Select the top m most repeated elements from array in Javascript

You could create a temporary object to hold the occurrences like key: count. Then use the array returned by Object.keys to sort the object values. Then slice the last "n" elements from that array in .reverse() order.

Here is an example utility function to get top "n" max occurrences. Note: As @n⁶ stated above in the comments, this ignores the ties.

Demo fiddle: https://jsfiddle.net/abhitalks/s43k6vsq/

Snippet:

var data1 = [1, 3, 2, 3, 5, 2, 2, 3, 6, 8, 9], 
    data2 = [4, 4, 4, 2, 2, 1, 5, 5, 8, 8, 8, 8, 3], 
    data3 = [3, 23, 23, 45, 54, 54, 54];

// Function parameters = Data Array and top "n" elements to find
function getMax(data, n) {
    var tmp = {}, tops = [];
  
    // Create object with count of occurances of each array element
    data.forEach(function(item) {
        tmp[item] = tmp[item] ? tmp[item]+1 : 1;
    });
  
    // Create an array of the sorted object properties
    tops = Object.keys(tmp).sort(function(a, b) { return tmp[a] - tmp[b] });
  
    // Return last n elements in reverse order
    return tops.slice(-(n)).reverse();
}

// Test with sample data and top "n" 
console.log(getMax(data1, 2));
console.log(getMax(data2, 3));
console.log(getMax(data3, 1));

a solution:

var arr = [1,2,3,41,11,2,3,4,1];
var sorted = [];

for (var i = 0; i < arr.length; i++) {
    var index = -1;

    for (var j = 0; j < sorted.length; j++) {
        if (sorted[j].val == arr[i]) index = j;
    }

    if (index == -1) {
        sorted.push({val: arr[i], count: 1})
    }

    else {
        sorted[index].count++;
    }

}

console.log(sorted.sort(function(a,b){

if (a.count > b.count) return -1;

if (a.count < b.count) return 1

    return 0

})[0])

FIDDLE