leetcode-problemset/leetcode-cn/originData/remove-invalid-parentheses.json

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            "categoryTitle": "Algorithms",
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            "title": "Remove Invalid Parentheses",
            "titleSlug": "remove-invalid-parentheses",
            "content": "<p>Given a string <code>s</code> that contains parentheses and letters, remove the minimum number of invalid parentheses to make the input string valid.</p>\n\n<p>Return <em>all the possible results</em>. You may return the answer in <strong>any order</strong>.</p>\n\n<p>&nbsp;</p>\n<p><strong>Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;()())()&quot;\n<strong>Output:</strong> [&quot;(())()&quot;,&quot;()()()&quot;]\n</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;(a)())()&quot;\n<strong>Output:</strong> [&quot;(a())()&quot;,&quot;(a)()()&quot;]\n</pre>\n\n<p><strong>Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;)(&quot;\n<strong>Output:</strong> [&quot;&quot;]\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= s.length &lt;= 25</code></li>\n\t<li><code>s</code> consists of lowercase English letters and parentheses <code>&#39;(&#39;</code> and <code>&#39;)&#39;</code>.</li>\n\t<li>There will be at most <code>20</code> parentheses in <code>s</code>.</li>\n</ul>\n",
            "translatedTitle": "删除无效的括号",
            "translatedContent": "<p>给你一个由若干括号和字母组成的字符串 <code>s</code> ，删除最小数量的无效括号，使得输入的字符串有效。</p>\n\n<p>返回所有可能的结果。答案可以按 <strong>任意顺序</strong> 返回。</p>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>s = \"()())()\"\n<strong>输出：</strong>[\"(())()\",\"()()()\"]\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>s = \"(a)())()\"\n<strong>输出：</strong>[\"(a())()\",\"(a)()()\"]\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong>s = \")(\"\n<strong>输出：</strong>[\"\"]\n</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 <= s.length <= 25</code></li>\n\t<li><code>s</code> 由小写英文字母以及括号 <code>'('</code> 和 <code>')'</code> 组成</li>\n\t<li><code>s</code> 中至多含 <code>20</code> 个括号</li>\n</ul>\n",
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                    "code": "\n\n/**\n * Note: The returned array must be malloced, assume caller calls free().\n */\nchar ** removeInvalidParentheses(char * s, int* returnSize){\n\n}",
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                    "code": "class Solution {\n\n    /**\n     * @param String $s\n     * @return String[]\n     */\n    function removeInvalidParentheses($s) {\n\n    }\n}",
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                    "code": "defmodule Solution do\n  @spec remove_invalid_parentheses(s :: String.t) :: [String.t]\n  def remove_invalid_parentheses(s) do\n\n  end\nend",
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            "hints": [
                "Since we don't know which of the brackets can possibly be removed, we try out all the options!",
                "We can use recursion to try out all possibilities for the given expression. For each of the brackets, we have 2 options:\r\n\r\n<ol>\r\n<li> We keep the bracket and add it to the expression that we are building on the fly during recursion.</li>\r\n<li> OR, we can discard the bracket and move on.\r\n</ol>",
                "The one thing all these valid expressions have in common is that they will all be of the same length i.e. as compared to the original expression, all of these expressions will have the same number of characters removed. \r\n\r\nCan we somehow find the number of misplaced parentheses and use it in our solution?",
                "For every left parenthesis, we should have a corresponding right parenthesis. We can make use of two counters which keep track of misplaced left and right parenthesis and in one iteration we can find out these two values. \r\n\r\n<pre>\r\n0 1 2 3 4 5 6 7\r\n( ) ) ) ( ( ( )  \r\ni = 0, left = 1, right = 0\r\ni = 1, left = 0, right = 0\r\ni = 2, left = 0, right = 1\r\ni = 3, left = 0, right = 2\r\ni = 4, left = 1, right = 2\r\ni = 5, left = 2, right = 2\r\ni = 6, left = 3, right = 2\r\ni = 7, left = 2, right = 2\r\n</pre>\r\n\r\nWe have 2 misplaced left and 2 misplaced right parentheses.",
                "We found out that the exact number of left and right parenthesis that has to be removed to get a valid expression. So, e.g. in a 1000 parentheses string, if there are 2 misplaced left and 2 misplaced right parentheses, after we are done discarding 2 left and 2 right parentheses, we will have only one option per remaining character in the expression i.e. to consider them. We can't discard them."
            ],
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