Print all the possible strings a telephone number can represent.

Solution #1: The below code doesn't take into account the fact that keys 7 and 9 represent 4 alphabets and also it just prints ' ' and '+' for key 1 and 0 respectively. The code is just to explain the logic for printing the combinations and additional logic can be added to handle these cases. Here is the code,

char digitToChar(char x, int index)
{
    char ch;

    switch(x)
    {
        case '0':
            ch = '+'; break;
        case '1':
            ch = ' '; break;
        case '2':
            ch = "abc"[index]; break;
        case '3':
            ch = "def"[index]; break;
        case '4':
            ch = "ghi"[index]; break;
        case '5':
            ch = "jkl"[index]; break;
        case '6':
            ch = "mno"[index]; break;
        case '7':
            ch = "pqrs"[index]; break;
        case '8':
            ch = "tuv"[index]; break;
        case '9':
            ch = "wxyz"[index]; break;
        default:
            ch = ' ';

    }

    return ch;
}

void printAll(char* num, char* arr, int len, int x)
{
    int i=0;

    if(x == len)
    {
        printf("%s\n", arr);
        return;
    }

    for(i=0; i<3; i++)
    {
        arr[x] = digitToChar(num[x], i);
        printAll(num, arr, len, x+1);
    }
}

int main()
{
    char* num = "8017393450";
    char arr[20] = {0,};

    printAll(num, arr, 10, 0);

    return 0;
}