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.

How to Solve

The relative distance between each letter of the string and the first character is equal.

For example, abc and efg are mutually offset. For abc, the distance between b and a is 1, and the distance between c and a is 2. For efg, the distance between f and e is 1, and the distance between g and e is 2.

Let’s look at another example. The distance between az and yx, z and a is 25, and the distance between x and y is also 25 (direct subtraction is -1, which is the reason for adding 26 and taking the remainder).

Then, in this case, all strings that are offset from each other have a unique distance difference. According to this, the mapping can be well grouped.

Test Cases

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.

Solution

        
class Solution {
public:
    vector<vector<string>> groupStrings(vector<string>& strings) {
        unordered_map<string, vector<string>> mp;
        for (auto& s : strings) {
            int diff = s[0] - 'a';
            string t = s;
            for (int i = 0; i < t.size(); ++i) {
                char d = t[i] - diff;
                if (d < 'a') d += 26;
                t[i] = d;
            }
            cout << t << endl;
            mp[t].push_back(s);
        }
        vector<vector<string>> ans;
        for (auto& e : mp)
            ans.push_back(e.second);
        return ans;
    }
};

    

        
func groupStrings(strings []string) [][]string {
	mp := make(map[string][]string)
	for _, s := range strings {
		k := ""
		for i := range s {
			k += string((s[i]-s[0]+26)%26 + 'a')
		}
		mp[k] = append(mp[k], s)
	}
	var ans [][]string
	for _, v := range mp {
		ans = append(ans, v)
	}
	return ans
}

    

        
class Solution {
    public List<List<String>> groupStrings(String[] strings) {
        Map<String, List<String>> mp = new HashMap<>();
        for (String s : strings) {
            int diff = s.charAt(0) - 'a';
            char[] t = s.toCharArray();
            for (int i = 0; i < t.length; ++i) {
                char d = (char) (t[i] - diff);
                if (d < 'a') {
                    d += 26;
                }
                t[i] = d;
            }
            String key = new String(t);
            mp.computeIfAbsent(key, k -> new ArrayList<>()).add(s);
        }
        return new ArrayList<>(mp.values());
    }
}

    

        
from collections import defaultdict

class Solution:
    def groupStrings(self, strings: List[str]) -> List[List[str]]:
        mp = defaultdict(list)
        for s in strings:
            t = []
            diff = ord(s[0]) - ord('a')
            for c in s:
                d = ord(c) - diff
                if d < ord('a'):
                    d += 26
                t.append(chr(d))
            k = ''.join(t)
            mp[k].append(s)
        return list(mp.values())

    

Group Shifted Strings

Topics

Companies

Platforms

How to Solve

Test Cases

Solution

Time Complexity: O(n)

Space Complexity: O(n)