Group Shifted Strings

Posted 2020-08-11 amazingzoe

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.

For example, given: ["abc", "bcd", "acef", "xyz", "az", "ba", "a", "z"], 
A solution is:

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

 

Analyse: keep the diff between each character in the string. Put the strings with same diff in the same vector.

Runtime: 6ms

 1 class Solution {
 2 public:
 3     vector<vector<string>> groupStrings(vector<string>& strings) {
 4         unordered_map<string, vector<string> > groups;
 5         for (string s : strings) {
 6             if (s.size() == 1) {
 7                 groups["-"].push_back(s);
 8             } else {
 9                 string tempGap;
10                 for (int i = 1; i < s.size(); i++) {
11                     if (s[i] > s[i - 1])
12                         tempGap += to_string(s[i] - s[i - 1]);
13                     else
14                         tempGap += to_string(26 + (s[i] - s[i - 1]));
15                 }
16                 groups[tempGap].push_back(s);
17             }
18         }
19         
20         vector<vector<string> > result;
21         for (auto ite = groups.begin(); ite != groups.end(); ite++) {
22             vector<string> tempStrings = ite->second;
23             result.push_back(vector<string>());
24             for (auto s : tempStrings) {
25                 result.back().push_back(s);
26             }
27         }
28         return result;
29     }
30 };