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   "source": [
    "# [Maximum Path Sum II](https://projecteuler.net/problem=67)\n",
    "\n",
    "Since we took the time to find an efficient method in [problem 18](https://projecteuler.net/problem=18), we can just reuse that method here.\n",
    "\n",
    "So just read the triangle into a list-of-lists."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "id": "194eef78",
   "metadata": {},
   "outputs": [],
   "source": [
    "with open(\"txt/0067_triangle.txt\") as f:\n",
    "    triangle = [[int(n) for n in line.split(' ')] for line in f]"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "b39b979e",
   "metadata": {},
   "source": [
    "Then we'll use the bottom-up method from before (you could also use the top-down method)."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "id": "fa8a353f",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "7273"
      ]
     },
     "execution_count": 2,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "def max_path_sum(tri):\n",
    "    for i in reversed(range(0, len(tri) - 1)):\n",
    "        for j in range(0, len(tri[i])):\n",
    "            tri[i][j] += max(tri[i+1][j], tri[i+1][j+1])\n",
    "            \n",
    "    return tri[0][0]\n",
    "\n",
    "max_path_sum(triangle)"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "f06b0164",
   "metadata": {},
   "source": [
    "#### Copyright (C) 2025 filifa\n",
    "\n",
    "This work is licensed under the [Creative Commons Attribution-ShareAlike 4.0 International license](https://creativecommons.org/licenses/by-sa/4.0/) and the [BSD Zero Clause license](https://spdx.org/licenses/0BSD.html)."
   ]
  }
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