leetcode-problemset/leetcode-cn/originData/minimum-number-of-steps-to-make-two-strings-anagram-ii.json

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            "title": "Minimum Number of Steps to Make Two Strings Anagram II",
            "titleSlug": "minimum-number-of-steps-to-make-two-strings-anagram-ii",
            "content": "<p>You are given two strings <code>s</code> and <code>t</code>. In one step, you can append <strong>any character</strong> to either <code>s</code> or <code>t</code>.</p>\n\n<p>Return <em>the minimum number of steps to make </em><code>s</code><em> and </em><code>t</code><em> <strong>anagrams</strong> of each other.</em></p>\n\n<p>An <strong>anagram</strong> of a string is a string that contains the same characters with a different (or the same) ordering.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;<strong><u>lee</u></strong>tco<u><strong>de</strong></u>&quot;, t = &quot;co<u><strong>a</strong></u>t<u><strong>s</strong></u>&quot;\n<strong>Output:</strong> 7\n<strong>Explanation:</strong> \n- In 2 steps, we can append the letters in &quot;as&quot; onto s = &quot;leetcode&quot;, forming s = &quot;leetcode<strong><u>as</u></strong>&quot;.\n- In 5 steps, we can append the letters in &quot;leede&quot; onto t = &quot;coats&quot;, forming t = &quot;coats<u><strong>leede</strong></u>&quot;.\n&quot;leetcodeas&quot; and &quot;coatsleede&quot; are now anagrams of each other.\nWe used a total of 2 + 5 = 7 steps.\nIt can be shown that there is no way to make them anagrams of each other with less than 7 steps.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;night&quot;, t = &quot;thing&quot;\n<strong>Output:</strong> 0\n<strong>Explanation:</strong> The given strings are already anagrams of each other. Thus, we do not need any further steps.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= s.length, t.length &lt;= 2 * 10<sup>5</sup></code></li>\n\t<li><code>s</code> and <code>t</code> consist of lowercase English letters.</li>\n</ul>\n",
            "translatedTitle": "使两字符串互为字母异位词的最少步骤数",
            "translatedContent": "<p>给你两个字符串 <code>s</code> 和 <code>t</code> 。在一步操作中，你可以给 <code>s</code> 或者 <code>t</code> 追加 <strong>任一字符</strong> 。</p>\n\n<p>返回使 <code>s</code> 和 <code>t</code> 互为 <strong>字母异位词</strong> 所需的最少步骤数<em>。</em></p>\n\n<p><strong>字母异位词 </strong>指字母相同但是顺序不同（或者相同）的字符串。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre><strong>输入：</strong>s = \"<em><strong>lee</strong>t</em>co<em><strong>de</strong></em>\", t = \"co<em><strong>a</strong></em>t<em><strong>s</strong></em>\"\n<strong>输出：</strong>7\n<strong>解释：</strong>\n- 执行 2 步操作，将 \"as\" 追加到 s = \"leetcode\" 中，得到 s = \"leetcode<em><strong>as</strong></em>\" 。\n- 执行 5 步操作，将 \"leede\" 追加到 t = \"coats\" 中，得到 t = \"coats<em><strong>leede</strong></em>\" 。\n\"leetcodeas\" 和 \"coatsleede\" 互为字母异位词。\n总共用去 2 + 5 = 7 步。\n可以证明，无法用少于 7 步操作使这两个字符串互为字母异位词。</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre><strong>输入：</strong>s = \"night\", t = \"thing\"\n<strong>输出：</strong>0\n<strong>解释：</strong>给出的字符串已经互为字母异位词。因此，不需要任何进一步操作。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= s.length, t.length &lt;= 2 * 10<sup>5</sup></code></li>\n\t<li><code>s</code> 和 <code>t</code> 由小写英文字符组成</li>\n</ul>\n",
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                "Notice that for anagrams, the order of the letters is irrelevant.",
                "For each letter, we can count its frequency in s and t.",
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