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Updated n-pairs parentheses solution information

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= 2015-08-10 01:27:33 +05:30
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recursion_dynamic/n_pairs_parentheses/n_pairs_parentheses_solution.ipynb
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@ -41,7 +41,7 @@
"source": [
"## Test Cases\n",
"\n",
"* 0 -> ''\n",
"* 0 -> ' '\n",
"* 1 -> ()\n",
"* 2 -> (()), ()()\n",
"* 3 -> ((())), (()()), (())(), ()(()), ()()()"
@ -52,13 +52,26 @@
"metadata": {},
"source": [
"# Algorithm\n",
"* If no_of_left == 0 and no_of right == 0, add pair_string to result\n",
"* Else \n",
" * If (no_left > 0), parentheses_util(no_left - 1, no_right, pair_string + '(', result)\n",
" * If (no_right > no_left), parentheses_util(no_left, no_right - 1, pair_string + ')', result)\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)) "
"* 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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