Group Shifted Strings

 We can shift a string by shifting each of its letters to its successive letter.

• For example, "abc" can be shifted to be "bcd".

We can keep shifting the string to form a sequence.

• For example, we can keep shifting "abc" to form the sequence: "abc" -> "bcd" -> ... -> "xyz".

Given an array of strings strings, group all strings[i] that belong to the same shifting sequence. You may return the answer in any order.

 

Example 1:

Input: strings = ["abc","bcd","acef","xyz","az","ba","a","z"]
Output: [["acef"],["a","z"],["abc","bcd","xyz"],["az","ba"]]

Example 2:

Input: strings = ["a"]
Output: [["a"]]

 

Constraints:

• 1 <= strings.length <= 200
• 1 <= strings[i].length <= 50
• strings[i] consists of lowercase English letters.

class Solution {

public:

    vector<vector<string>> groupStrings(vector<string>& strings) {

        // base cases later

        unordered_map<string, vector<string>> mp;

        vector<vector<string>> ans;

        for (auto str : strings) {

            mp[Hash(str)].push_back(str);

        }

        for (auto entry : mp) {

            ans.push_back(entry.second);

        }

        

        return ans;

    }

    

    string Hash(string str) {

        int size = str.size();

        if (size == 0) {

            return "-";

        }

        if (size == 1) {

            return "";

        }

        string code;

        for (int i = 0; i < size - 1; i++) {

            int diff = str[i + 1] - str[i];

            diff = (diff + 26) % 26;

            if (code.empty()) {

                code = to_string(diff);

            } else {

                code += "-" + to_string(diff);

            }

        }

        return code;

    }

};