Given a positive integer n, the task is to find the n^th string in the following infinite list of all possible strings over two symbols a and b sorted lexicographically (Dictionary). 

a, b, aa, ab, ba, bb, aaa, aab, aba, abb, baa, bab, bba, bbb, aaaa, ... 
 

Examples:  

Input: n = 6 
Output: bb

Input: n = 11 
Output: baa 

A simple approach is to generate all strings up to n and then determine the n^th string. However, the approach is not suitable for large values of n.

An efficient approach is based on the fact that the number of length k strings that can be generated using 2 symbols is 2^k. Based on this we can calculate the relative index from the actual index(n) with respect to the length of the string in the list. The string at n^th index can then be determined easily using binary form of the relative index as the list is sorted. The following formula is used for calculation, 

relative index = n + 1 - 2^{floor(log(n + 1))} 
 

Consider the following example:  

Let n = 11 then floor(log(n + 1)) = 3. 
This suggests that index n consists of a length 3 string and length 3 strings start from (2³ - 1) = 7th index and 7th index contains the string "aaa". 
Therefore, relative index = 11 + 1 - 2³ = 4. 
This is the index relative to 7. Now, the string at index n = 11 can be simply obtained from the binary interpretation of the relative index 4. 
Here 0 means a and 1 means b. The table below illustrates this: 
 

Hence the string present at 11th index (relative index 4) is "baa"

Below is the implementation of the above approach:

// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
#define ll long long int

// Function to return the nth string in the required sequence
string obtain_str(ll n)
{

    // Length of the resultant string
    ll len = (int)log2(n + 1);

    // Relative index
    ll rel_ind = n + 1 - pow(2, len);

    ll i = 0;
    string str = "";
    for (i = 0; i < len; i++) {

        // Initial string of length len consists of
        // all a's since the list is sorted
        str += 'a';
    }

    i = 0;

    // Convert relative index to Binary form and set
    // 0 = a and 1 = b
    while (rel_ind > 0) {
        if (rel_ind % 2 == 1)
            str[i] = 'b';
        rel_ind /= 2;
        i++;
    }

    // Reverse and return the string
    reverse(str.begin(), str.end());
    return str;
}

// Driver function
int main()
{
    ll n = 11;
    cout << obtain_str(n);

    return 0;
}

// Java Implementation of the above approach
import java.io.*;
import java.util.*;
class Gfg {

    // Function to return the nth string in the required sequence
    static String obtain_str(int n)
    {
        // Length of the resultant string
        int len = (int)Math.floor((Math.log(n + 1) / Math.log(2)));

        // Relative Index
        int rel_ind = n + 1 - (int)Math.pow(2, len);

        int i = 0;
        StringBuilder str = new StringBuilder();
        for (i = 0; i < len; i++) {

            // Initial string of length len consists of
            // all a's since the list is sorted
            str.append('a');
        }

        i = 0;

        // Convert relative index to Binary form and set
        // 0 = a and 1 = b
        while (rel_ind > 0) {
            if (rel_ind % 2 == 1)
                str.setCharAt(i, 'b');
            rel_ind /= 2;
            i++;
        }

        // Reverse and return the string
        str = str.reverse();
        return str.toString();
    }

    // Driver function
    public static void main(String args[])
    {
        int n = 11;
        System.out.print(obtain_str(n));
    }
}

# Python3 implementation of the
# above approach 

# from math lib import log2 function
from math import log2

# Function to return the nth string 
# in the required sequence 
def obtain_str(n) :

    # Length of the resultant string 
    length = int(log2(n + 1)) 

    # Relative index 
    rel_ind = n + 1 - pow(2, length)

    i = 0
    string = "" 
    
    for i in range(length) :

        # Initial string of length len consists 
        # of all a's since the list is sorted 
        string += 'a'

    i = 0
    
    string_list = list(string)
    
