# Print a number as string of ‘A’ and ‘B’ in lexicographic order

Given a number N, the task is to print the string of ‘A’ and ‘B’ corresponding to that number.

If we represent all numbers as a string of ‘A’ and ‘B’ as follows,

```1 = A
2 = B
3 = AA
4 = AB
5 = BA
6 = BB
7 = AAA
8 = AAB
9 = ABA
10 = ABB
.....
.....
1000000 = BBBABAAAABAABAAAAAB
```

Examples:

```Input: N = 30
Output: BBBB

Input: N = 55
Output: BBAAA

Input: N = 100
Output: BAABAB
```

Approach:

1. This representation is a little bit related to binary representation.
2. First, we have to find the number of characters required to print the string corresponding to the given number, That is the length of the resultant string.
3. There are 2 numbers of length 1, 4 numbers of length 2, 8 numbers of length 3 and so on…
4. Therefore, k length string of ‘A’ and ‘B’ can represent pow(2, k) numbers from (pow(2, k)-2)+1 to pow(2, k+1)-2, that is AA…A (k times) to BB…B (k times).
5. Therefore, for printing the corresponding string, first, calculate the length of the string, let it be k. Now calculate,
```    N = M - (pow(2, k)-2);
```
6. Now run the following loop to print the corresponding string.
```while (k>0)
{
num = pow(2, k-1);

if (num >= N)
cout <<"A";
else{
cout <<"B";
N -= num;
}
k--;
}
```

Below is the implementation of the above approach:

## C++

 `// C++ program to implement the above approach ` ` `  `#include ` `#include ` `using` `namespace` `std; ` ` `  `// Function to calculate number of characters ` `// in corresponding string of 'A' and 'B' ` `int` `no_of_characters(``int` `M) ` `{ ` ` `  `    ``// Since the minimum number ` `    ``// of characters will be 1 ` `    ``int` `k = 1; ` ` `  `    ``// Calculating number of characters ` `    ``while` `(``true``) { ` ` `  `        ``// Since k length string can ` `        ``// represent at most pow(2, k+1)-2 ` `        ``// that is if k = 4, it can ` `        ``// represent at most pow(2, 4+1)-2 = 30 ` `        ``// so we have to calculate the ` `        ``// length of the corresponding string ` `        ``if` `(``pow``(2, k + 1) - 2 < M) ` `            ``k++; ` `        ``else` `            ``break``; ` `    ``} ` ` `  `    ``// return the length of ` `    ``// the corresponding string ` `    ``return` `k; ` `} ` ` `  `// Function to print corresponding ` `// string of 'A' and 'B' ` `void` `print_string(``int` `M) ` `{ ` `    ``int` `k, num, N; ` ` `  `    ``// Find length of string ` `    ``k = no_of_characters(M); ` ` `  `    ``// Since the first number that can be represented ` `    ``// by k length string will be (pow(2, k)-2)+1 ` `    ``// and it will be AAA...A, k times, ` `    ``// therefore, N will store that ` `    ``// how much we have to print ` `    ``N = M - (``pow``(2, k) - 2); ` ` `  `    ``// At a particular time, ` `    ``// we have to decide whether ` `    ``// we have to print 'A' or 'B', ` `    ``// this can be check by calculating ` `    ``// the value of pow(2, k-1) ` `    ``while` `(k > 0) { ` `        ``num = ``pow``(2, k - 1); ` ` `  `        ``if` `(num >= N) ` `            ``cout << ``"A"``; ` `        ``else` `{ ` `            ``cout << ``"B"``; ` `            ``N -= num; ` `        ``} ` `        ``k--; ` `    ``} ` ` `  `    ``// Print new line ` `    ``cout << endl; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` ` `  `    ``int` `M; ` ` `  `    ``M = 30; ` `    ``print_string(M); ` ` `  `    ``M = 55; ` `    ``print_string(M); ` ` `  `    ``M = 100; ` `    ``print_string(M); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach  ` `import` `java.util.*; ` ` `  `class` `GFG ` `{ ` `     `  `// Function to calculate number of characters  ` `// in corresponding string of 'A' and 'B'  ` `static` `int` `no_of_characters(``int` `M)  ` `{  ` ` `  `    ``// Since the minimum number  ` `    ``// of characters will be 1  ` `    ``int` `k = ``1``;  ` ` `  `    ``// Calculating number of characters  ` `    ``while` `(``true``) ` `    ``{  ` ` `  `        ``// Since k length string can  ` `        ``// represent at most pow(2, k+1)-2  ` `        ``// that is if k = 4, it can  ` `        ``// represent at most pow(2, 4+1)-2 = 30  ` `        ``// so we have to calculate the  ` `        ``// length of the corresponding string  ` `        ``if` `((``int``)Math.pow(``2``, k + ``1``) - ``2` `< M)  ` `            ``k++;  ` `        ``else` `            ``break``;  ` `    ``}  ` ` `  `    ``// return the length of  ` `    ``// the corresponding string  ` `    ``return` `k;  ` `}  ` ` `  `// Function to print corresponding  ` `// string of 'A' and 'B'  ` `static` `void` `print_string(``int` `M)  ` `{  ` `    ``int` `k, num, N;  ` ` `  `    ``// Find length of string  ` `    ``k = no_of_characters(M);  ` ` `  `    ``// Since the first number that can be represented  ` `    ``// by k length string will be (pow(2, k)-2)+1  ` `    ``// and it will be AAA...A, k times,  ` `    ``// therefore, N will store that  ` `    ``// how much we have to print  ` `    ``N = M - ((``int``)Math.pow(``2``, k) - ``2``);  ` ` `  `    ``// At a particular time,  ` `    ``// we have to decide whether  ` `    ``// we have to print 'A' or 'B',  ` `    ``// this can be check by calculating  ` `    ``// the value of pow(2, k-1)  ` `    ``while` `(k > ``0``)  ` `    ``{  ` `        ``num = (``int``)Math.pow(``2``, k - ``1``);  ` ` `  `        ``if` `(num >= N)  ` `            ``System.out.print(``"A"``);  ` `        ``else`  `        ``{  ` `            ``System.out.print( ``"B"``);  ` `            ``N -= num;  ` `        ``}  ` `        ``k--;  ` `    ``}  ` ` `  `    ``// Print new line  ` `    ``System.out.println();  ` `}  ` ` `  `// Driver code  ` `public` `static` `void` `main(String args[]) ` `{  ` ` `  `    ``int` `M;  ` ` `  `    ``M = ``30``;  ` `    ``print_string(M);  ` ` `  `    ``M = ``55``;  ` `    ``print_string(M);  ` ` `  `    ``M = ``100``;  ` `    ``print_string(M);  ` ` `  `}  ` `} ` ` `  `// This code is contributed by Arnab Kundu `

