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"categoryTitle": "Algorithms",
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"title": "Check if a Parentheses String Can Be Valid",
"titleSlug": "check-if-a-parentheses-string-can-be-valid",
"content": "<p>A parentheses string is a <strong>non-empty</strong> string consisting only of <code>'('</code> and <code>')'</code>. It is valid if <strong>any</strong> of the following conditions is <strong>true</strong>:</p>\n\n<ul>\n\t<li>It is <code>()</code>.</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 parentheses strings.</li>\n\t<li>It can be written as <code>(A)</code>, where <code>A</code> is a valid parentheses string.</li>\n</ul>\n\n<p>You are given a parentheses string <code>s</code> and a string <code>locked</code>, both of length <code>n</code>. <code>locked</code> is a binary string consisting only of <code>'0'</code>s and <code>'1'</code>s. For <strong>each</strong> index <code>i</code> of <code>locked</code>,</p>\n\n<ul>\n\t<li>If <code>locked[i]</code> is <code>'1'</code>, you <strong>cannot</strong> change <code>s[i]</code>.</li>\n\t<li>But if <code>locked[i]</code> is <code>'0'</code>, you <strong>can</strong> change <code>s[i]</code> to either <code>'('</code> or <code>')'</code>.</li>\n</ul>\n\n<p>Return <code>true</code> <em>if you can make <code>s</code> a valid parentheses string</em>. Otherwise, return <code>false</code>.</p>\n\n<p> </p>\n<p><strong>Example 1:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/11/06/eg1.png\" style=\"width: 311px; height: 101px;\" />\n<pre>\n<strong>Input:</strong> s = "))()))", locked = "010100"\n<strong>Output:</strong> true\n<strong>Explanation:</strong> locked[1] == '1' and locked[3] == '1', so we cannot change s[1] or s[3].\nWe change s[0] and s[4] to '(' while leaving s[2] and s[5] unchanged to make s valid.</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = "()()", locked = "0000"\n<strong>Output:</strong> true\n<strong>Explanation:</strong> We do not need to make any changes because s is already valid.\n</pre>\n\n<p><strong>Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = ")", locked = "0"\n<strong>Output:</strong> false\n<strong>Explanation:</strong> locked permits us to change s[0]. \nChanging s[0] to either '(' or ')' will not make s valid.\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>n == s.length == locked.length</code></li>\n\t<li><code>1 <= n <= 10<sup>5</sup></code></li>\n\t<li><code>s[i]</code> is either <code>'('</code> or <code>')'</code>.</li>\n\t<li><code>locked[i]</code> is either <code>'0'</code> or <code>'1'</code>.</li>\n</ul>\n",
"translatedTitle": "判断一个括号字符串是否有效",
"translatedContent": "<p>一个括号字符串是只由 <code>'('</code> 和 <code>')'</code> 组成的 <strong>非空</strong> 字符串。如果一个字符串满足下面 <b>任意</b> 一个条件,那么它就是有效的:</p>\n\n<ul>\n\t<li>字符串为 <code>()</code>.</li>\n\t<li>它可以表示为 <code>AB</code><span style=\"\">(</span><code>A</code> 与 <code>B</code> 连接),其中<code>A</code> 和 <code>B</code> 都是有效括号字符串。</li>\n\t<li>它可以表示为 <code>(A)</code> ,其中 <code>A</code> 是一个有效括号字符串。</li>\n</ul>\n\n<p>给你一个括号字符串 <code>s</code> 和一个字符串 <code>locked</code> ,两者长度都为 <code>n</code> 。<code>locked</code> 是一个二进制字符串,只包含 <code>'0'</code> 和 <code>'1'</code> 。对于 <code>locked</code> 中 <strong>每一个</strong> 下标 <code>i</code> :</p>\n\n<ul>\n\t<li>如果 <code>locked[i]</code> 是 <code>'1'</code> ,你 <strong>不能</strong> 改变 <code>s[i]</code> 。</li>\n\t<li>如果 <code>locked[i]</code> 是 <code>'0'</code> ,你 <strong>可以</strong> 将 <code>s[i]</code> 变为 <code>'('</code> 或者 <code>')'</code> 。</li>\n</ul>\n\n<p>如果你可以将 <code>s</code> 变为有效括号字符串,请你返回 <code>true</code> ,否则返回 <code>false</code> 。</p>\n\n<p> </p>\n\n<p><strong>示例 1:</strong></p>\n\n<p><img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/11/06/eg1.png\" style=\"width: 311px; height: 101px;\" /></p>\n\n<pre>\n<b>输入:</b>s = \"))()))\", locked = \"010100\"\n<b>输出:</b>true\n<b>解释:</b>locked[1] == '1' 和 locked[3] == '1' ,所以我们无法改变 s[1] 或者 s[3] 。\n我们可以将 s[0] 和 s[4] 变为 '(' ,不改变 s[2] 和 s[5] ,使 s 变为有效字符串。</pre>\n\n<p><strong>示例 2:</strong></p>\n\n<pre>\n<b>输入:</b>s = \"()()\", locked = \"0000\"\n<b>输出:</b>true\n<b>解释:</b>我们不需要做任何改变,因为 s 已经是有效字符串了。\n</pre>\n\n<p><strong>示例 3:</strong></p>\n\n<pre>\n<b>输入:</b>s = \")\", locked = \"0\"\n<b>输出:</b>false\n<b>解释:</b>locked 允许改变 s[0] 。\n但无论将 s[0] 变为 '(' 或者 ')' 都无法使 s 变为有效字符串。\n</pre>\n\n<p> </p>\n\n<p><strong>提示:</strong></p>\n\n<ul>\n\t<li><code>n == s.length == locked.length</code></li>\n\t<li><code>1 <= n <= 10<sup>5</sup></code></li>\n\t<li><code>s[i]</code> 要么是 <code>'('</code> 要么是 <code>')'</code> 。</li>\n\t<li><code>locked[i]</code> 要么是 <code>'0'</code> 要么是 <code>'1'</code> 。</li>\n</ul>\n",
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"From left to right, if a locked ')' is encountered, it must be balanced with either a locked '(' or an unlocked index on its left. If neither exist, what conclusion can be drawn? If both exist, which one is more preferable to use?",
"After the above, we may have locked indices of '(' and additional unlocked indices. How can you balance out the locked '(' now? What if you cannot balance any locked '('?"
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