Return Keypad Code #

int keypad(int num, string output[])
{
    /* Insert all the possible combinations of the integer number into the output string array. You do not need to
    print anything, just return the number of strings inserted into the array.
    */
    if (num <= 1)
    {
        output[0] = ""; // base case
        return 1;
    }
    // we can either do n%10 or n/multplier. The first is good.
    // as we are taking the last place it will need to be printed at the end of the branch. So a head recursion would suffice, I hope.

    int count = keypad(num / 10, output);

    // assume that the output is now filled.
    // we need to append the output with all possible values of the given string.

    int a = num % 10;
    string toPut = "";

    switch (a)
    {
    case 2:
        toPut = "abc";
        break;
    case 3:
        toPut = "def";
        break;
    case 4:
        toPut = "ghi";
        break;
    case 5:
        toPut = "jkl";
        break;
    case 6:
        toPut = "mno";
        break;
    case 7:
        toPut = "pqrs";
        break;
    case 8:
        toPut = "tuv";
        break;
    case 9:
        toPut = "wxyz";
        break;
    // default: return 1; // end of everything
    }

    // we need to make the copies of the output
    a = toPut.size();   // done with a
    int jumper = count; // jumper

    for (int i = 0; i < a - 1; i++)
    {
        for (int j = 0; j < count; j++)
        {
            output[jumper + j] = output[j];
        }
        jumper += count;
    }

    jumper = 0;
    // append the value
    for (int i = 0; i < a; i++)
    {
        char app = toPut[i];
        // cout << app << endl;
        for (int j = 0; j < count; j++)
        {
            output[jumper + j] += app;
        }
        jumper += count;
    }

    return a * count;
}

// 1 ""
// 2 abc
// 3 def
// 4 ghi
// 5 jkl
// 6 mno
// 7 pqrs
// 8 tuv
// 9 wxyz