Generate combinations by order and fixed length.

Question

Suppose there is array

main[] = [a,b,c,....z];
a[] = [a1,a2,a3,....] ,
b[] = [b1,b2,b3,....]

and so on.

So if we generate combination in main.length of ordered then it looks like .....
a1 b1 c1, a1 b1 c2, a1 b1 c3 ......
a1 b2 c1, a1 b3 c1, .....
a2 b1 c1, a3 b1 c1, .....

and total count will be a.length * b.length * c.length ........ = ?

So write a code in any language to generate those combination. all the best.

alex

Answer 1

var combine = function(a) {
  var fn = function(n, src, got, all) {
    if (n == 0) {
      if (got.length > 0) {
        all[all.length] = got;
      }
      return;
    }
    for (var j = 0; j < src.length; j++) {
      fn(n - 1, src.slice(j + 1), got.concat([src[j]]), all);
    }
    return;
  }
  var all = [];
  for (var i=0; i < a.length; i++) {
    fn(i, a, [], all);
  }
  all.push(a);
  return all;
}

above code generates combine([1,2]) == [[1], [2], [1,2]];
but this does not meet my requirement is describe above.

Answer 2

function combination(x,selRow,currCol)
{
    if(currCol==x.length-1)
    {
        for(var i = 0; i < x[currCol].length; i++)
        {
            g_com.push(g_temItem.join("") + x[currCol][i]);
        }
        g_temItem.splice(g_temItem.length -1,1)

        return;
    }
    else
    {
        for(var i = 0; i < x[currCol].length; i++)
        {
            g_temItem.push(x[currCol][i]);
            combination(x,selRow,currCol + 1);
        }
        g_temItem.splice(g_temItem.length -1,1)
    }


}

here is my required code. recursion help me to solve the problem.

I have a comments for you people. you can comments what ever you want but those questions are technical not literary.