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Lexicographically smallest string obtained after concatenating array

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  • Difficulty Level : Medium
  • Last Updated : 07 Jul, 2022

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++




// 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;
}

Java




// 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
            // lexicographically 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("lexicographically smallest string = "
                                      + lexsmallest(a, n));
 
}
}
 
// This code is contributed by Arnab Kundu

Python 3




# Python 3 code to find the lexicographically
# smallest string
def lexSmallest(a, n):
  # Sort strings using above compare()
  for i in range(0,n):
    for j in range(i+1,n):
      if(a[i]+a[j]>a[j]+a[i]):
        s=a[i]
        a[i]=a[j]
        a[j]=s
 
  # 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 vibhu karnwal
 
 
   

C#




// 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
            // lexicographically 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("lexicographically smallest string = "
                                 + lexsmallest(a, n));
 
}
}
 
// This code is contributed by nitin mittal

Javascript




<script>
// Javascript code to find the lexicographically
// smallest string
 
// function to sort the
// array of string
function sort(a,n)
{
 
    // sort the array
    for(let i = 0;i < n;i++)
    {
        for(let j = i + 1;j < n;j++)
        {
               
            // comparing which of the
            // two concatenation causes
            // lexicographically smaller
            // string
            if((a[i] + a[j])>(a[j] + a[i]) )
            {
                let s = a[i];
                a[i] = a[j];
                a[j] = s;
            }
        }
    }
}
 
 
function lexsmallest(a,n)
{
    // Sort strings
    sort(a,n);
   
    // Concatenating sorted strings
    let answer = "";
    for (let i = 0; i < n; i++)
        answer += a[i];
   
    return answer;
}
 
// Driver code
let a=["c", "cb", "cba"];
let n = 3;
document.write("lexicographically smallest string = "
                                      + lexsmallest(a, n));
       
    // This code is contributed by rag2127
</script>

Output: 

cbacbc

 

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

Auxiliary Space: O(n)

This article is contributed by Aarti_Rathi. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 
 


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