Find largest number in array of arrays

Question

JS Function:

function largestOfFour(arr) {


var biggest = 0;
var newArray = [];

for(var i = 0; i < arr.sort().reverse().length; i++){
  for(var j = 0; j < arr[i].length; j++){

    if(biggest < arr[i][j]){
      biggest = arr[i][j]; // problem after this
      //newArray.push(biggest);  this will add the first element in each sub-array
      // result [4,5,13,27,32,35,37,39,1000,1001]
    }

  }
}

return newArray;

}

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


//  Expected Output:

// [5, 27, 39, 1001]

I am having a hard time with this. How would I add or change anything to what I have to get the biggest number from each sub-array and add that number to newArray using for-loops?

add 'newArray.push (biggest)' after your second loop inside the first loop — krisph
– krisph, Commented Jan 31, 2017 at 15:57
for(var i = 0; i < arr.length; i++) { newArray.push(Math.max.apply(null, arr[i])); } — Dom
– Dom, Commented Jan 31, 2017 at 16:02

Seiyria · Accepted Answer · 2017-01-31 16:01:50Z

4

Since you'd like to use a for loop:

var largestOfFour = (arr) => {
  let largest = [];
  for(let i = 0; i < arr.length; i++) {
    largest[i] = arr[i].sort((a,b) => b-a)[0];
  }
  return largest;
}

What this will do is sort the inner arrays and grab the first one out of there.

edited Jan 31, 2017 at 16:01

answered Jan 31, 2017 at 16:00

Seiyria

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2 Comments

mr rogers Over a year ago

you could save a little time by doing b-a in the comparator and skip the reverse()

Seiyria Over a year ago

Sure. I'll make that change. I forgot JS doesn't sort arrays of numbers numerically so I just sorta patched that in, thanks.

Nenad Vracar · Accepted Answer · 2017-01-31 15:58:01Z

3

You can use map() instead of for loops.

var arr = [[4, 5, 1, 3], [13, 27, 18, 26], [32, 35, 37, 39], [1000, 1001, 857, 1]]

function arrMax(data) {
  return data.map(e => Math.max.apply(null, e))
}
console.log(arrMax(arr))

answered Jan 31, 2017 at 15:58

Nenad Vracar

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4 Comments

Grizzly Over a year ago

In my question I stated that I would like to use for loops

baao Over a year ago

Yeah, why use a ferrari if you can have a lada... @BviLLe_Kid

Grizzly Over a year ago

@baao I know that this answer is probably the better answer.. but this is just to test myself with loops

Dom Over a year ago

@BviLLe_Kid then for loop over the arr, then use Math.max.apply(null, arr[i]) on the sub arrays. However, I'd argue .map would be the preferable approach since it returns out a new array.

Nina Scholz · Accepted Answer · 2017-01-31 16:02:26Z

1

You could use the spread syntax for Math.max.

function arrMax(data) {
    return data.map(a => Math.max(...a))
}

var arr = [[4, 5, 1, 3], [13, 27, 18, 26], [32, 35, 37, 39], [1000, 1001, 857, 1]]

console.log(arrMax(arr));

answered Jan 31, 2017 at 16:02

Nina Scholz

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Comments

Olivier Boissé · Accepted Answer · 2017-01-31 16:10:19Z

0

Maybe not the best but this should work

function largestOfFour(series) {
   return series.map(function (serie) {
      return Math.max.apply(null, serie);
      // return Math.max(...serie);
   });
}

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

edited Jan 31, 2017 at 16:10

answered Jan 31, 2017 at 16:04

Olivier Boissé

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1 Comment

Seiyria Over a year ago

You can't just do .sort(), you need to pass in a comparator when dealing with numbers. If you try this code, the last array will return 857 as the highest number, which is not correct.

Redu · Accepted Answer · 2017-01-31 16:47:00Z

0

For very large sizes of arrays and sub array items i guess this would be the most efficient..

var arr = [[4, 5, 1, 3], [13, 27, 18, 26], [32, 35, 37, 39], [1000, 1001, 857, 1]],
    res = arr.map(a => a.reduce((p,c) => p > c ? p : c));
console.log(res);

answered Jan 31, 2017 at 16:47

Redu

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Comments

Vladyslav Olshevskyi · Accepted Answer · 2022-09-25 18:53:43Z

0

I did it like this, it works with the negative numbers as well as with positives

function largestOfFour(arr) {
  var maxvalarr = [];
  for(let i = 0; i < arr.length ; i++){
    let maxvalue = -10000000;
    for(let g = 0; g < arr[i].length; g++)
      if(arr[i][g] > maxvalue)
         maxvalue = arr[i][g];
      
    maxvalarr[i] = maxvalue;
  }
console.log(maxvalarr)

  return maxvalarr;
}

largestOfFour([[4, 5, 1, 3], [13, 27, 18, 26], [32, 35, 37, 39], [1000, 1001, 857, 1]]);
largestOfFour([[4, 9, 1, 3], [13, 35, 18, 26], [32, 35, 97, 39], [1000000, 1001, 857, 1]]);
largestOfFour([[17, 23, 25, 12], [25, 7, 34, 48], [4, -10, 18, 21], [-72, -3, -17, -10]])

edited Sep 25, 2022 at 18:53

answered Sep 25, 2022 at 18:50

Vladyslav Olshevskyi

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