leetcode-problemset/leetcode-cn/originData/find-mirror-score-of-a-string.json

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            "title": "Find Mirror Score of a String",
            "titleSlug": "find-mirror-score-of-a-string",
            "content": "<p>You are given a string <code>s</code>.</p>\n\n<p>We define the <strong>mirror</strong> of a letter in the English alphabet as its corresponding letter when the alphabet is reversed. For example, the mirror of <code>&#39;a&#39;</code> is <code>&#39;z&#39;</code>, and the mirror of <code>&#39;y&#39;</code> is <code>&#39;b&#39;</code>.</p>\n\n<p>Initially, all characters in the string <code>s</code> are <strong>unmarked</strong>.</p>\n\n<p>You start with a score of 0, and you perform the following process on the string <code>s</code>:</p>\n\n<ul>\n\t<li>Iterate through the string from left to right.</li>\n\t<li>At each index <code>i</code>, find the closest <strong>unmarked</strong> index <code>j</code> such that <code>j &lt; i</code> and <code>s[j]</code> is the mirror of <code>s[i]</code>. Then, <strong>mark</strong> both indices <code>i</code> and <code>j</code>, and add the value <code>i - j</code> to the total score.</li>\n\t<li>If no such index <code>j</code> exists for the index <code>i</code>, move on to the next index without making any changes.</li>\n</ul>\n\n<p>Return the total score at the end of the process.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<div class=\"example-block\">\n<p><strong>Input:</strong> <span class=\"example-io\">s = &quot;aczzx&quot;</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">5</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<ul>\n\t<li><code>i = 0</code>. There is no index <code>j</code> that satisfies the conditions, so we skip.</li>\n\t<li><code>i = 1</code>. There is no index <code>j</code> that satisfies the conditions, so we skip.</li>\n\t<li><code>i = 2</code>. The closest index <code>j</code> that satisfies the conditions is <code>j = 0</code>, so we mark both indices 0 and 2, and then add <code>2 - 0 = 2</code> to the score.</li>\n\t<li><code>i = 3</code>. There is no index <code>j</code> that satisfies the conditions, so we skip.</li>\n\t<li><code>i = 4</code>. The closest index <code>j</code> that satisfies the conditions is <code>j = 1</code>, so we mark both indices 1 and 4, and then add <code>4 - 1 = 3</code> to the score.</li>\n</ul>\n</div>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<div class=\"example-block\">\n<p><strong>Input:</strong> <span class=\"example-io\">s = &quot;abcdef&quot;</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">0</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<p>For each index <code>i</code>, there is no index <code>j</code> that satisfies the conditions.</p>\n</div>\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</code> consists only of lowercase English letters.</li>\n</ul>\n",
            "translatedTitle": "计算字符串的镜像分数",
            "translatedContent": "<p>给你一个字符串 <code>s</code>。</p>\n\n<p>英文字母中每个字母的&nbsp;<strong>镜像&nbsp;</strong>定义为反转字母表之后对应位置上的字母。例如，<code>'a'</code> 的镜像是 <code>'z'</code>，<code>'y'</code> 的镜像是 <code>'b'</code>。</p>\n\n<p>最初，字符串 <code>s</code> 中的所有字符都&nbsp;<strong>未标记&nbsp;</strong>。</p>\n\n<p>字符串 <code>s</code>&nbsp;的初始分数为 0 ，你需要对其执行以下过程：</p>\n\n<ul>\n\t<li>从左到右遍历字符串。</li>\n\t<li>对于每个下标&nbsp;<code>i&nbsp;</code>，找到距离最近的&nbsp;<strong>未标记</strong> 下标&nbsp;<code>j</code>，下标 <code>j</code> 需要满足&nbsp;<code>j &lt; i</code> 且 <code>s[j]</code> 是 <code>s[i]</code> 的镜像。然后&nbsp;<strong>标记</strong> 下标&nbsp;<code>i</code> 和 <code>j</code>，总分加上&nbsp;<code>i - j</code>&nbsp;的值。</li>\n\t<li>如果对于下标&nbsp;<code>i</code>，不存在满足条件的下标&nbsp;<code>j</code>，则跳过该下标，继续处理下一个下标，不需要进行标记。</li>\n</ul>\n\n<p>返回最终的总分。</p>\n\n<p>&nbsp;</p>\n\n<p><strong class=\"example\">示例 1：</strong></p>\n\n<div class=\"example-block\">\n<p><strong>输入：</strong> <span class=\"example-io\">s = \"aczzx\"</span></p>\n\n<p><strong>输出：</strong> <span class=\"example-io\">5</span></p>\n\n<p><strong>解释：</strong></p>\n\n<ul>\n\t<li><code>i = 0</code>。没有符合条件的下标&nbsp;<code>j</code>，跳过。</li>\n\t<li><code>i = 1</code>。没有符合条件的下标&nbsp;<code>j</code>，跳过。</li>\n\t<li><code>i = 2</code>。距离最近的符合条件的下标是 <code>j = 0</code>，因此标记下标&nbsp;0 和 2，然后将总分加上&nbsp;<code>2 - 0 = 2</code>&nbsp;。</li>\n\t<li><code>i = 3</code>。没有符合条件的下标&nbsp;<code>j</code>，跳过。</li>\n\t<li><code>i = 4</code>。距离最近的符合条件的下标是 <code>j = 1</code>，因此标记下标&nbsp;1 和 4，然后将总分加上&nbsp;<code>4 - 1 = 3</code>&nbsp;。</li>\n</ul>\n</div>\n\n<p><strong class=\"example\">示例 2：</strong></p>\n\n<div class=\"example-block\">\n<p><strong>输入：</strong> <span class=\"example-io\">s = \"abcdef\"</span></p>\n\n<p><strong>输出：</strong> <span class=\"example-io\">0</span></p>\n\n<p><strong>解释：</strong></p>\n\n<p>对于每个下标&nbsp;<code>i</code>，都不存在满足条件的下标&nbsp;<code>j</code>。</p>\n</div>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</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</code> 仅由小写英文字母组成。</li>\n</ul>\n",
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