# Lexicographically next greater string using same character set

Given a number K and a string S, We have to Find the lexicographically smallest string str of length K such that it’s set of letters is a subset of the set of letters of S and S is lexicographically smaller than str.

Examples:

```Input :k = 3
s = zbf
Output: zbz
Explanation: zbz is greater than zbf and it is
smaller than any other lexicographically greater
string than zbf

Input :k = 3
s = gi
Output: gig
Explanation: gig > gi and size is 3.
```

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

Approach: If size of string is less than k, we should simply add k – s.size() minimum symbols from s.
If the size of the string is greater then or equal to k then we need to replace all symbols in the suffix of first k symbols of string smallest symbols and replace the character before this suffix with next greater character existing in the string.
Note: If k – 1 index in string holds the greatest character then we move one index back and we move back until we find a character which is not equal to the greatest character.

## C++

 `// C++ implementation of above algorithm. ` `#include ` `using` `namespace` `std; ` ` `  `// function to print output ` `void` `lexoString(string s, ``int` `k) ` `{ ` `     `  `    ``int` `n = s.size(); ` `     `  `    ``// to store unique characters of the string ` `    ``vector<``char``> v; ` `     `  `    ``// to check uniqueness ` `    ``map<``char``, ``int``> mp;  ` `     `  `    ``for` `(``int` `i = 0; i < s.size(); i++) { ` `         `  `        ``if` `(mp[s[i]] == 0) { ` `             `  `            ``// if mp[s[i]] = 0 then it is ` `            ``// first time ` `            ``mp[s[i]] = 1; ` `            ``v.push_back(s[i]); ` `        ``} ` `    ``} ` `     `  `    ``// sort the unique characters ` `    ``sort(v.begin(), v.end()); ` `     `  `    ``// simply add n-k smallest characters ` `    ``if` `(k > n) ` `    ``{ ` `        ``cout << s; ` `        ``for` `(``int` `i = n; i < k; i++) { ` `            ``cout << v[0]; ` `        ``} ` `         `  `        ``return``; ``// end the program ` `    ``} ` `     `  `    ``// searching the first character left of ` `    ``// index k and not equal to greatest ` `    ``// character of the string ` `    ``for` `(``int` `i = k - 1; i >= 0; i--) { ` `        ``if` `(s[i] != v[v.size() - 1]) { ` `            ``for` `(``int` `j = 0; j < i; j++) { ` `                ``cout << s[j]; ` `            ``} ` `             `  `            ``// finding the just next greater ` `            ``// character than s[i] ` `            ``for` `(``int` `j = 0; j < v.size(); j++) { ` `                ``if` `(v[j] > s[i]) { ` `                    ``cout << v[j]; ` `                    ``break``; ` `                ``} ` `            ``} ` `             `  `            ``// suffix with smallest character ` `            ``for` `(``int` `j = i + 1; j < k; j++)  ` `                ``cout << v[0];          ` `             `  `            ``return``; ` `        ``} ` `    ``} ` `     `  `    ``// if we reach here then all indices to the left  ` `    ``// of k had the greatest character ` `    ``cout << ``"No lexicographically greater string of length "` `         ``<< k << ``" possible here."``; ` `     `  `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``string s = ``"gi"``; ` `    ``int` `k = 3; ` ` `  `    ``// Function call ` `    ``lexoString(s, k); ` `    ``return` `0; ` `} `

