Rotate String - Practice Coding Problems

Rotate String - Problem

Given two strings s and goal, return true if and only if s can become goal after some number of shifts on s.

A shift on s consists of moving the leftmost character of s to the rightmost position. For example, if s = "abcde", then it will be "bcdea" after one shift.

Input & Output

Example 1 — Basic Rotation

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

› Output: true

💡 Note: We can rotate "abcde" by moving "ab" from front to back: "abcde" → "cdeab"

Example 2 — No Valid Rotation

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

› Output: false

💡 Note: No amount of rotation can transform "abcde" to "abced" because they have different character arrangements

Example 3 — Same String

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

› Output: true

💡 Note: The strings are identical, so goal is a rotation of s (0 rotations needed)

Constraints

1 ≤ s.length, goal.length ≤ 100
s and goal consist of lowercase English letters

Visualization

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Understanding the Visualization

Input

Two strings s and goal

Process

Check if goal is a rotation of s

Output

Return true if rotation possible, false otherwise

Key Takeaway

🎯 Key Insight: If goal is a rotation of s, then goal must appear as substring in s+s

Asked in

G Google 25 a Amazon 18 M Microsoft 12

The key insight is that if goal is a rotation of s, then goal must appear as a substring in s+s. Best approach is concatenation check with Time: O(n), Space: O(n)

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force Rotation	O(n²)	O(n)	Try all possible rotations of string s to see if any matches goal
Concatenation Check	O(n)	O(n)	Check if goal exists as substring in s+s (doubled string)

Brute Force Rotation — Algorithm Steps

Check if lengths are equal (if not, return false)
For each position i from 0 to n-1, create rotation
Compare each rotation with goal string

Visualization

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Step-by-Step Walkthrough

Original

Start with string s = "abcde"

Rotate

Try each rotation: "bcdea", "cdeab", etc.

Compare

Check if any rotation matches goal

Code -

solution.c — C

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

int solution(char* s, char* goal) {
    int sLen = strlen(s);
    int goalLen = strlen(goal);
    
    if (sLen != goalLen) {
        return 0;
    }
    
    char* rotated = malloc((sLen + 1) * sizeof(char));
    for (int i = 0; i < sLen; i++) {
        strcpy(rotated, s + i);
        strncat(rotated, s, i);
        rotated[sLen] = '\0';
        
        if (strcmp(rotated, goal) == 0) {
            free(rotated);
            return 1;
        }
    }
    
    free(rotated);
    return 0;
}

int main() {
    char s[1001], goal[1001];
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0;
    fgets(goal, sizeof(goal), stdin);
    goal[strcspn(goal, "\n")] = 0;
    
    int result = solution(s, goal);
    printf(result ? "true\n" : "false\n");
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

We perform n rotations, each taking O(n) time to create substring

⚠ Quadratic Growth

Space Complexity

O(n)

Each rotation creates a new string of length n

⚡ Linearithmic Space

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Medium Frequency

~15 min Avg. Time

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force Rotation — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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