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. 

Steps to solve this problem:

1. sort the string a from start index to end index.

2. declare an answer string as blank.

3. iterate through i=0 till n:

       *update answer to answer+arr[i].

4. return answer.

Implementation:

// 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 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 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# 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

<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>

cbacbc

Complexity Analysis:

• 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)