    # Convert relative index to Binary 
    # form and set 0 = a and 1 = b 
    while (rel_ind > 0) :
        if (rel_ind % 2 == 1) : 
            string_list[i] = 'b'
            
        rel_ind //= 2
        i += 1
    
    # Reverse and return the string 
    string_list.reverse()
    string = "".join(string_list)
    
    return string

# Driver Code 
if __name__ == "__main__" : 

    n = 11
    print(obtain_str(n))

# This code is contributed by Ryuga

// C# Implementation of the above approach
using System;
using System.Text;

class GFG
{

    // Function to return the nth string
    // in the required sequence
    static String obtain_str(int n)
    {
        // Length of the resultant string
        int len = (int)Math.Floor((Math.Log(n + 1) / 
                                    Math.Log(2)));

        // Relative Index
        int rel_ind = n + 1 - (int)Math.Pow(2, len);

        int i = 0;
        StringBuilder str = new StringBuilder();
        for (i = 0; i < len; i++)
        {

            // Initial string of length len consists of
            // all a's since the list is sorted
            str.Append('a');
        }

        i = 0;

        // Convert relative index to Binary form and set
        // 0 = a and 1 = b
        while (rel_ind > 0) 
        {
            if (rel_ind % 2 == 1)
                str[i]='b';
            rel_ind /= 2;
            i++;
        }

        // Reverse and return the string
        return reverse(str.ToString());
    }
    
    static String reverse(String input) 
    {
        char[] a = input.ToCharArray();
        int l, r = a.Length - 1;
        for (l = 0; l < r; l++, r--) 
        {
            char temp = a[l];
            a[l] = a[r];
            a[r] = temp;
        }
        return String.Join("", a);
    }
    
    // Driver function
    public static void Main(String []args)
    {
        int n = 11;
        Console.Write(obtain_str(n));
    }
}

// This code is contributed by PrinciRaj1992

<script>

// Javascript Implementation of the above approach

// Function to return the nth string
// in the required sequence
function obtain_str(n)
{
    
    // Length of the resultant string
    let len = Math.floor((Math.log(n + 1) /
                          Math.log(2)));

    // Relative Index
    let rel_ind = n + 1 - Math.pow(2, len);

    let i = 0;
    let str = [];
    
    for(i = 0; i < len; i++)
    {
        
        // Initial string of length len consists of
        // all a's since the list is sorted
        str.push('a');
    }

    i = 0;

    // Convert relative index to Binary
    // form and set 0 = a and 1 = b
    while (rel_ind > 0) 
    {
        if (rel_ind % 2 == 1)
            str[i]='b';
            
        rel_ind = parseInt(rel_ind / 2, 10);
        i++;
    }

    // Reverse and return the string
    return reverse(str.join(""));
}
  
function reverse(input) 
{
    let a = input.split('');
    let l, r = a.length - 1;
    
    for(l = 0; l < r; l++, r--) 
    {
        let temp = a[l];
        a[l] = a[r];
        a[r] = temp;
    }
    return a.join("");
}

// Driver code
let n = 11;

document.write(obtain_str(n));

// This code is contributed by rameshtravel07

</script>

<?php
// PHP implementation of the above approach

// Function to return the nth string 
// in the required sequence
function obtain_str($n)
{

    // Length of the resultant string
    $len = (int)(log($n + 1) / log(2));

    // Relative index
    $rel_ind = $n + 1 - pow(2, $len);

    $i = 0;
    $str = "";
    for ($i = 0; $i < $len; $i++) 
    {

        // Initial string of length len 
        // consists of all a's since the 
        // list is sorted
        $str .= 'a';
    }

    $i = 0;

    // Convert relative index to Binary 
    // form and set 0 = a and 1 = b
    while ($rel_ind > 0)
    {
        if ($rel_ind % 2 == 1)
            $str[$i] = 'b';
        $rel_ind = (int)($rel_ind / 2);
        $i++;
    }

    // Reverse and return the string
    return strrev($str);
}

// Driver Code
$n = 11;
echo obtain_str($n);

// This code is contributed 
// by chandan_jnu
?>

baa

Time complexity : O(logn)

Auxiliary Space: O(1)