## Python3

 `# Python 3 program to implement ` `# the above approach ` `from` `math ``import` `pow` ` `  `# Function to calculate number of characters ` `# in corresponding string of 'A' and 'B' ` `def` `no_of_characters(M): ` `     `  `    ``# Since the minimum number ` `    ``# of characters will be 1 ` `    ``k ``=` `1` ` `  `    ``# Calculating number of characters ` `    ``while` `(``True``): ` `         `  `        ``# Since k length string can ` `        ``# represent at most pow(2, k+1)-2 ` `        ``# that is if k = 4, it can ` `        ``# represent at most pow(2, 4+1)-2 = 30 ` `        ``# so we have to calculate the ` `        ``# length of the corresponding string ` `        ``if` `(``pow``(``2``, k ``+` `1``) ``-` `2` `< M): ` `            ``k ``+``=` `1` `        ``else``: ` `            ``break` ` `  `    ``# return the length of ` `    ``# the corresponding string ` `    ``return` `k ` ` `  `# Function to print corresponding ` `# string of 'A' and 'B' ` `def` `print_string(M): ` `     `  `    ``# Find length of string ` `    ``k ``=` `no_of_characters(M) ` ` `  `    ``# Since the first number that can be  ` `    ``# represented by k length string will  ` `    ``# be (pow(2, k)-2)+1 and it will be  ` `    ``# AAA...A, k times, therefore, N will  ` `    ``# store that how much we have to print ` `    ``N ``=` `M ``-` `(``pow``(``2``, k) ``-` `2``) ` ` `  `    ``# At a particular time, ` `    ``# we have to decide whether ` `    ``# we have to print 'A' or 'B', ` `    ``# this can be check by calculating ` `    ``# the value of pow(2, k - 1) ` `    ``while` `(k > ``0``): ` `        ``num ``=` `pow``(``2``, k ``-` `1``) ` ` `  `        ``if` `(num >``=` `N): ` `            ``print``(``"A"``, end ``=` `"") ` `        ``else``: ` `            ``print``(``"B"``, end ``=` `"") ` `            ``N ``-``=` `num ` `        ``k ``-``=` `1` `         `  `    ``print``(``"\n"``, end ``=` `"") ` ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``M ``=` `30``; ` `    ``print_string(M) ` ` `  `    ``M ``=` `55` `    ``print_string(M) ` ` `  `    ``M ``=` `100` `    ``print_string(M) ` `     `  `# This code is contributed by ` `# Surendra_Gangwar `

