使用相同字符集按字典顺序排列的下一个更大的字符串

给定一个数字 K 和一个字符串S，我们必须找到长度为 K 的字典上最小的字符串str，使得它的字母集是 S 的字母集的子集，并且 S 在字典上小于 str。
例子：

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.

方法：如果字符串的大小小于 k，我们应该简单地从 s 中添加 k – s.size() 最小符号。
如果字符串的大小大于或等于k，那么我们需要替换字符串最小符号的前k个符号的后缀中的所有符号，并将该后缀之前的字符替换为字符串中存在的下一个更大的字符。
注意：如果字符串中的 k – 1 索引包含最大字符，那么我们将向后移动一个索引并向后移动，直到找到一个不等于最大字符的字符。

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 v;
     
    // to check uniqueness
    map 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 v = new List();
 
        // to check uniqueness
        Dictionary mp = new Dictionary();
 
        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

Javascript

      // JavaScript implementation of above algorithm.
      // function to print output
      function lexoString(s, k) {
        var n = s.length;
 
        // to store unique characters of the string
        var v = [];
 
        // to check uniqueness
        var mp = {};
 
        for (var i = 0; i < s.length; i++) {
          if (!mp.hasOwnProperty(s[i])) {
            // if mp[s[i]] = 0 then it is
            // first time
            mp[s[i]] = 1;
            v.push(s[i]);
          }
        }
 
        // sort the unique characters
        v.sort((a, b) => a - b);
 
        // simply add n-k smallest characters
        if (k > n) {
          document.write(s.join(""));
          for (var i = n; i < k; i++) {
            document.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 (var i = k - 1; i >= 0; i--) {
          if (s[i] !== v[v.length - 1]) {
            for (var j = 0; j < i; j++) {
              document.write(s[j]);
            }
 
            // finding the just next greater
            // character than s[i]
            for (var j = 0; j < v.length; j++) {
              if (v[j] > s[i]) {
                document.write(v[j]);
                break;
              }
            }
 
            // suffix with smallest character
            for (var j = i + 1; j < k; j++) {
              document.write(v[0]);
            }
            return;
          }
        }
 
        // if we reach here then all indices to the left
        // of k had the greatest character
        document.write(
          "No lexicographically greater " +
            "string of length " +
            k +
            " possible here."
        );
      }
 
      // Driver code
      var s = "gi";
      var k = 3;
 
      // Function call
      lexoString(s.split(""), k);
    

输出：

gig

时间复杂度： O(k + s.size())