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    "# In how many ways can you split a string into all possible substrings?\n",
    "\n",
    "I recently chatted with a friend about the following question. Let's say that you have a string, say \"sky\". In how many ways can you split it into substrings. All possibilities are:\n",
    "\n",
    "```\n",
    "sky\n",
    "sk y\n",
    "s ky\n",
    "s k y\n",
    "```\n",
    "\n",
    "Intuitively, I suspected an answer similar to $ 2^{n} $, but I didn't sit down to figure out the exact answer.\n",
    "\n",
    "So let's solve this problem. If we start with a string which's length is N then each split can be represented in a binary form of length N. For example:\n",
    "\n",
    "```\n",
    "sk y = 01 1\n",
    "```\n",
    "\n",
    "A `1` shows that we are ending a string with the character corresponding to this `1` and a `0` shows that we're continuing a string. If we have a `0` in the beginning, we're just continuing the empty string. So in our example, `01` corresonds to `sk` and the ending `1` corresponds to `y`. We have to notice that a valid split always ends with `1`, because we're always ending at the end of the original word. Furthermore, all binary representations of the first `N-1` characters corresponds to a possible split.\n",
    "\n",
    "We can now see that the answer to the question is $ 2^{N-1} $.\n",
    "\n",
    "Knowing this mapping between splits and binary numbers, we can even write some code that uses this property. It looks like this:\n"
   ]
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     "text": [
      "0001 rain\n",
      "1001 r ain\n",
      "0101 ra in\n",
      "1101 r a in\n",
      "0011 rai n\n",
      "1011 r ai n\n",
      "0111 ra i n\n",
      "1111 r a i n\n"
     ]
    }
   ],
   "source": [
    "word = \"rain\"\n",
    "N = len(word)\n",
    "\n",
    "for b in range(2**(N-1)):\n",
    "  split = '{0:03b}'.format(b)[::-1] + '1 '\n",
    "  for i in range(len(word)):\n",
    "    split += word[i]\n",
    "    if b & (1 << i):\n",
    "      split += ' '\n",
    "  print split"
   ]
  },
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   "metadata": {},
   "outputs": [],
   "source": []
  }
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