249. Group Shifted Strings 🔒

Description

Perform the following shift operations on a string:

Right shift: Replace every letter with the successive letter of the English alphabet, where 'z' is replaced by 'a'. For example, "abc" can be right-shifted to "bcd" or "xyz" can be right-shifted to "yza".
Left shift: Replace every letter with the preceding letter of the English alphabet, where 'a' is replaced by 'z'. For example, "bcd" can be left-shifted to "abc" or "yza" can be left-shifted to "xyz".

We can keep shifting the string in both directions to form an endless shifting sequence.

For example, shift "abc" to form the sequence: ... <-> "abc" <-> "bcd" <-> ... <-> "xyz" <-> "yza" <-> .... <-> "zab" <-> "abc" <-> ...

You are given an array of strings strings, group together 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.

Solutions

Solution 1: Hash Table

We use a hash table $g$ to store each string after shifting and with the first character as 'a'. That is, $g[t]$ represents the set of all strings that become $t$ after shifting.

We iterate through each string. For each string, we calculate its shifted string $t$, and then add it to $g[t]$.

Finally, we take out all the values in $g$, which is the answer.

The time complexity is $O(L)$ and the space complexity is $O(L)$, where $L$ is the sum of the lengths of all strings.

Python3JavaC++Go

class Solution:
    def groupStrings(self, strings: List[str]) -> List[List[str]]:
        g = defaultdict(list)
        for s in strings:
            diff = ord(s[0]) - ord("a")
            t = []
            for c in s:
                c = ord(c) - diff
                if c < ord("a"):
                    c += 26
                t.append(chr(c))
            g["".join(t)].append(s)
        return list(g.values())

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

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

func groupStrings(strings []string) [][]string {
    g := make(map[string][]string)
    for _, s := range strings {
        t := []byte(s)
        diff := t[0] - 'a'
        for i := range t {
            t[i] -= diff
            if t[i] < 'a' {
                t[i] += 26
            }
        }
        g[string(t)] = append(g[string(t)], s)
    }
    ans := make([][]string, 0, len(g))
    for _, v := range g {
        ans = append(ans, v)
    }
    return ans
}

249. Group Shifted Strings 🔒

Description

Solutions

Solution 1: Hash Table

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