Lexicographically smallest string obtained after concatenating array

Given n strings, concatenate them in an order that produces the lexicographically smallest possible string.

Examples:

Input :  a[] = ["c", "cb", "cba"]
Output : cbacbc
Possible strings are ccbcba, ccbacb, 
cbccba, cbcbac, cbacbc and cbaccb. 
Among all these strings, cbacbc is 
the lexicographically smallest.

Input :  a[] = ["aa", "ab", "aaa"]
Output : aaaaaab

One might think that sorting the given strings in the lexicographical order and then concatenating them produces the correct output. This approach produces the correct output for inputs like [“a”, “ab”, “abc”]. However, applying this method on [“c”, “cb”, “cba”] produces the wrong input and hence this approach is incorrect.

The correct approach is to use a regular sorting algorithm. When two strings a and b are compared to decide if they have to be swapped or not, do not check if a is lexicographically smaller than b or not. Instead check if appending b at the end of a produces a lexicographically smaller string or appending a at the end of b does. This approach works because we want the concatenated string to be lexicographically small, not the individual strings to be in the lexicographical order.

C++

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// CPP code to find the lexicographically
// smallest string
#include <bits/stdc++.h>
using namespace std;
  
// Compares two strings by checking if 
// which of the two concatenations causes
// lexicographically smaller string.
bool compare(string a, string b)
{
    return (a+b < b+a);
}
  
string lexSmallest(string a[], int n)
{
    // Sort strings using above compare()
    sort(a, a+n, compare);
  
    // Concatenating sorted strings
    string answer = "";
    for (int i = 0; i < n; i++)
        answer += a[i];
  
    return answer;
}
  
// Driver code
int main()
{
    string a[] = { "c", "cb", "cba" };
    int n = sizeof(a)/sizeof(a[0]);
    cout << lexSmallest(a, n);
    return 0;
}

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Java














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// Java code to find the lexicographically 


// smallest string 


  


class GFG { 


      


// function to sort the 


// array of string 


static void sort(String a[], int n) 


{ 


      


    //sort the array 


    for(int i = 0;i < n;i++) 


    { 


        for(int j = i + 1;j < n;j++) 


        { 


              


            // comparing which of the 


            // two concatenation causes 


            // lexiographically smaller 


            // string 


            if((a[i] + a[j]).compareTo(a[j] + a[i]) > 0) 


            { 


                String s = a[i]; 


                a[i] = a[j]; 


                a[j] = s; 


            } 


        } 


    } 


} 


      


static String lexsmallest(String a[], int n) 


{ 


      


    // Sort strings 


    sort(a,n); 


  


    // Concatenating sorted strings 


    String answer = ""; 


    for (int i = 0; i < n; i++) 


        answer += a[i]; 


  


    return answer; 


} 


  


// Driver code 


public static void main(String args[]) 


{ 


    String a[] = {"c", "cb", "cba"}; 


    int n = 3; 


    System.out.println("lexiographically smallest string = "


                                      + lexsmallest(a, n)); 


  


} 


} 


  


// This code is contributed by Arnab Kundu 



















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Python 3





# Python 3 code to find the lexicographically

# smallest string


def lexSmallest(a, n):


	# Sort strings using above compare()

	a.sort(reverse = True)


	# Concatenating sorted strings

	answer = “”

	for i in range( n):

		answer += a[i]


	return answer


# Driver code

if __name__ == “__main__”:


	a = [ “c”, “cb”, “cba” ]

	n = len(a)

	print(lexSmallest(a, n))


# This code is contributed by ita_c



C#














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// C# code to find  


// the lexicographically 


// smallest string 


using System; 


  


class GFG { 


      


// function to sort the 


// array of string 


static void sort(String []a, int n) 


{ 


      


    //sort the array 


    for(int i = 0;i < n;i++) 


    { 


        for(int j = i + 1;j < n;j++) 


        { 


              


            // comparing which of the 


            // two concatenation causes 


            // lexiographically smaller 


            // string 


            if((a[i] + a[j]).CompareTo(a[j] + 


                                  a[i]) > 0) 


            { 


                String s = a[i]; 


                a[i] = a[j]; 


                a[j] = s; 


            } 


        } 


    } 


} 


      


static String lexsmallest(String []a, int n) 


{ 


      


    // Sort strings 


    sort(a,n); 


  


    // Concatenating sorted  


    // strings 


    String answer = ""; 


    for (int i = 0; i < n; i++) 


        answer += a[i]; 


  


    return answer; 


} 


  


// Driver code 


public static void Main() 


{ 


    String []a = {"c", "cb", "cba"}; 


    int n = 3; 


    Console.Write("lexiographically smallest string = "


                                 + lexsmallest(a, n)); 


  


} 


} 


  


// This code is contributed by nitin mittal 



















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Output:


cbacbc





Time complexity : The above code runs inO(M * N * logN) where N is number of strings and M is maximum length of a string.



          
          
          
          

            

    

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