Returning an Array of Largest Number in sub-arrays of an Array

Question

I am trying to get the largest number of each sub-array in the array I sorted them but I can't seem to get my result. The result should be an array of the largest number of each sub-array.

function largestOfFour(arr) {
  for(var i = 0; i < arr.length; i+=1){
    for(var n = 0; n < arr[i].length; n+=1 ){
      arr[n].sort(function(a, b){return b-a});
    }
      console.log(arr[i][0]);
  }
  return arr[i][0];
}

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

A Paul · Accepted Answer · 2020-12-27 04:10:50Z

1

Try something like this. You do not need the second loop, also create a temp array and put the values there.

function largestOfFour(arr) {
var tmparr = [];
  for(var i = 0; i < arr.length; i+=1){
    //for(var n = 0; n < arr[i].length; n+=1 ){
      arr[i].sort(function(a, b){return b-a});
   // }
      tmparr.push(arr[i][0]);
  }
  console.log(tmparr)
  return tmparr;
}

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

answered Dec 27, 2020 at 4:10

A Paul

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Comments

Jonathan Gray · Accepted Answer · 2020-12-27 04:35:49Z

1

function largestInArrays(... arrays) {
  return arrays.map(a => Math.max(...a));
}

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

This works because Math.max combined with the ... argument trick will return the largest numerical value within an array and .map allows you to create a new array from the old array by performing an operation on each element of the array.

edited Dec 27, 2020 at 4:35

answered Dec 27, 2020 at 4:16

Jonathan Gray

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2 Comments

Sunday Ikpe Over a year ago

While I added an upvote to this answer, you don't really need the ellipsis (...) in the function definition, you only need it in the return line. This answer is simple and efficient.

Jonathan Gray Over a year ago

@SundayIkpe The reason the ellipsis are in the function definition is because I want all of the arguments to be treated as a single array (so the outer brackets aren't necessary when calling the function). Hence the name change from largestOfFour to largestInArrays. However looking back at the original question it seems I might have mistaken the specific wanted functionality here.

Caleb Mabry · Accepted Answer · 2020-12-27 04:59:35Z

1

You only have to loop through the array once and create a temporary array to store the largest numbers. Using Math.max() allows you to get the largest number within each array. There's no need to sort the arrays when Math.max() will already find the largest number for us.

function largestOfFour(arr) {
    let largestNumbers = []
    arr.forEach(innerArr => {
        let largestNumberOfInnerArray = (Math.max(...innerArr))
        largestNumbers.push(largestNumberOfInnerArray)
    });
    return largestNumbers
  }
   console.log(largestOfFour([[4, 5, 1, 3], [13, 27, 18, 26], [32, 35, 37, 39], [1000, 1001, 857, 1]]));

edited Dec 27, 2020 at 4:59

answered Dec 27, 2020 at 4:13

Caleb Mabry

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2 Comments

Okiemute Gold Over a year ago

Thanks alot. This worked, but after i added var, like this: var largestNumbers = [];

Caleb Mabry Over a year ago

Thanks for catching that. I've updated the post

Okiemute Gold · Accepted Answer · 2020-12-27 05:39:42Z

1

function largestOfFour(arr) {
  const largestNumb = [];
  // loop through the array
  for(var i = 0; i < arr.length; i+=1){
  // loop through each sub-array (of grabbed array), and sort the numbers in descending order.
      arr[i].sort(function(a, b){return b-a});
      console.log(arr[i][0]);
  // save each of the first value (largest num) in an array
      largestNumb.push(arr[i][0]);
  }
  // console.log(largestNumb);
  return largestNumb;
}

edited Dec 27, 2020 at 5:39

answered Dec 27, 2020 at 5:11

Okiemute Gold

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Comments

Sunday Ikpe · Accepted Answer · 2020-12-27 04:50:40Z

0

You could use underscore.js to achieve this:

<script src="https://cdnjs.cloudflare.com/ajax/libs/underscore.js/1.12.0/underscore-min.js"></script>
<script>
    function largestOfFour(arr){
      return arr.map(a => _.max(a));
    }
</script>

Tryining

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

your result will be:

 [5, 27, 39, 1001]

edited Dec 27, 2020 at 4:50

answered Dec 27, 2020 at 4:33

Sunday Ikpe

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2 Comments

Jonathan Gray Over a year ago

Regular sort does a string comparison not a numerical comparison

Sunday Ikpe Over a year ago

Yea, I have updated with underscoreJS. That way it behaves the way you want. Its the simplest answer I could think of.

Jonathan Gray · Accepted Answer · 2020-12-27 05:06:42Z

Functions in javascript have variables sometimes. A variable saves a value to use whenever coders want. But functions have specific scopes so variables can be defined twice in a function. For example

let food = 'banana';
console.log(food) // it shows 'banana'
function digestion() {
let food = 'poop'
console.log(food) // it shows 'poop'
}
console.log(food) // it still shows 'banana'

So scopes can define values differently. But in largestOfFour function arr[i][0] seems out of for loof. Seems there is a little mistake. Without the line the function works.

function largestOfFour(arr) {
  for(var i = 0; i < arr.length; i+=1){
    for(var n = 0; n < arr[i].length; n+=1 ){
      arr[n].sort(function(a, b){return b-a});
    }
      console.log(arr[i][0]);
  }
  //return arr[i][0];
}

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

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