按排序顺序找到第n个二进制字符串

给定正整数n ，任务是在按照字典顺序(词典)排序的两个符号a和b上，在所有可能的字符串的以下无限列表中找到第n^个字符串。

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

例子：

Input: n = 6
Output: bb

Input: n = 11
Output: baa

一种简单的方法是生成直到n的所有字符串，然后确定第n^个字符串。但是，该方法不适用于较大的n值。

一种有效的方法基于以下事实：可以使用2个符号生成的长度为k的字符串数为2 ^k 。基于此，我们可以根据列表中字符串的长度，根据实际索引(n)计算相对索引。然后，在对列表进行排序时，可以使用相对索引的二进制形式轻松确定^第n^个索引处的字符串。以下公式用于计算，

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

考虑以下示例：

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 form (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

因此，位于第11个索引(相对索引4 )的字符串为“ baa”

下面是上述方法的实现：

C++

// C++ implementation of the above approach
#include 
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

// 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 funtion
    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 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#

// 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 funtion
    public static void Main(String []args)
    {
        int n = 11;
        Console.Write(obtain_str(n));
    }
}
  
// This code is contributed by PrinciRaj1992

PHP

 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
?>

输出：

baa