leetcode-problemset/leetcode-cn/originData/sum-of-digits-of-string-after-convert.json

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        "question": {
            "questionId": "2076",
            "questionFrontendId": "1945",
            "categoryTitle": "Algorithms",
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            "title": "Sum of Digits of String After Convert",
            "titleSlug": "sum-of-digits-of-string-after-convert",
            "content": "<p>You are given a string <code>s</code> consisting of lowercase English letters, and an integer <code>k</code>. Your task is to <em>convert</em> the string into an integer by a special process, and then <em>transform</em> it by summing its digits repeatedly <code>k</code> times. More specifically, perform the following steps:</p>\n\n<ol>\n\t<li><strong>Convert</strong> <code>s</code> into an integer by replacing each letter with its position in the alphabet (i.e.&nbsp;replace <code>&#39;a&#39;</code> with <code>1</code>, <code>&#39;b&#39;</code> with <code>2</code>, ..., <code>&#39;z&#39;</code> with <code>26</code>).</li>\n\t<li><strong>T</strong><strong>ransform</strong> the integer by replacing it with the <strong>sum of its digits</strong>.</li>\n\t<li>Repeat the <strong>transform</strong> operation (step 2) <code>k</code><strong> times</strong> in total.</li>\n</ol>\n\n<p>For example, if <code>s = &quot;zbax&quot;</code> and <code>k = 2</code>, then the resulting integer would be <code>8</code> by the following operations:</p>\n\n<ol>\n\t<li><strong>Convert</strong>: <code>&quot;zbax&quot; ➝ &quot;(26)(2)(1)(24)&quot; ➝ &quot;262124&quot; ➝ 262124</code></li>\n\t<li><strong>Transform #1</strong>: <code>262124 ➝ 2 + 6 + 2 + 1 + 2 + 4 ➝ 17</code></li>\n\t<li><strong>Transform #2</strong>: <code>17 ➝ 1 + 7 ➝ 8</code></li>\n</ol>\n\n<p>Return the <strong>resulting</strong> <strong>integer</strong> after performing the <strong>operations</strong> described above.</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;iiii&quot;, k = 1</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">36</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<p>The operations are as follows:<br />\n- Convert: &quot;iiii&quot; ➝ &quot;(9)(9)(9)(9)&quot; ➝ &quot;9999&quot; ➝ 9999<br />\n- Transform #1: 9999 ➝ 9 + 9 + 9 + 9 ➝ 36<br />\nThus the resulting integer is 36.</p>\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;leetcode&quot;, k = 2</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">6</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<p>The operations are as follows:<br />\n- Convert: &quot;leetcode&quot; ➝ &quot;(12)(5)(5)(20)(3)(15)(4)(5)&quot; ➝ &quot;12552031545&quot; ➝ 12552031545<br />\n- Transform #1: 12552031545 ➝ 1 + 2 + 5 + 5 + 2 + 0 + 3 + 1 + 5 + 4 + 5 ➝ 33<br />\n- Transform #2: 33 ➝ 3 + 3 ➝ 6<br />\nThus the resulting integer is 6.</p>\n</div>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<div class=\"example-block\">\n<p><strong>Input:</strong> <span class=\"example-io\">s = &quot;zbax&quot;, k = 2</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">8</span></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;= 100</code></li>\n\t<li><code>1 &lt;= k &lt;= 10</code></li>\n\t<li><code>s</code> consists of lowercase English letters.</li>\n</ul>\n",
            "translatedTitle": "字符串转化后的各位数字之和",