## C#

 `// C# implementation of the approach  ` `using` `System; ` ` `  `class` `GFG  ` `{  ` `     `  `    ``// Function to calculate number of characters  ` `    ``// in corresponding string of 'A' and 'B'  ` `    ``static` `int` `no_of_characters(``int` `M)  ` `    ``{  ` `     `  `        ``// Since the minimum number  ` `        ``// of characters will be 1  ` `        ``int` `k = 1;  ` `     `  `        ``// Calculating number of characters  ` `        ``while` `(``true``)  ` `        ``{  ` `     `  `            ``// Since k length string can  ` `            ``// represent at most pow(2, k+1)-2  ` `            ``// that is if k = 4, it can  ` `            ``// represent at most pow(2, 4+1)-2 = 30  ` `            ``// so we have to calculate the  ` `            ``// length of the corresponding string  ` `            ``if` `((``int``)Math.Pow(2, k + 1) - 2 < M)  ` `                ``k++;  ` `            ``else` `                ``break``;  ` `        ``}  ` `     `  `        ``// return the length of  ` `        ``// the corresponding string  ` `        ``return` `k;  ` `    ``}  ` `     `  `    ``// Function to print corresponding  ` `    ``// string of 'A' and 'B'  ` `    ``static` `void` `print_string(``int` `M)  ` `    ``{  ` `        ``int` `k, num, N;  ` `     `  `        ``// Find length of string  ` `        ``k = no_of_characters(M);  ` `     `  `        ``// Since the first number that can be represented  ` `        ``// by k length string will be (pow(2, k)-2)+1  ` `        ``// and it will be AAA...A, k times,  ` `        ``// therefore, N will store that  ` `        ``// how much we have to print  ` `        ``N = M - ((``int``)Math.Pow(2, k) - 2);  ` `     `  `        ``// At a particular time,  ` `        ``// we have to decide whether  ` `        ``// we have to print 'A' or 'B',  ` `        ``// this can be check by calculating  ` `        ``// the value of pow(2, k-1)  ` `        ``while` `(k > 0)  ` `        ``{  ` `            ``num = (``int``)Math.Pow(2, k - 1);  ` `     `  `            ``if` `(num >= N)  ` `                ``Console.Write(``"A"``);  ` `            ``else` `            ``{  ` `                ``Console.Write( ``"B"``);  ` `                ``N -= num;  ` `            ``}  ` `            ``k--;  ` `        ``}  ` `     `  `        ``// Print new line  ` `        ``Console.WriteLine();  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `Main()  ` `    ``{  ` `     `  `        ``int` `M;  ` `     `  `        ``M = 30;  ` `        ``print_string(M);  ` `     `  `        ``M = 55;  ` `        ``print_string(M);  ` `     `  `        ``M = 100;  ` `        ``print_string(M);  ` `     `  `    ``}  ` `}  ` ` `  `// This code is contributed by Ryuga `

## PHP

 ` 0)  ` `    ``{ ` `        ``\$num` `= pow(2, ``\$k` `- 1); ` ` `  `        ``if` `(``\$num` `>= ``\$N``) ` `            ``echo` `"A"``; ` `        ``else` `{ ` `            ``echo` `"B"``; ` `            ``\$N` `-= ``\$num``; ` `        ``} ` `        ``\$k``--; ` `    ``} ` ` `  `    ``// Print new line ` `    ``echo` `"\n"``; ` `} ` ` `  `// Driver code ` `\$M``; ` ` `  `\$M` `= 30; ` `print_string(``\$M``); ` ` `  `\$M` `= 55; ` `print_string(``\$M``); ` ` `  `\$M` `= 100; ` `print_string(``\$M``); ` ` `  `// This code is contributed  ` `// by Akanksha Rai ` `?> `

Output:

```BBBB
BBAAA
BAABAB
```

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