Finding the 1-based index of a character in alphabets using JavaScript

We are required to write a JavaScript function that takes in a lowercase English alphabet character. Our function should return the character's 1-based index in the alphabets.

Problem

Given a lowercase English alphabet character, we need to find its position in the alphabet where 'a' = 1, 'b' = 2, 'c' = 3, and so on.

Method 1: Using String indexOf()

We can create a string containing all alphabets and use indexOf() to find the position:

const char = 'j';
const findCharIndex = (char = '') => {
    const legend = ' abcdefghijklmnopqrstuvwxyz';
    if(!char || !legend.includes(char) || char.length !== 1){
        return -1;
    };
    return legend.indexOf(char);
};
console.log(findCharIndex(char));

10

Method 2: Using charCodeAt()

A more efficient approach uses ASCII values. The character 'a' has ASCII value 97, so we subtract 96 to get 1-based indexing:

const findCharIndexASCII = (char) => {
    if (!char || char.length !== 1 || char < 'a' || char > 'z') {
        return -1;
    }
    return char.charCodeAt(0) - 96;
};

console.log(findCharIndexASCII('a'));  // 1
console.log(findCharIndexASCII('j'));  // 10
console.log(findCharIndexASCII('z'));  // 26

1
10
26

Method 3: Using Alphabet Array

We can also use an array and indexOf():

const findCharIndexArray = (char) => {
    const alphabet = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z'];
    const index = alphabet.indexOf(char);
    return index === -1 ? -1 : index + 1;
};

console.log(findCharIndexArray('f'));  // 6
console.log(findCharIndexArray('m'));  // 13

6
13

Comparison

Conclusion

The charCodeAt() method is the most efficient approach for finding alphabet positions. It uses ASCII arithmetic and provides O(1) time complexity with minimal memory usage.