get the number of the reverse order of a given series

Question

I want to get the number of the reverse order of a given series in javascript. If I have the following series:

[1,2,2,2,5,5,7,8,8,10]

then if the input is number 8 the output should be 2, since:

 1  = 10
 2  = 7
 2  = 7
 2  = 7
 5  = 5
 5  = 5
 7  = 4
[8  = 2]
[8  = 2]
 10 = 1

//--> [1 , 2,3,4, 5,6, 7, 8,9, 10]
  --> [1 , 2,2,2, 5,5, 7, 8,8, 10]
      [10, 7,7,7, 5,5, 4, 2,2, 1 ] <-- // ==> [1,2,2,4,5,5,7,7,7,10]

Here is what I have done so far:

function getReverseNumber(arr, num)
{
    var newArr = new Array(arr.length);
    var temp;
    var counter = 1;
    for(var i = arr.length; i > 0; i--)
    {
        if(temp === arr[i])
        {
            newArr[arr.length - i] = counter;
            continue;
        }
        newArr[arr.length - i] = counter;
        temp = arr[i];
        counter++;
    }
    return newArr[num - 1];
}

but it doesn't work as expected:

getReverseNumber(new Array(1,2,2,2,5,5,7,8,8,10), 8) // returns 5 not 2

what is wrong in my function?

Answer 1

I think that you are overcomplicating it. You are only increasing counter by one when you increase it, and you place the numbers in revese order, so newArr end up as [1,2,2,3,4,4,5,5,5,6] instead of [10,7,7,7,5,5,4,2,2,1].

There is no need to calculate all those numbers and keep in an array. Just loop from 1 and up, and calculate which position that is in the array. Return the index when you find the value:

function getReverseNumber(arr, num) {
  for (var i = 1; i <= arr.length; i++) {
    if (arr[arr.length - i] == num) return i;
  }
  return -1; // not found
}

Demo: http://jsfiddle.net/Xedz6/

Answer 2

fiddle tested in IE 8 and 7 enjoy

if (!Array.prototype.indexOf) {
  Array.prototype.indexOf = function (searchElement /*, fromIndex */ ) {
    "use strict";
    if (this === void 0 || this === null) {
        throw new TypeError();
    }
    var t = Object(this);
    var len = t.length >>> 0;
    if (len === 0) {
        return -1;
    }
    var n = 0;
    if (arguments.length > 0) {
        n = Number(arguments[1]);
        if (n !== n) { // shortcut for verifying if it's NaN
            n = 0;
        } else if (n !== 0 && n !== (1 / 0) && n !== -(1 / 0)) {
            n = (n > 0 || -1) * Math.floor(Math.abs(n));
        }
    }
    if (n >= len) {
        return -1;
    }
    var k = n >= 0 ? n : Math.max(len - Math.abs(n), 0);
    for (; k < len; k++) {
        if (k in t && t[k] === searchElement) {
            return k;
       }

    }

    return -1;

  }

}


...

arr[ Math.Abs(arr.indexOf(num)-arr.length-1)]

sorry about the formatting. I'm on my phone.

Answer 3

Works in IE6 even ;)

function getReverseNumber(arr,num){
    alert(arr[ arr.length + arr.indexOf(num) * -1  ]);
}

getReverseNumber(new Array(1,2,2,2,5,5,7,8,8,10), 8); // alerts 2

Working demo: http://jsfiddle.net/AlienWebguy/7qvzE/

Answer 4

check out lastIndexOf

a = [1,2,2,2,5,5,7,8,8,10]
n = a.lastIndexOf(8)
alert(a.length - n)

Regarding these "IE8" remarks - even if dealing with obsolete browsers, this is not an excuse for reinventing the wheel. Use standard documented javascript library functions and include compatibility/degradation layers when appropriate.