# Group Shifted String

Given an array of strings (all lowercase letters), the task is to group them in such a way that all strings in a group are shifted versions of each other. Two string S and T are called shifted if,

```S.length = T.length
and
S[i] = T[i] + K for
1 <= i <= S.length  for a constant integer K
```

For example strings {acd, dfg, wyz, yab, mop} are shifted versions of each other.

```Input  : str[] = {"acd", "dfg", "wyz", "yab", "mop",
"bdfh", "a", "x", "moqs"};

Output : a x
acd dfg wyz yab mop
bdfh moqs
All shifted strings are grouped together.
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

We can see a pattern among string of one group, the difference between consecutive characters for all character of string are equal. As in above example take acd, dfg and mop

a c d -> 2 1
d f g -> 2 1
m o p -> 2 1

Since the differences are same, we can use this to identify strings that belong to same group. The idea is to form a string of differences as key. If a string with same difference string is found, then this string also belongs to same group. For example, above three strings have same difference string, that is “21”.

In below implementation, we add ‘a’ to every difference and store 21 as “ba”.

 `/* C/C++ program to print groups of shifted strings ` `   ``together. */` `#include ` `using` `namespace` `std; ` `const` `int` `ALPHA = 26;   ``// Total lowercase letter ` ` `  `// Method to a difference string for a given string. ` `// If string is "adf" then difference string will be ` `// "cb" (first difference 3 then difference 2) ` `string getDiffString(string str) ` `{ ` `    ``string shift = ``""``; ` `    ``for` `(``int` `i = 1; i < str.length(); i++) ` `    ``{ ` `        ``int` `dif = str[i] - str[i-1]; ` `        ``if` `(dif < 0) ` `            ``dif += ALPHA; ` ` `  `        ``// Representing the difference as char ` `        ``shift += (dif + ``'a'``); ` `    ``} ` ` `  `    ``// This string will be 1 less length than str ` `    ``return` `shift; ` `} ` ` `  `// Method for grouping shifted string ` `void` `groupShiftedString(string str[], ``int` `n) ` `{ ` `    ``// map for storing indices of string which are ` `    ``// in same group ` `    ``map< string, vector<``int``> > groupMap; ` `    ``for` `(``int` `i = 0; i < n; i++) ` `    ``{ ` `        ``string diffStr = getDiffString(str[i]); ` `        ``groupMap[diffStr].push_back(i); ` `    ``} ` ` `  `    ``// iterating through map to print group ` `    ``for` `(``auto` `it=groupMap.begin(); it!=groupMap.end(); ` `                                                ``it++) ` `    ``{ ` `        ``vector<``int``> v = it->second; ` `        ``for` `(``int` `i = 0; i < v.size(); i++) ` `            ``cout << str[v[i]] << ``" "``; ` `        ``cout << endl; ` `    ``} ` `} ` ` `  `// Driver method to test above methods ` `int` `main() ` `{ ` `    ``string str[] = {``"acd"``, ``"dfg"``, ``"wyz"``, ``"yab"``, ``"mop"``, ` `                    ``"bdfh"``, ``"a"``, ``"x"``, ``"moqs"` `                   ``}; ` `    ``int` `n = ``sizeof``(str)/``sizeof``(str); ` `    ``groupShiftedString(str, n); ` `    ``return` `0; ` `} `

Output:

```a x
acd dfg wyz yab mop
bdfh moqs
```

This article is contributed by Utkarsh Trivedi. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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