Encuentre la string más pequeña con cuyos caracteres se pueden generar todas las strings dadas

Dada una array de strings arr[]. La tarea es generar la string que contiene todos los caracteres de todas las strings presentes en la array y de menor tamaño. Puede haber muchas strings posibles y cualquiera es aceptable.

Ejemplos :

Entrada: arr[] = {“tu”, “tú”, “o”, “yo”}
Salida: ruyo
Explicación: La string “ruyo” es la string que contiene todos los caracteres presentes en todas las strings de arr[] .
Puede haber muchas otras strings de tamaño 4, por ejemplo, «oury». Esos también son aceptables.

Entrada: arr[] = {“abm”, “bmt”, “cd”, “tca”}
Salida: abctdm

Enfoque : este problema se puede resolver utilizando la estructura de datos establecida . Set tiene la capacidad de eliminar duplicados, lo cual es necesario en este problema para minimizar el tamaño de la string. Agregue todos los caracteres del conjunto de todas las strings de la array arr[] y forme una string que contenga todos los caracteres restantes del conjunto, que es la respuesta requerida.

A continuación se muestra la implementación del enfoque anterior.

// C++ code for the above approach
#include <bits/stdc++.h>
using namespace std;
 
string minSubstr(vector<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.size(); i++)
    {
        str += s[i];
    }
 
    // Set to store the characters
    unordered_set<char> 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<string> arr = {"your", "you",
                          "or", "yo"};
 
    cout << (minSubstr(arr));
    return 0;
}
 
// This code is contributed by Potta Lokesh

// 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<Character> set =
            new HashSet<Character>();
 
        // 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<Character> 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));
    }
}

# 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# 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<char> set =
      new HashSet<char>();
 
    // 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

<script>
// 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
</script>

ruyo

Complejidad de tiempo : O(N*M), donde M es la longitud promedio de las strings en la array dada
Espacio auxiliar: O(1)