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.

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++

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// C++ implementation of above algorithm.
#include <bits/stdc++.h>
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 charcter 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;
}

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Java

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// 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<Character> v = new Vector<>();
  
        // to check uniqueness
        Map<Character, Integer> 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 charcter 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

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Python3

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# 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 charcter 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

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Output:

gig

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



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Improved By : Shashank_Sharma, Rajput-Ji



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