最小字符串，不是给定字符串

给定一个由小写字母组成的字符串str ，任务是找到最短的字符串，该字符串不是给定字符串。如果存在多个字符串，则打印其中任何一个。

例子：

Input: str = “abaabcc” 
Output: d 
Explanation: 
One of the possible shortest string which is not a subsequence of the given string is “d”. Therefore, the required output is “d”. 

Input: str = “abcdefghijklmnopqrstuvwxyzaabbccdd” 
Output: ze 

方法：可以使用贪婪技术解决问题。请按照以下步骤解决问题：

• 初始化字符串，说shortestString，存储这不是给定的字符串的一个子最短的字符串。
• 初始化一个Set(例如segments) ，以存储每个子段的所有可能字符。
• 遍历字符串和插入字符串的字符成段，并检查段包含了所有的小写字母或没有。如果发现是真的，则将当前字符附加到shortestString中，然后从segment中删除所有元素。
• 遍历[a – z]范围内的所有可能的小写字母，并检查该段中是否存在当前字符。如果发现为真，则将当前字符插入shortestString。
• 最后，打印shortestString的价值。

下面是上述方法的实现：

C++

// C++ program to implement
// the above approach
 
#include 
using namespace std;
 
// Function to find shortest string which
// not a subsequence of the given string
string ShortestSubsequenceNotPresent(string str)
{
 
    // Stores the shortest string which is
    // not a subsequence of the given string
    string shortestString;
 
    // Stores length of string
    int N = str.length();
 
    // Stores distinct character of subsegments
    unordered_set subsegments;
 
    // Traverse the given string
    for (int i = 0; i < N; i++) {
 
        // Insert current character
        // into subsegments
        subsegments.insert(str[i]);
 
        // If all lowercase alphabets
        // present in the subsegment
        if (subsegments.size() == 26) {
 
            // Insert the last character of
            // subsegment into shortestString
            shortestString.push_back(str[i]);
 
            // Remove all elements from
            // the current subsegment
            subsegments.clear();
        }
    }
 
    // Traverse all lowercase alphabets
    for (char ch = 'a'; ch <= 'z'; ch++) {
 
        // If current character is not
        // present in the subsegment
        if (subsegments.count(ch) == 0) {
            shortestString.push_back(ch);
 
            // Return shortestString
            return shortestString;
        }
    }
    return shortestString;
}
 
// Driver Code
int main()
{
 
    // Given String
    string str
        = "abcdefghijklmnopqrstuvwxyzaabbccdd";
 
    cout << ShortestSubsequenceNotPresent(str);
    return 0;
}

Java

// Java program to implement
// the above approach
 
import java.io.*;
import java.util.*;
 
class GFG {
    public static String
    ShortestSubsequenceNotPresent(String str)
    {
        // Stores the shortest string which is
        // not a subsequence of the given string
        String shortestString = "";
 
        // Stores length of string
        int N = str.length();
 
        // Stores distinct character of subsegments
        HashSet subsegments = new HashSet<>();
 
        // Traverse the given string
        for (int i = 0; i < N; i++) {
 
            // Insert current character
            // into subsegments
            subsegments.add(str.charAt(i));
 
            // If all lowercase alphabets
            // present in the subsegment
            if (subsegments.size() == 26) {
 
                // Insert the last character of
                // subsegment into shortestString
                shortestString
                    = shortestString + str.charAt(i);
 
                // Remove all elements from
                // the current subsegment
                subsegments.clear();
            }
        }
 
        // Traverse all lowercase alphabets
        for (char ch = 'a'; ch <= 'z'; ch++) {
 
            // If current character is not
            // present in the subsegment
            if (!subsegments.contains(ch)) {
                shortestString = shortestString + ch;
 
                // Return shortestString
                return shortestString;
            }
        }
        return shortestString;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        String str = "abcdefghijklmnopqrstuvwxyzaabbccdd";
 
        System.out.print(
            ShortestSubsequenceNotPresent(str));
    }
}
// This code is contributed by Manu Pathria

Python3

# Python3 program to implement
# the above approach
 
# Function to find shortest string which
# not a subsequence of the given string
def ShortestSubsequenceNotPresent(Str):
 
    # Stores the shortest string which is
    # not a subsequence of the given string
    shortestString = ""
 
    # Stores length of string
    N = len(Str)
 
    # Stores distinct character of subsegments
    subsegments = set()
 
    # Traverse the given string
    for i in range(N):
 
        # Insert current character
        # into subsegments
        subsegments.add(Str[i])
 
        # If all lowercase alphabets
        # present in the subsegment
        if (len(subsegments) == 26) :
 
            # Insert the last character of
            # subsegment into shortestString
            shortestString += Str[i]
 
            # Remove all elements from
            # the current subsegment
            subsegments.clear()
         
    # Traverse all lowercase alphabets
    for ch in range(int(26)):
         
        # If current character is not
        # present in the subsegment
        if (chr(ch + 97) not in subsegments) :
             
            shortestString += chr(ch + 97)
             
            # Return shortestString
            return shortestString
     
    return shortestString
 
# Driver code
# Given String
Str = "abcdefghijklmnopqrstuvwxyzaabbccdd"
 
print(ShortestSubsequenceNotPresent(Str))
 
# This code is contributed by divyeshrabadiya07

C#

// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
 
class GFG{
     
public static String ShortestSubsequenceNotPresent(
    String str)
{
     
    // Stores the shortest string which is
    // not a subsequence of the given string
    String shortestString = "";
 
    // Stores length of string
    int N = str.Length;
 
    // Stores distinct character of subsegments
    HashSet subsegments = new HashSet();
 
    // Traverse the given string
    for(int i = 0; i < N; i++)
    {
         
        // Insert current character
        // into subsegments
        subsegments.Add(str[i]);
 
        // If all lowercase alphabets
        // present in the subsegment
        if (subsegments.Count == 26)
        {
             
            // Insert the last character of
            // subsegment into shortestString
            shortestString = shortestString + str[i];
             
            // Remove all elements from
            // the current subsegment
            subsegments.Clear();
        }
    }
 
    // Traverse all lowercase alphabets
    for(char ch = 'a'; ch <= 'z'; ch++)
    {
         
        // If current character is not
        // present in the subsegment
        if (!subsegments.Contains(ch))
        {
            shortestString = shortestString + ch;
             
            // Return shortestString
            return shortestString;
        }
    }
    return shortestString;
}
 
// Driver Code
public static void Main(String[] args)
{
    String str = "abcdefghijklmnopqrstuvwxyzaabbccdd";
     
    Console.Write(
        ShortestSubsequenceNotPresent(str));
}
}
 
// This code is contributed by Rajput-Ji

输出

ze

时间复杂度： O(N)
辅助空间： O(N)