Question

If a user enters 5 numbers, lets say… 4, 4, 7, 7, 4. 4 occurred 3 (most number of) times. So the output should be 4.

How can I do this using JavaScript? Would much appreciate your help.
Thanks!

I’ve tried this so far. It works, but it’s too long, looking for something short and simple way.

P.S. This is not my homework!

    var n = parseInt(prompt("How many numbers do you like to enter?", ""));
    var num = new Array();

    for (i = 1; i <= n; i++) {
        num[i] = parseInt(prompt("Enter a number", ""));
        document.write("Entered numbers are: " + num[i] + "<br/>");
    }

    var total = new Array();
    for (i = 1; i <= n; i++) {
        var count = 1;
        for (j = i + 1; j <= n; j++) {
            if (num[i] == num[j]) {
                count++;
            }
            total[i] = count;
        }
    }

    var most = 0;
    for (i = 0; i < n; i++) {
        if (most < total[i]) {
            most = total[i];
        }
        var val = i;
    }
    document.write("<br/>" + num[val] + " is occurred " + most + " times");

Answer 1

Create an array a with a lot of numbers, using array literals:

var a = [1, 2, 3, 4, 4, 5, 1, 2, 3, 1, 2, 1, 1];

Create a plain object o, using object literals.

var o = {},          /* Creates a new object */
    i = 0,           /* Declare variable i, initialise value at 0*/
    m = {m:0,t:null},
    t,               /* Declare variable t */
    len = a.length;  /* Declare variable len, set value to the array's length */

Loop through array a using a for(;;)-loop and increment the counter. The counter is stored in a hashmap on object o.
(o[a[i]] || 0) is needed for the first occurrence of the key: When it’s not found, the value 0 is used instead of undefined. See also Short-circuit evaluation: Logical OR.

for ( ; i < len ; i++ ) {
    o[ a[i] ] = ( o[ a[i] ] || 0 ) + 1;
}

Then you have an object o which looks like:

o = {
    "1": 5,
    "2": 3,
    "3": 2,
    "4": 2,
    "5": 1
}

Then loop through o using a for(.. in ..)-loop and find max times presented.
At the bottom of the loop, the conditional ternary .. ? .. : .. operator is used:

for ( i in o ) {
    t = { 
        m: o[i], 
        i: i 
    };
    m = m.m < t.m ? t : m;
}

After this loop m is equal to:

m = { 
    i: "1", 
    m: "5"
};

And the maximum value can be captured using:

o[m];

witch gives you:

5

---

DEMO

http://jsbin.com/utiqey/

var a = [1, 2, 3, 4, 4, 5, 1, 2, 3, 1, 2, 1, 1]; 

var o = {}, 
    i = 0, 
    m = {m:0,t:null}, 
    t,
    len = a.length; 

for ( ; i < len ; i++ ) { 
    o[ a[i] ] = ( o[ a[i] ] || 0 ) + 1; 
} 


for ( i in o ) { 
    t = { 
        m: o[i], 
        i: i 
    };
    m = m.m < t.m ? t : m;
} 

alert(m.i + " is the highest presented " + m.m + " times");