# Find letter’s position in Alphabet using Bit operation

• Difficulty Level : Hard
• Last Updated : 26 Nov, 2021

Given a string of English alphabets. The task is, for every character in the string print its position in the English alphabets.
Note: The characters in the string are considered to be case-insensitive. That is, both ‘A’ and ‘a’ is at the first position.
Examples:

Input: “Geeks”
Output: 7 5 5 11 19
‘G’ is the 7th character of the alphabets
‘e’ is the 5th and so on…
Input: “Algorithms”
Output: 1 12 7 15 18 9 20 8 13 19

Approach:
A letter’s position in Alphabet can easily be found by performing logical AND operation with the number 31.
Note that this is only applicable to letters and not special characters.
Every letter has an ASCII value which can be represented in binary form. Performing the bitwise and of this value with the number 31 will give the letter’s position in the alphabets.
Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `const` `int` `NUM = 31;` `// Function to calculate the position``// of characters``void` `positions(string str, ``int` `n)``{``    ``for` `(``int` `i = 0; i < n; i++) {` `        ``// Performing AND operation``        ``// with number 31``        ``cout << (str[i] & NUM) << ``" "``;``    ``}``}` `// Driver code``int` `main()``{``    ``string str = ``"Geeks"``;``    ``int` `n = str.length();` `    ``positions(str, n);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``public` `class` `GFG {` `    ``public` `static` `final` `int` `NUM = ``31``;` `    ``// Function to calculate the position``    ``// of characters``    ``static` `void` `positions(String str, ``int` `n)``    ``{``        ``for` `(``int` `i = ``0``; i < n; i++) {` `            ``// Performing AND operation``            ``// with number 31``            ``System.out.print((str.charAt(i) & NUM) + ``" "``);``        ``}``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``String str = ``"Geeks"``;``        ``int` `n = str.length();``        ``positions(str, n);``    ``}``}`

## Python

 `# Python3 implementation of the approach`` ` `NUM ``=` `31`` ` `# Function to calculate the position``# of characters``def` `positions(``str``):``    ``for` `i ``in` `str``:``         ` `        ``# Performing AND operation``        ``# with number 31``        ``print``((``ord``(i) & NUM), end ``=``" "``)`` ` `# Driver code``str` `=` `"Geeks"``positions(``str``)`

## C#

 `// C# implementation of the approach``using` `System;` `class` `GFG {` `    ``public` `static` `int` `NUM = 31;` `    ``// Function to calculate the position``    ``// of characters``    ``static` `void` `positions(``string` `str, ``int` `n)``    ``{``        ``for` `(``int` `i = 0; i < n; i++) {` `            ``// Performing AND operation``            ``// with number 31``            ``Console.Write((str[i] & NUM) + ``" "``);``        ``}``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``        ``string` `str = ``"Geeks"``;``        ``int` `n = str.Length;``        ``positions(str, n);``    ``}``}` `// This code is contributed by AnkitRai01`

## PHP

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## Javascript

 ``

Output:

`7 5 5 11 19`

Time Complexity: O(n)

Auxiliary Space: O(1)

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