# Basic Algorithm Scripting - Return Largest Numbers in Arrays

Tell us what’s happening:
After implementing the code the largest numbers of each subarray where all displayed in a new array on the console but it didn’t still pass the test

``````function largestOfFour(arr) {
let newArr = [];
for (let i = 0; i < arr.length; i++) {
let largestNumbers = arr[0][i]
for (let j = 0; j < arr[i].length; j++) {
if (arr[i][j] > largestNumbers) {
largestNumbers = arr[i][j]
}
}
newArr.push(largestNumbers)
}
return newArr;
}

largestOfFour([[4, 5, 1, 3], [13, 27, 18, 26], [32, 35, 37, 39], [1000, 1001, 857, 1]]);
console.log(largestOfFour([[4, 5, 1, 3], [13, 27, 18, 26], [32, 35, 37, 39], [1000, 1001, 857, 1]]))
``````

User Agent is: `Mozilla/5.0 (Macintosh; Intel Mac OS X 10_15_7) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/111.0.0.0 Safari/537.36`

Challenge: Basic Algorithm Scripting - Return Largest Numbers in Arrays

Consider test case;
`([[13, 27, 18, 26], [4, 5, 1, 3], [32, 35, 37, 39], [1000, 1001, 857, 1]])` should return `[27, 5, 39, 1001]` .

`[ 27, 27, 39, 1001 ]`

For the case
`largestOfFour([[17, 23, 25, 12], [25, 7, 34, 48], [4, -10, 18, 21], [-72, -3, -17, -10]])` should return `[25, 48, 21, -3]`

`[ 25, 48, 25, 12 ]`

There is just one tiny mistake in your code. Take a look at this line:

``````let largestNumbers = arr[0][i]
``````

You are always grabbing the very first array and then `i` inside of that array. Those should be flipped around:

``````let largestNumbers = arr[i][0]
``````

So that it is always grabbing the current array at `i` and the first index of that current array instead!

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