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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 
            // 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 

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#

// 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 

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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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python 3

C#

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