tape_equilibrium.ipynb

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    "A non-empty array A consisting of N integers is given. Array A represents numbers on a tape.\n",
    "\n",
    "Any integer P, such that 0 < P < N, splits this tape into two non-empty parts: A[0], A[1], ..., A[P − 1] and A[P], A[P + 1], ..., A[N − 1].\n",
    "\n",
    "The difference between the two parts is the value of: |(A[0] + A[1] + ... + A[P − 1]) − (A[P] + A[P + 1] + ... + A[N − 1])|\n",
    "\n",
    "In other words, it is the absolute difference between the sum of the first part and the sum of the second part.\n",
    "\n",
    "For example, consider array A such that:\n",
    "\n",
    "  A[0] = 3\n",
    "  A[1] = 1\n",
    "  A[2] = 2\n",
    "  A[3] = 4\n",
    "  A[4] = 3\n",
    "We can split this tape in four places:\n",
    "\n",
    "P = 1, difference = |3 − 10| = 7 \n",
    "P = 2, difference = |4 − 9| = 5 \n",
    "P = 3, difference = |6 − 7| = 1 \n",
    "P = 4, difference = |10 − 3| = 7 \n",
    "Write a function:\n",
    "\n",
    "def solution(A)\n",
    "\n",
    "that, given a non-empty array A of N integers, returns the minimal difference that can be achieved.\n",
    "\n",
    "For example, given:\n",
    "\n",
    "  A[0] = 3\n",
    "  A[1] = 1\n",
    "  A[2] = 2\n",
    "  A[3] = 4\n",
    "  A[4] = 3\n",
    "the function should return 1, as explained above.\n",
    "\n",
    "Write an efficient algorithm for the following assumptions:\n",
    "\n",
    "N is an integer within the range [2..100,000];\n",
    "each element of array A is an integer within the range [−1,000..1,000]."
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   "cell_type": "code",
   "execution_count": 25,
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     "name": "stdout",
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     "text": [
      "[0, -9, -10, 6, -2, -10, 6, 7, 6]\n"
     ]
    }
   ],
   "source": [
    "from random import randint\n",
    "N = 9\n",
    "A = [randint(-10, 10) for x in range(1, N + 1)]\n",
    "print(A)"
   ]
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    "# you can write to stdout for debugging purposes, e.g.\n",
    "# print(\"this is a debug message\")\n",
    "\n",
    "def solution(A, stdout=True):\n",
    "    # write your code in Python 3.6\n",
    "    n = len(A) - 1\n",
    "    p = (n // 2)\n",
    "    history = []\n",
    "    while p not in [x.get('p') for x in history]:\n",
    "        left = A[:p]\n",
    "        sum_left = sum(left)\n",
    "        right = A[p:]\n",
    "        sum_right = sum(right)\n",
    "        diff = abs(sum_left - sum_right)\n",
    "        if stdout:\n",
    "            print('P = {}, difference = |{} − {}| = {}'.format(p, sum_left, sum_right, diff))\n",
    "        history.append({'p': p, 'diff': diff})\n",
    "        if sum_left > sum_right:\n",
    "            p -= 1\n",
    "        else:\n",
    "            p += 1\n",
    "    return min(x.get('diff') for x in history)"
   ]
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   "cell_type": "code",
   "execution_count": 84,
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     "name": "stdout",
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     "text": [
      "P = 2, difference = |4 − 9| = 5\n",
      "P = 3, difference = |6 − 7| = 1\n",
      "P = 4, difference = |10 − 3| = 7\n"
     ]
    },
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     "execution_count": 84,
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   ],
   "source": [
    "solution([3, 1, 2, 4, 3])"
   ]
  },
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   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": []
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