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JavaScript

Posted by Trevor Johnson
Trevor Johnson
14,427 Points

JS Return Largest Numbers in Array

So here is my code challenge everyone. I think I am really close to solving this problem, and I would like to figure it out on my own without anyone posting the answer unless I ask.

When my code runs it stores the two greatest numbers from each array and adds them to newArray. I want it to only store the greatest value from each array and add it to newArray. I kind of understand why it is doing this, but I don't know how to fix it. I appreciate any advice on this.

Challenge: Return an array consisting of the largest number from each provided sub-array. For simplicity, the provided array will contain exactly 4 sub-arrays. Remember, you can iterate through an array with a simple for loop, and access each member with array syntax arr[i].

function largestOfFour(arr) {
  var arrayValue = 0;
  var newArray = [];
  var oldArray = arr;

  for (i = 0; i < oldArray.length; i ++) {
    for (j = 0; j < oldArray[i].length; j ++) {
      if (oldArray[i][j] > arrayValue) {
        arrayValue = oldArray[i][j];
        newArray.push(arrayValue);
      }
    }
  }
  return newArray;
}

largestOfFour([[4, 5, 1, 3], [13, 27, 18, 26], [32, 35, 37, 39], [1000, 1001, 857, 1]]);
Vlad Legkowski
Vlad Legkowski
9,882 Points

Hi Trevor, I suggest you to look into Math.max() function.

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Math/max

I managed to solve your issue using above function and ES6 with much less code just now, let me know if you want to see it!

Trevor Johnson
Trevor Johnson
14,427 Points

Yeah post your code that would be great

Vlad Legkowski
Vlad Legkowski
9,882 Points 
function largestOfFour(arr) {
    let newArr = [];
    arr.forEach((a) => {
      newArr.push(Math.max.apply(null, a));
    });
    return newArr;
  }

lmk if you have any questions

Trevor Johnson
Trevor Johnson
14,427 Points

Your code is a bit advanced for me. I don't know what let is, or forEach, or =>

Vlad Legkowski
Vlad Legkowski
9,882 Points

Sure,

This is ES6, the updated syntax of JS (i think that's what it is in a nutshell), there are few courses on T3H about it, I strongly suggest you look into them.

Good luck with coding!

2 Answers

Tri Pham
Tri Pham
18,671 Points

Ah okay. You should scope arrayValue so that its inside the loop (because your not going to use it outside of the loop) and it gets reset after each loop. Also you should put var next to i and j so that you don't accidentally use it outside and get an unintended bug.

function largestOfFour(arr) {

  var newArray = [];
  var oldArray = arr;

  for (var i = 0; i < oldArray.length; i ++) {
   var arrayValue = 0;
    for (var j = 0; j < oldArray[i].length; j ++) {
      if (oldArray[i][j] > arrayValue) {
        arrayValue = oldArray[i][j];
      }
    }
        newArray.push(arrayValue);
  }
  return newArray;
}
Trevor Johnson
Trevor Johnson
14,427 Points

Why did moving the variable inside of the loop change the result like that?

Tri Pham
Tri Pham
18,671 Points

Technically you could have done it your way and just do arrayValue = 0 inside but this is much cleaner code .This is the concept of scope. Variables declared (using var) inside of a "scope" can't be access outside of the scope. Usually you can tell the scope by these brackets { }. 

function justAnyFunction(){
     var a = 3;
}
console.log(a);
//you get something like undefined because you're outside the brackets and a was declared inside.

if(4 > 3){
    var b = 3;
}
console.log(b);
//you get something like undefined because you're outside the brackets and b was declared inside.

var c; 
if(4 > 3){
    c = 3;
}
console.log(c);
//you get 3 because you're outside the brackets and c was declared outside the brackets too.

As for forloops, everything declared inside the forloop gets destroyed each time one iteration finishes so a new arrayValue gets created and set to 0. In yours the arrayValue does not get its value reset after going through each array because its declared in a higher scope (outside the brackets).

Tri Pham
Tri Pham
18,671 Points
Tri Pham
Tri Pham
18,671 Points

You can push the new Value at the end outside of the inner loop so that you only push a max value once.

  for (i = 0; i < oldArray.length; i ++) {
    for (j = 0; j < oldArray[i].length; j ++) {
      if (oldArray[i][j] > arrayValue) {
        arrayValue = oldArray[i][j];
      }
    }
        newArray.push(arrayValue);     
  }
Trevor Johnson
Trevor Johnson
14,427 Points

That helps but it still doesn't work. It doesn't pass when you switch the first and second sub arrays like this.

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

and returns this

[27, 27, 39, 1001]