Fizzbuzz - a Pyhton solution with timings

fizzbuzz.ipynb

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   "cell_type": "markdown",
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   "source": [
    "# Fizzbuzz\n",
    "\n",
    "The problem: print all numbers from 1 to a given limit, however if the number is:\n",
    "- a multiple of 3, prinf \"Fizz\"\n",
    "- a multiple of 5, print \"Buzz\"\n",
    "- a multiple of 3 and 5, print \"FizzBuzz\""
   ]
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   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## A brute solution\n",
    "For each number in the sequence, check if it is a multiple of 3 and 5"
   ]
  },
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   "execution_count": 1,
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   "source": [
    "def brute_fizzbuzz_aux(num):\n",
    "    multiple_3 = bool(num % 3 == 0)\n",
    "    multiple_5 = bool(num % 5 == 0)\n",
    "    \n",
    "    output = ''\n",
    "    if multiple_3:\n",
    "        output += 'Fizz'\n",
    "    if multiple_5:\n",
    "        output += 'Buzz'\n",
    "    \n",
    "    return output or num\n",
    "\n",
    "def brute_fizzbuzz(limit):\n",
    "    for n in range(0, limit):\n",
    "        print(brute_fizzbuzz_aux(n))"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## A slightly smarter solution\n",
    "The solution for each number cycles at every 15 numbers (since \"3*5 == 15\"). So, we can calculate and store the results for the first 15 numbers and just retrieve them when needed."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {},
   "outputs": [],
   "source": [
    "def soft_fizzbuzz_aux(num):\n",
    "    multiple_3 = bool(num % 3 == 0)\n",
    "    multiple_5 = bool(num % 5 == 0)\n",
    "    \n",
    "    output = ''\n",
    "    if multiple_3:\n",
    "        output += 'Fizz'\n",
    "    if multiple_5:\n",
    "        output += 'Buzz'\n",
    "    \n",
    "    return output or False\n",
    "\n",
    "def soft_fizzbuzz(limit):\n",
    "    current = 1\n",
    "    cycle = []\n",
    "    while current <= 15:\n",
    "        result = soft_fizzbuzz_aux(current)\n",
    "        cycle.append(result)\n",
    "        print(result or current)\n",
    "        current += 1\n",
    "        \n",
    "    while current < limit:\n",
    "        print(cycle[((current-1) % 15)] or current)\n",
    "        current += 1"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Time it\n",
    "When the limit is low (below 100), the optimization is barely noticiable."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "903 µs ± 219 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)\n"
     ]
    }
   ],
   "source": [
    "%%timeit\n",
    "%%capture\n",
    "brute_fizzbuzz(100)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "784 µs ± 147 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)\n"
     ]
    }
   ],
   "source": [
    "%%timeit\n",
    "%%capture\n",
    "soft_fizzbuzz(100)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "However, with greater values, we can see the improvement that the \"cycle\" approach has."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "9.73 s ± 995 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)\n"
     ]
    }
   ],
   "source": [
    "%%timeit\n",
    "%%capture\n",
    "brute_fizzbuzz(10000000)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "6.16 s ± 175 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)\n"
     ]
    }
   ],
   "source": [
    "%%timeit\n",
    "%%capture\n",
    "soft_fizzbuzz(10000000)"
   ]
  }
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