Dear Sir or Madame,

I had a different solution to the problem which I think might interest you.
Generally I try to solve every looping with the array functions .filter(), .map() and .reduce(). I am from Hungary so at some point I might made some mistakes in the explanation, feel free to correct it if you found my solution worth to post it as a sample solution.

Yours faithfully

function updateInventory(arr1, arr2) {
    // first of all map through the curInv array to either preserve or update the count of its elements
    let arr = => {
        return !arr2.some(element => element[1] === currInvElement[1]) // if there is no element in the newInv array, that matches the curInv array's element
                 ? currInvElement // then preserve the element as it is
                 : [ 
                    currInvElement[0] + arr2.filter(newInvElement => currInvElement[1] == newInvElement[1])[0][0], //otherwise find the matching elements in both arrays and sum their count as the 0th element of the array
                    currInvElement[1] // and take the name of the element as the 1st element of the returned array

    // now let's add the items which are only in the newInv array
    arr = arr.concat(arr2
        .filter(elem => !arr.some(el => el[1] == elem[1])))
        .sort((a, b) => { //then sort them alphabetically
        return a[1] === b[1]
                ? 0
                : a[1] < b[1]
                    ? -1
                    : 1 });
    console.log(arr); // check your result
    return arr;

// Example inventory lists
var curInv = [
    [21, "Bowling Ball"],
    [2, "Dirty Sock"], 
    [1, "Hair Pin"], 
    [5, "Microphone"]

var newInv = [
    [2, "Hair Pin"], 
    [3, "Half-Eaten Apple"], 
    [67, "Bowling Ball"], 
    [7, "Toothpaste"]

updateInventory(curInv, newInv);

Hi @mihaly.nandor95 !

I edited your post to include the challenge link and I formatted your code.
I also moved your post over to the #contributors section.

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