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Find smallest string with whose characters all given Strings can be generated

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  • Difficulty Level : Expert
  • Last Updated : 17 Jan, 2022

Given an array of strings arr[]. The task is to generate the string which contains all the characters of all the strings present in array and smallest in size. There can be many such possible strings and any one is acceptable.

Examples:

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

 

Approach: This problem can be solved by using the Set Data Structure. Set has the capability to remove duplicates, which is needed in this problem in order to minimize the string size. Add all the characters in the set from all the strings in the array arr[] and form a string containing all the characters remaining in the set, which is the required answer.

Below is the implementation of the above approach.

C++




// 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




// 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));
    }
}

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<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

Javascript




<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>

 
 

Output

ruyo

 

Time Complexity: O(N*M), where M is the average length of strings in the given array
Auxiliary Space: O(1)

 


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