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   "source": [
    "# Median of Two Sorted Arrays - Divide & Conquer - O(logn)\n",
    "1. Both arrays are of same length"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "def median(arr):\n",
    "    l = len(arr)\n",
    "    mid = int(l / 2)\n",
    "    \n",
    "    if l % 2 == 0:\n",
    "        return (arr[mid - 1] + arr[mid]) / 2\n",
    "    else:\n",
    "        return arr[mid]"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "def findMedianForArrays(arr1, arr2):\n",
    "    n = len(arr1) if len(arr1) == len(arr2) else 0\n",
    "    \n",
    "    if n <= 0:\n",
    "        return -1\n",
    "    if n == 1:\n",
    "        return (arr1[0] + arr2[0]) / 2\n",
    "    if n == 2:\n",
    "        return (max(arr1[0], arr2[0]) + min(arr1[1], arr2[1])) / 2\n",
    "\n",
    "    m1 = median(arr1)\n",
    "    m2 = median(arr2)\n",
    "    \n",
    "    if m1 == m2:\n",
    "        return m1\n",
    "    \n",
    "    mid = int(n / 2)\n",
    "    \n",
    "    if m1 < m2:\n",
    "        return findMedianForArrays(arr1[mid:], arr2[:mid + 1])\n",
    "    else:\n",
    "        return findMedianForArrays(arr1[:mid + 1], arr2[mid:])"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "16.0"
      ]
     },
     "execution_count": 3,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "arr1 = [1, 12, 15, 26, 38]\n",
    "arr2 = [2, 13, 17, 30, 45]\n",
    "\n",
    "findMedianForArrays(arr1, arr2)"
   ]
  }
 ],
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