Split string into array with alternate one & two digits

Question

Is it possible to split a string (possibly with regex) so that every other number is a pair(starting from the right hand end)?

//             1                [1]
//            12               [12]
//           123             [1,23]
//          1234           [1,2,34]
//         12345          [12,3,45]
//        123456        [1,23,4,56]
//       1234567      [1,2,34,5,67]
//      12345678     [12,3,45,6,78]
//     123456789   [1,23,4,56,7,89]
//    1234567890 [1,2,34,5,67,8,90]

I tried reversing the string & then adding blocks of alternating two and one characters till the end (front) of the string. Then reversing it again. That mainly worked but it was flawed (didn't work for all cases). I also tried the regex

(\d\d)(\d)(\d\d)(\d)(\d\d)(\d)

But that doesn't work either (only in the regex tester ironically) - it's too long but I'll need it to work for 10 digit numbers max.

In this case: // 1234 [1,2,34] maybe this // 1234 [12,34] ? — Jose Carlos Ramos Carmenates
– Jose Carlos Ramos Carmenates, Commented Feb 3, 2016 at 21:01
I'm breaking the digits down to words. 1 thousand, 2 hundred, thirty four. — Ghoul Fool
– Ghoul Fool, Commented Feb 3, 2016 at 21:03

Aminadav Glickshtein · Accepted Answer · 2016-02-03 21:28:43Z

3

No so hard:

Start from the right, I will take one time 2 digits, the second time 1 digits. Using slice. Then I will use unshift to push it in the begining of the array.

I'm using a flag, to know when to take only 1 parameter, and when to take two parameters (pair flag)

m(1)
m(12)
m(123)
m(1234)
m(12345)
m(123456)
m(12345678)
m(123456789)
m(1234567890)
function m(x){
    x=x.toString()
    var a=[]
    var v;
    var y=2
    while(x){
        v=x.slice(-y)
        x=x.slice(0,-y)        
        y=y==1? 2:1
        a.unshift(v)
    }
    console.log(a)
}

Result:

["1"]
["12"]
["1", "23"]
["1", "2", "34"]
["12", "3", "45"]
["1", "23", "4", "56"]
["12", "3", "45", "6", "78"]
["1", "23", "4", "56", "7", "89"]
["1", "2", "34", "5", "67", "8", "90"]

answered Feb 3, 2016 at 21:28

Aminadav Glickshtein

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1 Comment

Ghoul Fool Over a year ago

#Amina Your solution is more elegant and quicker - but I was determined to get there!

Ghoul Fool · Accepted Answer · 2016-02-05 20:12:07Z

I got there in the end.

//             1                [1]
//            12               [12]
//           123             [1,23]
//          1234           [1,2,34]
//         12345          [12,3,45]
//        123456        [1,23,4,56]
//       1234567      [1,2,34,5,67]
//      12345678     [12,3,45,6,78]
//     123456789   [1,23,4,56,7,89]
//    1234567890 [1,2,34,5,67,8,90]

var results = "";

pairUp(1)
pairUp(12)
pairUp(123)
pairUp(1234)
pairUp(12345)
pairUp(123456)
pairUp(12345678)
pairUp(123456789)
pairUp(1234567890)

alert(results);

function pairUp(num)
{
  var s = num.toString();
  s = s.split("").reverse().join("");
  var a = s.split("");

  var r = []; // our results array
  count = 0;

  for (var i = 0; i <= a.length -1; i++)
  {
    temp = a[count];

    if ((i % 2) == 0) // even (0, 2, 4)
    {
      var p = a[count+1];
      var q = a[count];

      if (p == undefined) p = "";
      if (q) r.push(p + q+ "");

      count+=2;
    }
    else 
    {
      if (temp != undefined) r.push(temp + "");
      count+=1;
    }
  } // end loop


  r = r.reverse();

  results+= r + "\n";

} // end pair up

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