include <stdbool.h>

bool rotateString(char* s, char* goal) {
    int len_s = strlen(s);
    int len_goal = strlen(goal);
    
    if (len_s != len_goal) {
        return false;
    }
    
    char rotated[201];
    for (int i = 0; i < len_s; i++) {
        strcpy(rotated, s + i);
        strncat(rotated, s, i);
        rotated[len_s] = '\0';
        if (strcmp(rotated, goal) == 0) {
            return true;
        }
    }
    
    return false;
}

int main() {
    char s[101], goal[101];
    scanf("%s %s", s, goal);
    printf("%s\n", rotateString(s, goal) ? "true" : "false");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

O(n) rotations × O(n) string comparison and creation

⚠ Quadratic Growth

Space Complexity

O(n)

Space for creating rotated strings

⚡ Linearithmic Space

Constraints

1 ≤ s.length, goal.length ≤ 100
s and goal consist of lowercase English letters
Both strings must have equal length for a valid transformation

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Smart Actions

💡 Explanation

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Input & Output

Visualization

Time & Space Complexity

Related Problems

Constraints

Common Approaches

Brute Force (Try All Rotations) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Constraints

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