## Java

 `// Java implementation of above algorithm. ` `import` `java.util.*; ` ` `  `class` `GFG  ` `{ ` ` `  `    ``// function to print output ` `    ``static` `void` `lexoString(``char``[] s, ``int` `k)  ` `    ``{ ` ` `  `        ``int` `n = s.length; ` ` `  `        ``// to store unique characters of the string ` `        ``Vector v = ``new` `Vector<>(); ` ` `  `        ``// to check uniqueness ` `        ``Map mp = ``new` `HashMap<>(); ` ` `  `        ``for` `(``int` `i = ``0``; i < s.length; i++)  ` `        ``{ ` ` `  `            ``if` `(!mp.containsKey(s[i])) ` `            ``{ ` ` `  `                ``// if mp[s[i]] = 0 then it is ` `                ``// first time ` `                ``mp.put(s[i], ``1``); ` `                ``v.add(s[i]); ` `            ``} ` `        ``} ` ` `  `        ``// sort the unique characters ` `        ``Collections.sort(v); ` ` `  `        ``// simply add n-k smallest characters ` `        ``if` `(k > n) ` `        ``{ ` `            ``System.out.print(String.valueOf(s)); ` `            ``for` `(``int` `i = n; i < k; i++)  ` `            ``{ ` `                ``System.out.print(v.get(``0``)); ` `            ``} ` ` `  `            ``return``; ``// end the program ` `        ``} ` ` `  `        ``// searching the first character left of ` `        ``// index k and not equal to greatest ` `        ``// character of the string ` `        ``for` `(``int` `i = k - ``1``; i >= ``0``; i--)  ` `        ``{ ` `            ``if` `(s[i] != v.get(v.size() - ``1``))  ` `            ``{ ` `                ``for` `(``int` `j = ``0``; j < i; j++)  ` `                ``{ ` `                    ``System.out.print(s[j]); ` `                ``} ` ` `  `                ``// finding the just next greater ` `                ``// character than s[i] ` `                ``for` `(``int` `j = ``0``; j < v.size(); j++)  ` `                ``{ ` `                    ``if` `(v.get(j) > s[i])  ` `                    ``{ ` `                        ``System.out.print(v.get(j)); ` `                        ``break``; ` `                    ``} ` `                ``} ` ` `  `                ``// suffix with smallest character ` `                ``for` `(``int` `j = i + ``1``; j < k; j++)  ` `                ``{ ` `                    ``System.out.print(v.get(``0``)); ` `                ``} ` ` `  `                ``return``; ` `            ``} ` `        ``} ` ` `  `        ``// if we reach here then all indices to the left  ` `        ``// of k had the greatest character ` `        ``System.out.print(``"No lexicographically greater string of length "` `                ``+ k + ``" possible here."``); ` ` `  `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String arr[])  ` `    ``{ ` `        ``String s = ``"gi"``; ` `        ``int` `k = ``3``; ` ` `  `        ``// Function call ` `        ``lexoString(s.toCharArray(), k); ` `    ``} ` `} ` ` `  `// This code contributed by Rajput-Ji `

## Python3

 `# Python 3 implementation of above algorithm. ` ` `  `# function to print output ` `def` `lexoString(s, k): ` `    ``n ``=` `len``(s) ` `     `  `    ``# to store unique characters ` `    ``# of the string ` `    ``v ``=` `[] ` `     `  `    ``# to check uniqueness ` `    ``mp ``=` `{s[i]:``0` `for` `i ``in` `range``(``len``(s))}  ` `     `  `    ``for` `i ``in` `range``(``len``(s)): ` `        ``if` `(mp[s[i]] ``=``=` `0``): ` `             `  `            ``# if mp[s[i]] = 0 then it is ` `            ``# first time ` `            ``mp[s[i]] ``=` `1` `            ``v.append(s[i]) ` `     `  `    ``# sort the unique characters ` `    ``v.sort(reverse ``=` `False``) ` `     `  `    ``# simply add n-k smallest characters ` `    ``if` `(k > n): ` `        ``print``(s, end ``=` `"") ` `        ``for` `i ``in` `range``(n, k, ``1``): ` `            ``print``(v[``0``], end ``=` `"") ` `         `  `        ``return``; ` `         `  `    ``# end the program ` `         `  `    ``# searching the first character left of ` `    ``# index k and not equal to greatest ` `    ``# character of the string ` `    ``i ``=` `k ``-` `1` `    ``while``(i >``=` `0``): ` `        ``if` `(s[i] !``=` `v[``len``(v) ``-` `1``]): ` `            ``for` `j ``in` `range``(``0``, i, ``1``): ` `                ``print``(s[j], end ``=` `" "``) ` `             `  `            ``# finding the just next greater ` `            ``# character than s[i] ` `            ``for` `j ``in` `range``(``0``, ``len``(v), ``1``): ` `                ``if` `(v[j] > s[i]): ` `                    ``print``(v[j], end ``=` `" "``) ` `                    ``break` `                         `  `            ``# suffix with smallest character ` `            ``for` `j ``in` `range``(i ``+` `1``, k, ``1``): ` `                ``print``(v[``0``], end ``=` `" "``)      ` `             `  `            ``re1turn ` `        ``i ``-``=` `1` `     `  `    ``# if we reach here then all indices to  ` `    ``# the left of k had the greatest character ` `    ``print``(``"No lexicographically greater"``,  ` `          ``"string of length"``, k, ``"possible here."``) ` `     `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``s ``=` `"gi"` `    ``k ``=` `3` ` `  `    ``# Function call ` `    ``lexoString(s, k) ` `     `  `# This code is contributed by ` `# Shashank_Sharma `

