How do you generate all 5 letter strings from 3 characters?

Given 3 characters (abc), I want to generate all possible 5-letter strings with them (aaaaa, aaaab, … ccccb, ccccc)

const s = 'byg';
const p = [];

for (let i = 0; i < 3; i++) {
    for (let j = 0; j < 3; j++) {
        for (let k = 0; k < 3; k++) {
            for (let l = 0; l < 3; l++) {
                for (let m = 0; m < 3; m++) {
                    p.push(s[i] + s[j] + s[k] + s[l] + s[m]);
                }
            }
        }
    }
}

console.log(p, p.length === 3 ** 5)

This feels like an inefficient way to do this, so is there a more elegant/efficient way to do this?

Answer

Your nested for loops does imply that code can be refactored either using recursion or as in my example below by creating a higher level loop.

This approach allows us to generate strings of any desired length.

let characters = "abc";
let desiredLength = 5;

let theSet = [""];
for ( let length = 0; length < desiredLength; length++){
    let extendedSet = [];
    theSet.forEach( (item) => 
        extendedSet.push( ...extendWith( item, characters) )
    )
    theSet = extendedSet; 
}

console.log('result ', theSet);
   
// given a strings and a set of characters
// generate an array of string extended with each
// of the characters. 
// "a" with "xy" generates
// [ "ax", "ay" ]
function extendWith( extendThis, characters){
    let result = [];
    [...characters].forEach( (c) => result.push(extendThis+c) );
    return result;
}

We can make that extendWith function more succinct like this

function extendWith( extendThis, characters){   
   return [...characters].map( c => extendThis + c );  
}

and as it’s now just a one line expression we can dispense with the utility function and simplify a bit more

for ( let length = 0; length < desiredLength; length++){

    theSet = theSet.flatMap( (item) => 
        [...characters].map( c => item + c )
    );

}