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# Lexicographically next greater string using same character set

• Difficulty Level : Medium
• Last Updated : 28 May, 2021

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.```

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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;``        ``}``        ` `        ``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;        ``            ` `            ``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);``            ``}` `            ``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);``                ``}``                ``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`

## Javascript

 ``
Output:
`gig`

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

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