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leetcode-problemset/算法题(国外版)/minimum-remove-to-make-valid-parentheses.json
2022-03-27 18:52:06 +08:00

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{
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"title": "Minimum Remove to Make Valid Parentheses",
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"content": "<p>Given a string <font face=\"monospace\">s</font> of <code>&#39;(&#39;</code> , <code>&#39;)&#39;</code> and lowercase English characters.</p>\n\n<p>Your task is to remove the minimum number of parentheses ( <code>&#39;(&#39;</code> or <code>&#39;)&#39;</code>, in any positions ) so that the resulting <em>parentheses string</em> is valid and return <strong>any</strong> valid string.</p>\n\n<p>Formally, a <em>parentheses string</em> is valid if and only if:</p>\n\n<ul>\n\t<li>It is the empty string, contains only lowercase characters, or</li>\n\t<li>It can be written as <code>AB</code> (<code>A</code> concatenated with <code>B</code>), where <code>A</code> and <code>B</code> are valid strings, or</li>\n\t<li>It can be written as <code>(A)</code>, where <code>A</code> is a valid string.</li>\n</ul>\n\n<p>&nbsp;</p>\n<p><strong>Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;lee(t(c)o)de)&quot;\n<strong>Output:</strong> &quot;lee(t(c)o)de&quot;\n<strong>Explanation:</strong> &quot;lee(t(co)de)&quot; , &quot;lee(t(c)ode)&quot; would also be accepted.\n</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;a)b(c)d&quot;\n<strong>Output:</strong> &quot;ab(c)d&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<strong>Explanation:</strong> An empty string is also valid.\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;= 10<sup>5</sup></code></li>\n\t<li><code>s[i]</code> is either<code>&#39;(&#39;</code> , <code>&#39;)&#39;</code>, or lowercase English letter<code>.</code></li>\n</ul>\n",
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"Each prefix of a balanced parentheses has a number of open parentheses greater or equal than closed parentheses, similar idea with each suffix.",
"Check the array from left to right, remove characters that do not meet the property mentioned above, same idea in backward way."
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