## C#

 `// C# implementation of above algorithm. ` `using` `System; ` `using` `System.Collections.Generic; ` ` `  `class` `GFG  ` `{ ` ` `  `    ``// function to print output ` `    ``static` `void` `lexoString(``char``[] s, ``int` `k)  ` `    ``{ ` ` `  `        ``int` `n = s.Length; ` ` `  `        ``// to store unique characters of the string ` `        ``List<``char``> v = ``new` `List<``char``>(); ` ` `  `        ``// to check uniqueness ` `        ``Dictionary<``char``,  ` `                   ``int``> mp = ``new` `Dictionary<``char``,  ` `                                            ``int``>(); ` ` `  `        ``for` `(``int` `i = 0; i < s.Length; i++)  ` `        ``{ ` `            ``if` `(!mp.ContainsKey(s[i])) ` `            ``{ ` ` `  `                ``// if mp[s[i]] = 0 then it is ` `                ``// first time ` `                ``mp.Add(s[i], 1); ` `                ``v.Add(s[i]); ` `            ``} ` `        ``} ` ` `  `        ``// sort the unique characters ` `        ``v.Sort(); ` ` `  `        ``// simply add n-k smallest characters ` `        ``if` `(k > n) ` `        ``{ ` `            ``Console.Write(String.Join(``""``, s)); ` `            ``for` `(``int` `i = n; i < k; i++)  ` `            ``{ ` `                ``Console.Write(v[0]); ` `            ``} ` ` `  `            ``return``; ``// end the program ` `        ``} ` ` `  `        ``// searching the first character left of ` `        ``// index k and not equal to greatest ` `        ``// character of the string ` `        ``for` `(``int` `i = k - 1; i >= 0; i--)  ` `        ``{ ` `            ``if` `(s[i] != v[v.Count - 1])  ` `            ``{ ` `                ``for` `(``int` `j = 0; j < i; j++)  ` `                ``{ ` `                    ``Console.Write(s[j]); ` `                ``} ` ` `  `                ``// finding the just next greater ` `                ``// character than s[i] ` `                ``for` `(``int` `j = 0; j < v.Count; j++)  ` `                ``{ ` `                    ``if` `(v[j] > s[i])  ` `                    ``{ ` `                        ``Console.Write(v[j]); ` `                        ``break``; ` `                    ``} ` `                ``} ` ` `  `                ``// suffix with smallest character ` `                ``for` `(``int` `j = i + 1; j < k; j++)  ` `                ``{ ` `                    ``Console.Write(v[0]); ` `                ``} ` `                ``return``; ` `            ``} ` `        ``} ` ` `  `        ``// if we reach here then all indices to the left  ` `        ``// of k had the greatest character ` `        ``Console.Write(``"No lexicographically greater "` `+  ` `                              ``"string of length "` `+ k + ` `                                    ``" possible here."``); ` ` `  `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main(String []arr)  ` `    ``{ ` `        ``String s = ``"gi"``; ` `        ``int` `k = 3; ` ` `  `        ``// Function call ` `        ``lexoString(s.ToCharArray(), k); ` `    ``} ` `} ` ` `  `// This code is contributed by Princi Singh `

Output:

```gig
```

Time Complexity : O(k + s.size())

My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

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.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Article Tags :
Practice Tags :

Be the First to upvote.

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.