Max value of a multidimensional array javascript

Regardless of the dimension of the array, i believe this is the way to get the max of all primitives involved.

function getMax(a){
  return Math.max(...a.map(e => Array.isArray(e) ? getMax(e) : e));
}

var arr = [[1,  37.8, 80.8, 41.8],
           [2,  30.9, 69.5, 32.4],
           [3,  25.4,   57, 25.7],
           [4,  11.7, 18.8, 10.5],
           [5,  11.9, 17.6, 10.4],
           [6,   8.8, 13.6,  7.7],
           [7,   7.6, 12.3,  9.6],
           [8,  12.3, 29.2, 10.6],
           [9,  16.9, 42.9, 14.8],
           [10, 12.8, 30.9, 11.6],
           [11,  5.3,  7.9,  4.7],
           [12,  6.6,  8.4,  5.2],
           [13,  4.8,  6.3,  3.6],
           [14,  4.2,  6.2,  3.4]];
 console.log(getMax(arr));

It should work on arrays with indefinite dimension.

function getMax(a){
  return Math.max(...a.map(e => Array.isArray(e) ? getMax(e) : e));
}

var arr = [[1,  37.8, 80.8, 41.8],
           [2,  30.9, 69.5, 32.4],
           [3,  25.4,   57, 25.7],
           [4,  11.7, 18.8, 10.5],
           [5,  11.9, 17.6, 10.4],
           [6,   8.8, 13.6,  7.7],
           [7,   7.6, 12.3,  9.6],
           [8,  12.3, 29.2, 10.6],
           [9,  16.9, 42.9, 14.8],
           [10, 12.8, 30.9, 11.6],
           [11,  5.3,  [6.1,[56.7,[98.55]]],  4.7],
           [12,  6.6,  8.4,  5.2],
           [13,  4.8,  6.3,  3.6],
           [14,  4.2,  6.2,  3.4]];
 console.log(getMax(arr));

Simplest Solution

Flatten and use the usual Math.max with spread operator.

Math.max(...arr.flat())

const arr = [[1,  37.8, 80.8, 41.8],
           [2,  30.9, 69.5, 32.4],
           [3,  25.4,   57, 25.7],
           [4,  11.7, 18.8, 10.5],
           [5,  11.9, 17.6, 10.4],
           [6,   8.8, 13.6,  7.7],
           [7,   7.6, 12.3,  9.6],
           [8,  12.3, 29.2, 10.6],
           [9,  16.9, 42.9, 14.8],
           [10, 12.8, 30.9, 11.6],
           [11,  5.3,  7.9,  4.7],
           [12,  6.6,  8.4,  5.2],
           [13,  4.8,  6.3,  3.6],
           [14,  4.2,  6.2,  3.4]];

Math.max(...arr.flat())

You can get max value of multi dimension array using following method:

var arr = [[1, 5,6], [4, 7,8], [3, 8,20], [2, 3,1],[12, 4,5]];

console.log(Math.max.apply(Math, arr.map(function (i) {
    return i[0]+i[1]+i[2];
})));

It first use array.map() to convert the multi-dimensional array to a flat one, and after that use Math.max().

Most answers here use apply or the spread operator ... to call the Math.max function with all the elements of an array as parameters.

For large arrays it is safer to use reduce:

// returns maximum of an array
const getArrayMax = array => array.reduce((a, b) => Math.max(a, b));

// returns maximum of a 2D array
const getArrayMax2d = array2d => getArrayMax(array2d.map(getArrayMax));

based on this answer, you can do it in 1 line (assuming ES6):

const arr = [[12,45,75], [54,45,2],[23,54,75,2]];

const max = Math.max(...[].concat(...arr));

const min = Math.min(...[].concat(...arr));

console.log(max);

console.log(min);

The solution using Array.prototype.push, Math.min and Math.max methods:

// arr is your initial array
var flattened = [], minValue, maxValue;
arr.forEach(function (v) {
    Array.prototype.push.apply(flattened, v);
});

minValue = Math.min.apply(null, flattened);
maxValue = Math.max.apply(null, flattened);

console.log('min: ' + minValue, 'max: ' + maxValue);  // min: 1 max: 80.8

DEMO link

You can also achieve this with a reduction (it's slower but if the array is not huge, it doesn't matter) resulting in a object containing both the min and the max like so:

matrix.reduce(function (res, item) {
    item.forEach(function (val) {
        if (!res.hasOwnProperty('max') || val > res.max) res.max = val;
        if (!res.hasOwnProperty('min') || val < res.min) res.min = val;
    });
    return res;
}, {});

If i could i would comment this on DankMasterDan's answer

let a = [[12,45,75], [54,45,2],[23,54,75,2]];
let max = Math.max(...a.map(r=>Math.max(...r)));

this may avoid the stack size exceeded error. David mentioned since its is batching each row with the map first.

I found this way to your solution very simple way

function largestOfFour(arr) {
  let maxValue;
  let arrMax = []
for(let i = 0 ; i < arr.length ; i++){
maxValue = Math.max.apply(null ,arr[i])
arrMax.push(maxValue)
}
  return arrMax;
}
largestOfFour([[4, 5, 1, 3], [13, 27, 18, 26], [32, 35, 37, 39], [1000, 1001, 857, 1]]);
 return --> [ 5, 27, 39, 1001 ]