 # Algorithm Scripting, Mutations

Tell us what’s happening:

Hi guys. i 'm sure there is more elegant/shorter way to get done with this, i just cant get to that. Anyway most of requirements checks expect last one as i assumed, because it count every “o” . And when i try to test “===” with my ending if statement then hello fails because it has more than the length of second word. Is there way just to count letters that repeat just once, L for example, or something like that, i mean is there solution for this code, or i should try something else?

ps. Sorry for reposting same question already asked, just wanted to check if somehow my code can be “fixed”.

``````
function mutation(arr) {
let counter = 0;                         //  ----for later check

let temp1 = arr.slice(0, 1).join().toLowerCase();     //  ----- string-ing, lowerCase-ing
let temp2 = arr.slice(1, 2).join().toLowerCase();

for (let i = 0; i < temp2.length; i++) {            //    ----- loops, checking letters and "adding" ones that exist
for (let j = 0; j < temp1.length; j++) {
if (temp2.charAt(i) === temp1.charAt(j)) {
counter++;
}
}
}

if (counter >= temp2.length) {                           // ----- checking if it has  enough to pass true
return true;
}
else {
return false;
}

}

mutation(["hello", "Hello"]);
``````

Solved with:

``````function mutation(arr) {
let word1 = arr.toLowerCase();
let word2 = arr.toLowerCase();

for (let i = 0; i < word2.length; i++) {
if (!word1.includes(word2[i])) {
return false;
}
}
return true;
}
``````

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