Frequent number in Javascript

Question

In my task I have to write a program that finds the most frequent number in an array and how many time it is repeated. I wrote something, but only prints the max number of repeating time. So my question is how can I print the value of this element(max number)(in my case is 4)? :)

var array = ['13', '4', '1', '1', '4', '2', '3', '4', '4', '1', '2', '4', '9', '3'];

function frequentNumber(arr) {
   var result = [],
       result2 = [];

   sorted_arr = arr.sort();

    console.log(sorted_arr);
    n=1;
    buffer = [];

    for (var i = 0; i < sorted_arr.length; i++) {

        if (sorted_arr[i + 1] === sorted_arr[i]) {
            n++;
        }else{

            if(buffer.length != 0 ){
                for(var j = 0; j < buffer.length; j++){

                    if(buffer[j] < n){
                        result = sorted_arr[j] + " is " + n;
                    }
                }
            }else{buffer.push(n)}

            n=1;
        }
    }

//    console.log(Math.max.apply(Math, buffer));
    console.log(buffer);
}

frequentNumber(array); 

Answer 1

You can use .reduce to create an object with the numbers and the times they're listed - then simply iterate the object and find the highest value:

var ranks = array.reduce(function(totals, num) {
    if (!totals[num]) totals[num] = 0;
    totals[num]++;

    return totals;
}, {});

//Iterate and find
var max = 0;
Object.keys(ranks).forEach(function(num) {
    if (ranks[num] > max) {
        max = num;
    }
});

console.log(max); //4 - number
console.log(ranks[max]); //5 - times repeated

This doesn't take numbers with the same count into play - whichever number is iterated in the object first with the highest count will be the result - and since objects are unordered, a same count could have different results over multiple executions.