interactive-coding-challenges/recursion_dynamic/n_pairs_parentheses/n_pairs_parentheses_solution.ipynb

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   "source": [
    "<small><i>This notebook was prepared by [Rishi Rajasekaran](https://github.com/rishihot55). Source and license info is on [GitHub](https://github.com/donnemartin/interactive-coding-challenges).</i></small>"
   ]
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   "source": [
    "# Solution Notebook"
   ]
  },
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   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Problem: Print all valid combinations of n-pairs of parentheses\n",
    "\n",
    "* [Constraints](#Constraints)\n",
    "* [Test Cases](#Test-Cases)\n",
    "* [Algorithm](#Algorithm)\n",
    "* [Code](#Code)\n",
    "* [Unit Test](#Unit-Test)"
   ]
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   "source": [
    "## Constraints\n",
    "* None"
   ]
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   "source": [
    "## Test Cases\n",
    "\n",
    "* 0 -> ' '\n",
    "* 1 -> ()\n",
    "* 2 -> (()), ()()\n",
    "* 3 -> ((())), (()()), (())(), ()(()), ()()()"
   ]
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    "# Algorithm\n",
    "\n",
    "Let `l` and `r` denote the number of left and right parentheses remaining at any given point. \n",
    "The algorithm makes use of the following conditions applied recursively:\n",
    "* Left braces can be inserted any time, as long as we do not exhaust them i.e. `l > 0`.\n",
    "* Right braces can be inserted, as long as the number of right braces remaining is greater than the left braces remaining i.e. `r > l`. Violation of the aforementioned condition produces an unbalanced string of parentheses.\n",
    "* If both left and right braces have been exhausted i.e. `l = 0 and r = 0`, then the resultant string produced is balanced.\n",
    "\n",
    "The algorithm can be rephrased as:\n",
    "* Base case: `l = 0 and r = 0`\n",
    "    - Add the string generated to the result set\n",
    "* Case 1: `l > 0`\n",
    "    - Add a left parenthesis to the parentheses string.\n",
    "    - Call parentheses_util(l - 1, r, new_string, result_set)\n",
    "* Case 2: `r > l`\n",
    "    - Add a right parenthesis to the parentheses string.\n",
    "    - Call parentheses_util(l, r - 1, new_string, result_set)\n",
    "\n",
    "Complexity:\n",
    "* Time: `O(4^n/n^(3/2))`. See [Catalan numbers](https://en.wikipedia.org/wiki/Catalan_number#Applications_in_combinatorics)\n",
    "* Space complexity: `O(n)` (Due to the implicit call stack storing a maximum of 2n function calls)"
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   "metadata": {},
   "source": [
    "## Code"
   ]
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  {
   "cell_type": "code",
   "execution_count": 1,
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   "source": [
    "def parentheses_util(no_left, no_right, pair_string, result):\n",
    "    if no_left == 0 and no_right == 0:\n",
    "        result.add(pair_string)\n",
    "    else:\n",
    "        if no_left > 0:\n",
    "            parentheses_util(no_left - 1, no_right, pair_string + '(', result)\n",
    "        if no_right > no_left:\n",
    "            parentheses_util(no_left, no_right - 1, pair_string + ')', result)\n",
    "\n",
    "\n",
    "def pair_parentheses(n):\n",
    "    result_set = set()\n",
    "    if n == 0:\n",
    "        return result_set\n",
    "    parentheses_util(n, n, '', result_set)\n",
    "    return result_set\n"
   ]
  },
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   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Unit Test"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {
    "collapsed": false
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   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Success: test_pair_parentheses\n"
     ]
    }
   ],
   "source": [
    "# %load test_n_pairs_parentheses.py\n",
    "from nose.tools import assert_equal\n",
    "\n",
    "class TestPairParentheses(object):\n",
    "    \n",
    "    def test_pair_parentheses(self, solution):\n",
    "        assert_equal(solution(0), set([]))\n",
    "        assert_equal(solution(1), set(['()']))\n",
    "        assert_equal(solution(2), set(['(())', '()()']))\n",
    "        assert_equal(solution(3), set(['((()))','(()())', '(())()', '()(())', '()()()']))\n",
    "        print('Success: test_pair_parentheses')\n",
    "\n",
    "def main():\n",
    "    test = TestPairParentheses()\n",
    "    test.test_pair_parentheses(pair_parentheses)\n",
    "\n",
    "if __name__ == '__main__':\n",
    "    main()"
   ]
  }
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