Day 7 - Strings Rotations of Each Other.md

Difficulty	Source	Tags
Easy	160 Days of Problem Solving	Strings

🚀 Day 7. Strings Rotations of Each Other 🧠

The problem can be found at the following link: Problem Link

💡 Problem Description:

You are given two strings of equal lengths, s1 and s2. The task is to check if s2 is a rotated version of s1.
Rotation means that the string s2 can be formed by shifting the characters of s1 cyclically.

Note: The characters in the strings are in lowercase.

🔍 Example Walkthrough:

Input:
s1 = "abcd", s2 = "cdab"
Output:
true

Explanation:
After two right rotations, s1 becomes equal to s2.

Input:
s1 = "aab", s2 = "aba"
Output:
true

Explanation:
After one left rotation, s1 becomes equal to s2.

Input:
s1 = "abcd", s2 = "acbd"
Output:
false

Explanation:
The strings are not rotations of each other.

Constraints:

• $(1 \leq \text{s1.size(), s2.size()} \leq 10^5)$

🎯 My Approach:

1. String Concatenation and Substring Check:

   • The solution relies on the fact that a rotated version of s1 will always be a substring of s1 + s1.
   • By concatenating s1 with itself, all possible rotations of s1 are included in the resulting string.
   • Then, we check if s2 is a substring of the concatenated string.

2. Edge Cases:

   • If the lengths of s1 and s2 are different, they cannot be rotations.
   • Strings with only one character are trivially rotations if they are equal.

3. Steps:

   • Check if the lengths of s1 and s2 are equal.
   • Concatenate s1 with itself to create a new string.
   • Use efficient substring search techniques like strstr or KMP to check if s2 is present in the concatenated string.

🕒 Time and Auxiliary Space Complexity

• Expected Time Complexity: (O(n)), where (n) is the length of the string. The concatenation takes (O(n)), and the substring search also runs in (O(n)) using algorithms like KMP.
• Expected Auxiliary Space Complexity: (O(n)), as concatenating the string requires additional space for storing (s1 + s1).

📝 Solution Code

Code (C)

int areRotations(char* s1, char* s2) {
    if (strlen(s1) != strlen(s2)) return 0;
    char* temp = (char*)malloc(2 * strlen(s1) + 1);
    strcpy(temp, s1);
    strcat(temp, s1);
    int result = (strstr(temp, s2) != NULL);

    free(temp); 
    return result;
}

Code (Cpp)

class Solution {
  public:
    bool areRotations(string &s1, string &s2) {
        if (s1.length() != s2.length()) return false;
        string temp = s1 + s1;
        return (temp.find(s2) != string::npos);
    }
};

Code (Java)

class Solution {
    public static boolean areRotations(String s1, String s2) {
        if (s1.length() != s2.length()) return false;
        String temp = s1 + s1;
        return kmpSearch(temp, s2);
    }

    private static boolean kmpSearch(String text, String pattern) {
        int[] lps = computeLPSArray(pattern); 
        int i = 0, j = 0;

        while (i < text.length()) {
            if (text.charAt(i) == pattern.charAt(j)) {
                i++;
                j++;

                if (j == pattern.length()) {
                    return true; 
                }
            } else {
                if (j != 0) {
                    j = lps[j - 1];
                } else {
                    i++;
                }
            }
        }
        return false;
    }
    private static int[] computeLPSArray(String pattern) {
        int[] lps = new int[pattern.length()];
        int len = 0, i = 1;

        while (i < pattern.length()) {
            if (pattern.charAt(i) == pattern.charAt(len)) {
                len++;
                lps[i] = len;
                i++;
            } else {
                if (len != 0) {
                    len = lps[len - 1];
                } else {
                    lps[i] = 0;
                    i++;
                }
            }
        }
        return lps;
    }
}

Code (Python)

class Solution:
    def areRotations(self, s1, s2):
        if len(s1) != len(s2):
            return False
        temp = s1 + s1
        return s2 in temp

🎯 Contribution and Support:

For discussions, questions, or doubts related to this solution, feel free to connect on LinkedIn: Any Questions. Let’s make this learning journey more collaborative!

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