找到可以生成所有给定字符的最小字符串

给定一个字符串数组arr[]。任务是生成包含数组中所有字符串的所有字符且大小最小的字符串。可能有许多这样的可能字符串，任何一个都是可以接受的。

例子：

Input: arr[] = {“your”, “you”, “or”, “yo”}
Output: ruyo
Explanation: The string “ruyo” is the string which contains all the characters present in all the strings in arr[].
There can be many other strings of size 4 e.g. “oury”. Those are also acceptable.

Input: arr[] = {“abm”, “bmt”, “cd”, “tca”}
Output: abctdm

方法：这个问题可以通过使用Set Data Structure来解决。 Set 具有删除重复项的能力，这在此问题中是必需的，以最小化字符串大小。从数组arr[]中的所有字符串中添加集合中的所有字符，并形成一个包含集合中剩余的所有字符的字符串，这就是所需的答案。

下面是上述方法的实现。

C++

// C++ code for the above approach
#include 
using namespace std;
 
string minSubstr(vector s)
{
   
    // Stores the concatenated string
    // of all the given strings
    string str = "";
 
    // Loop to iterate through all
    // the given strings
    for (int i = 0; i < s.size(); i++)
    {
        str += s[i];
    }
 
    // Set to store the characters
    unordered_set set;
 
    // Loop to iterate over all
    // the characters in str
    for (int i = 0; i < str.length(); i++)
    {
        set.insert(str[i]);
    }
    string res = "";
   
    // Loop to iterate over the set
    for (auto itr = set.begin(); itr != set.end(); itr++)
    {
        res = res + (*itr);
    }
 
    // Return Answer
    return res;
}
 
// Driver Code
int main()
{
    vector arr = {"your", "you",
                          "or", "yo"};
 
    cout << (minSubstr(arr));
    return 0;
}
 
// This code is contributed by Potta Lokesh

Java

// Java program to implement above approach
import java.util.*;
 
public class GfG {
    public static String minSubstr(String s[])
    {
        // Stores the concatenated string
        // of all the given strings
        String str = "";
 
        // Loop to iterate through all
        // the given strings
        for (int i = 0; i < s.length; i++) {
            str += s[i];
        }
 
        // Set to store the characters
        Set set =
            new HashSet();
 
        // Loop to iterate over all
        // the characters in str
        for (int i = 0; i < str.length();
             i++) {
            set.add(str.charAt(i));
        }
 
        // Stores the required answer
        String res = "";
        Iterator itr =
            set.iterator();
 
        // Loop to iterate over the set
        while (itr.hasNext()) {
            res += itr.next();
        }
 
        // Return Answer
        return res;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        String arr[]
            = new String[] { "your", "you",
                            "or", "yo" };
 
        System.out.println(minSubstr(arr));
    }
}

Python3

# Python program to implement above approach
def minSubstr(s):
 
      # Stores the concatenated string
    # of all the given strings
    str = "";
 
    # Loop to iterate through all
    # the given strings
    for i in range(len(s)):
        str += s[i];
 
    # Set to store the characters
    setv = set();
 
    # Loop to iterate over all
    # the characters in str
    for i in range(len(str)):
        setv.add(str[i]);
 
    # Stores the required answer
    res = "";
 
    # Loop to iterate over the set
    for itr in setv:
        res += itr;
 
    # Return Answer
    return res;
 
# Driver Code
if __name__ == '__main__':
    arr = ["your", "you", "or", "yo"];
 
    print(minSubstr(arr));
 
# This code is contributed by 29AjayKumar

C#

// C# program to implement above approach
using System;
using System.Collections.Generic;
 
public class GfG {
  public static String minSubstr(String []s)
  {
     
    // Stores the concatenated string
    // of all the given strings
    String str = "";
 
    // Loop to iterate through all
    // the given strings
    for (int i = 0; i < s.Length; i++) {
      str += s[i];
    }
 
    // Set to store the characters
    HashSet set =
      new HashSet();
 
    // Loop to iterate over all
    // the characters in str
    for (int i = 0; i < str.Length;
         i++) {
      set.Add(str[i]);
    }
 
    // Stores the required answer
    String res = "";
    // Loop to iterate over the set
    foreach (char itr in set) {
      res += itr;
    }
 
    // Return Answer
    return res;
  }
 
  // Driver Code
  public static void Main(String[] args)
  {
    String []arr
      = new String[] { "your", "you",
                      "or", "yo" };
 
    Console.WriteLine(minSubstr(arr));
  }
}
 
// This code is contributed by 29AjayKumar

Javascript

// javascript program to implement above approach
public class GfG
{
    function minSubstr(s)
    {
     
        // Stores the concatenated string
        // of all the given strings
        var str = "";
 
        // Loop to iterate through all
        // the given strings
        for (var i = 0; i < s.length; i++) {
            str += s[i];
        }
 
        // Set to store the characters
        var set = new Set();
 
        // Loop to iterate over all
        // the characters in str
        for (var i = 0; i < str.length;
             i++) {
            set.add(str.charAt(i));
        }
 
        // Stores the required answer
        var res = "";
 
        // Loop to iterate over the set
        for(let itr of set){
            res += itr;
        }
 
        // Return Answer
        return res;
    }
 
    // Driver Code
var arr = [ "your", "you",
                            "or", "yo" ];
document.write(minSubstr(arr));
 
// This code is contributed by 29AjayKumar

输出

ruyo

时间复杂度： O(N*M)，其中 M 是给定数组中字符串的平均长度
辅助空间： O(1)