Given a positive integer n, the task is to find the nth 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: bbInput: n = 11
Output: baa
A simple approach is to generate all strings up to n and then determine the nth 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 2k. 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 nth 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 – 2floor(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:
| Relative Index | Binary | String |
|---|---|---|
| 0 | 000 | aaa |
| 1 | 001 | aab |
| 2 | 010 | aba |
| 3 | 011 | abb |
| 4 | 100 | baa |
| 5 | 101 | bab |
| 6 | 110 | bba |
| 7 | 111 | bbb |
Hence the string present at 11th index (relative index 4) is “baa”
Below is the implementation of the above approach:
C++
// 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 sequencestring 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 functionint main(){ ll n = 11; cout << obtain_str(n); return 0;} |
Java
// Java Implementation of the above approachimport 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
# Python3 implementation of the# above approach # from math lib import log2 functionfrom 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#
// C# Implementation of the above approachusing 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 |
Javascript
<script>// Javascript Implementation of the above approach// Function to return the nth string// in the required sequencefunction 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 codelet n = 11;document.write(obtain_str(n));// This code is contributed by rameshtravel07</script> |
PHP
<?php// PHP implementation of the above approach// Function to return the nth string // in the required sequencefunction 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)
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