LESSON 157 OF 410 · LOOPING OVER ARRAY INDICES

Top K Brands V3 - Max in Array in JavaScript

Practice

Many brands sell the same product, stored in two parallel arrays: ratings[i] is the rating of brands[i]. Find the highest-rated brand by scanning with a loop.

You start with:

const brands = ["TechPro", "MegaCorp", "Quantum", "NeoByte", "Zenith"];
const ratings = [4, 5, 3, 5, 4];

Build:

1. MegaCorp (5)

Rules:

Keep the position of the best brand so far in bestIndex, starting at 0, and scan the positions from 1 to the end with a loop.
Update bestIndex only on a strictly higher rating. If two brands share the top rating, the one listed first wins (MegaCorp at position 1, not NeoByte at position 3).
Format the winner as "1. Brand (rating)" and print it with console.log(...).

Walk the positions with for (let i = 1; i < ratings.length; i++) { and compare each rating with the best one so far.

JavaScript turns the number into text for you when you join it to a string with +.

const brands = ["TechPro", "MegaCorp", "Quantum", "NeoByte", "Zenith"];
const ratings = [4, 5, 3, 5, 4];

let bestIndex = 0;
for (let i = 1; i < ratings.length; i++) {
    if (ratings[i] > ratings[bestIndex]) {
        bestIndex = i;
    }
}
console.log("1. " + brands[bestIndex] + " (" + ratings[bestIndex] + ")");

STEP 1 OF 5

Guided

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Top K Brands V3 - Max in Array in JavaScript

Top K Brands V3 in JavaScript