Group Shifted Strings

Given a string, we can “shift” each of its letter to its successive letter, for example: “abc” -> “bcd”. We can keep “shifting” which forms the sequence:

"abc" -> "bcd" -> ... -> "xyz" Given a list of strings which contains only lowercase alphabets, group all strings that belong to the same shifting sequence.

Example,

given: ["abc", "bcd", "acef", "xyz", "az", "ba", "a", "z"], Return:

[
  ["abc","bcd","xyz"],
  ["az","ba"],
  ["acef"],
  ["a","z"]
]

Note

For the return value, each inner list’s elements must follow the lexicographic order.

Think

• There is a regular pattern: if two words are shifted, the distance between two adjacent characters are the same.

    2   3        2   3
    /\ /\        /\ /\
   a  c  f - >  e  g  j

• So all we need to do is figure out the distance between each character in each word and make these distance as a code to represent the word.

Solution

public class GroupShiftedString {
    public List<List<String>> groupStrings(String[] strings) {
        List<List<String>> res = new ArrayList<>();
        HashMap<String, List<String>> map = new HashMap<>();
        for (String str : strings) {
            String code = getCode(str);
            List<String> val;
            if (!map.containsKey(code)) {
                val = new ArrayList<>();
            } else {
                val = map.get(code);
            }
            val.add(str);
            map.put(code, val);
        }

        for (Map.Entry<String, List<String>> entry : map.entrySet()) {
            List<String> val = entry.getValue();
            Collections.sort(val);
            res.add(val);
        }
        return res;
    }

    private String getCode(String s) {
        StringBuilder sb = new StringBuilder();
        sb.append("#");
        for (int i = 1; i < s.length(); i++) {
            int tmp = ((s.charAt(i) - s.charAt(i - 1)) + 26) % 26;
            sb.append(tmp).append("#");
        }
        return sb.toString();
    }
}