            "translatedContent": "<p>给你一个由小写字母组成的字符串 <code>s</code> ，以及一个整数 <code>k</code> 。你的任务是通过一种特殊处理将字符串转为整数，然后通过重复对它的数位求和&nbsp;<code>k</code> 次来进行转换。更具体地说，执行以下步骤：</p>\n\n<ol>\n\t<li>用字母在字母表中的位置&nbsp;<strong>替换&nbsp;</strong>该字母，将 <code>s</code> <strong>转化</strong> 为一个整数（也就是，<code>'a'</code> 用 <code>1</code> 替换，<code>'b'</code> 用 <code>2</code> 替换，... <code>'z'</code> 用 <code>26</code> 替换）。</li>\n\t<li>接着，将整数 <strong>转换</strong> 为其 <strong>各位数字之和</strong> 。</li>\n\t<li>共重复 <strong>转换</strong> 操作（第 2 步）&nbsp;<code>k</code><strong> 次</strong> 。</li>\n</ol>\n\n<p>例如，如果 <code>s = \"zbax\"</code> 且 <code>k = 2</code> ，那么执行下述步骤后得到的结果是整数 <code>8</code> ：</p>\n\n<ul>\n\t<li><strong>转化：</strong><code>\"zbax\" ➝ \"(26)(2)(1)(24)\" ➝ \"262124\" ➝ 262124</code></li>\n\t<li><strong>转换 #1</strong>：<code>262124&nbsp;➝ 2 + 6 + 2 + 1 + 2 + 4&nbsp;➝ 17</code></li>\n\t<li><strong>转换 #2</strong>：<code>17 ➝ 1 + 7 ➝ 8</code></li>\n</ul>\n\n<p>返回执行上述 <strong>操作</strong> 后得到的 <strong>结果整数</strong>。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<div class=\"example-block\"><strong>输入：</strong>s = \"iiii\", k = 1</div>\n\n<div class=\"example-block\"><strong>输出：</strong>36</div>\n\n<div class=\"example-block\"><strong>解释：</strong></div>\n\n<div class=\"example-block\">操作如下：</div>\n\n<ul>\n\t<li class=\"example-block\">转化：\"iiii\" ➝ \"(9)(9)(9)(9)\" ➝ \"9999\" ➝ 9999</li>\n\t<li class=\"example-block\">转换 #1：9999 ➝ 9 + 9 + 9 + 9 ➝ 36</li>\n</ul>\n\n<div class=\"example-block\">因此，结果整数为 36 。</div>\n\n<div class=\"example-block\">&nbsp;</div>\n\n<p><strong>示例 2：</strong></p>\n\n<div class=\"example-block\"><strong>输入：</strong>s = \"leetcode\", k = 2</div>\n\n<div class=\"example-block\"><strong>输出：</strong>6</div>\n\n<div class=\"example-block\"><strong>解释：</strong></div>\n\n<div class=\"example-block\">操作如下：</div>\n\n<ul>\n\t<li class=\"example-block\">转化：\"leetcode\" ➝ \"(12)(5)(5)(20)(3)(15)(4)(5)\" ➝ \"12552031545\" ➝ 12552031545</li>\n\t<li class=\"example-block\">转换 #1：12552031545 ➝ 1 + 2 + 5 + 5 + 2 + 0 + 3 + 1 + 5 + 4 + 5 ➝ 33</li>\n\t<li class=\"example-block\">转换 #2：33 ➝ 3 + 3 ➝ 6</li>\n</ul>\n\n<p class=\"example-block\">因此，结果整数为 6 。</p>\n\n<p class=\"example-block\">&nbsp;</p>\n\n<p><strong class=\"example\">示例 3：</strong></p>\n\n<div class=\"example-block\">\n<p><strong>输入：</strong><span class=\"example-io\">s = \"zbax\", k = 2</span></p>\n\n<p><span class=\"example-io\"><b>输出：</b>8</span></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;= 100</code></li>\n\t<li><code>1 &lt;= k &lt;= 10</code></li>\n\t<li><code>s</code> 由小写英文字母组成</li>\n</ul>\n",
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href=\\\"https:\\/\\/docs.racket-lang.org\\/guide\\/performance.html#%28tech._c%29\\\" target=\\\"_blank\\\">Racket CS<\\/a> v8.15<\\/p>\\r\\n\\r\\n<p>\\u4f7f\\u7528 #lang racket<\\/p>\\r\\n\\r\\n<p>\\u5df2\\u9884\\u5148 (require data\\/gvector data\\/queue data\\/order data\\/heap). \\u82e5\\u9700\\u4f7f\\u7528\\u5176\\u5b83\\u6570\\u636e\\u7ed3\\u6784\\uff0c\\u53ef\\u81ea\\u884c require\\u3002<\\/p>\"],\"erlang\":[\"Erlang\",\"Erlang\\/OTP 26\"],\"elixir\":[\"Elixir\",\"Elixir 1.17 with Erlang\\/OTP 26\"],\"dart\":[\"Dart\",\"<p>Dart 3.2\\u3002\\u60a8\\u53ef\\u4ee5\\u4f7f\\u7528 <a href=\\\"https:\\/\\/pub.dev\\/packages\\/collection\\/versions\\/1.18.0\\\" target=\\\"_blank\\\">collection<\\/a> \\u5305<\\/p>\\r\\n\\r\\n<p>\\u60a8\\u7684\\u4ee3\\u7801\\u5c06\\u4f1a\\u88ab\\u4e0d\\u7f16\\u8bd1\\u76f4\\u63a5\\u8fd0\\u884c<\\/p>\"],\"cangjie\":[\"Cangjie\",\"<p>\\u7248\\u672c\\uff1a0.53.11 (cjnative)<\\/p>\\r\\n\\r\\n<p>\\u7f16\\u8bd1\\u53c2\\u6570\\uff1a<code>-O2 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