Can someone walk me through this exercise? Write a JavaScript program to find the most frequent item of an array.
var arr1 = [3, 'a', 'a', 'a', 2, 3, 'a', 3, 'a', 2, 4, 9, 3];
var mf = 1;
var m = 0;
var item;
for (var i = 0; i < arr1.length; i++) {
for (var j = i; j < arr1.length; j++) {
if (arr1[i] == arr1[j]) m++;
if (mf < m) {
mf = m;
item = arr1[i];
}
}
m = 0;
}
alert(item + " ( " + mf + " times ) ");
I've been checking out some similar questions on stackoverflow, just can't find the answers that I want.
My questions are:
I don't understand why there needs to have two for loops.
Why the need for mf
and m
. Seems a bit confusing.
Is there other way of solution on this?
I really do not think there is a need for 2 loops in this solution. You can look at this prototyping code which uses a simple data structure called map:
var arr=[3, 'a', 'a', 'a', 2, 3, 'a', 3, 'a', 2, 4, 9, 3];
var map = {};
var mostFrequentElement = arr[0];
function findMostFrequent(){
for(var i = 0; i<arr.length; i++){
if(!map[arr[i]]){
map[arr[i]]=1;
}else{
++map[arr[i]];
if(map[arr[i]]>map[mostFrequentElement]){
mostFrequentElement = arr[i];
}
}
}
alert(mostFrequentElement);
}
User want's explanation of the code:
Here they select the 1st element of the array and compare it with every element then on.
Then they increment the counter m
every time the same elements occur again, frequency of that element.
Also a variable mf
is kept to keep track of maximum frequency. Compare the elements frequency with maximum frequency and update item
and mf
as per present element's frequency.
var arr1=[3, 'a', 'a', 'a', 2, 3, 'a', 3, 'a', 2, 4, 9, 3]; //array
var mf = 1; //default maximum frequency
var m = 0; //counter
var item; //to store item with maximum frequency
for (var i=0; i<arr1.length; i++) //select element (current element)
{
for (var j=i; j<arr1.length; j++) //loop through next elements in array to compare calculate frequency of current element
{
if (arr1[i] == arr1[j]) //see if element occurs again in the array
m++; //increment counter if it does
if (mf<m) //compare current items frequency with maximum frequency
{
mf=m; //if m>mf store m in mf for upcoming elements
item = arr1[i]; // store the current element.
}
}
m=0; // make counter 0 for next element.
}
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