leetcode-problemset/leetcode-cn/originData/rearrange-characters-to-make-target-string.json

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        "question": {
            "questionId": "2372",
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            "categoryTitle": "Algorithms",
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            "title": "Rearrange Characters to Make Target String",
            "titleSlug": "rearrange-characters-to-make-target-string",
            "content": "<p>You are given two <strong>0-indexed</strong> strings <code>s</code> and <code>target</code>. You can take some letters from <code>s</code> and rearrange them to form new strings.</p>\n\n<p>Return<em> the <strong>maximum</strong> number of copies of </em><code>target</code><em> that can be formed by taking letters from </em><code>s</code><em> and rearranging them.</em></p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;ilovecodingonleetcode&quot;, target = &quot;code&quot;\n<strong>Output:</strong> 2\n<strong>Explanation:</strong>\nFor the first copy of &quot;code&quot;, take the letters at indices 4, 5, 6, and 7.\nFor the second copy of &quot;code&quot;, take the letters at indices 17, 18, 19, and 20.\nThe strings that are formed are &quot;ecod&quot; and &quot;code&quot; which can both be rearranged into &quot;code&quot;.\nWe can make at most two copies of &quot;code&quot;, so we return 2.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;abcba&quot;, target = &quot;abc&quot;\n<strong>Output:</strong> 1\n<strong>Explanation:</strong>\nWe can make one copy of &quot;abc&quot; by taking the letters at indices 0, 1, and 2.\nWe can make at most one copy of &quot;abc&quot;, so we return 1.\nNote that while there is an extra &#39;a&#39; and &#39;b&#39; at indices 3 and 4, we cannot reuse the letter &#39;c&#39; at index 2, so we cannot make a second copy of &quot;abc&quot;.\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;abbaccaddaeea&quot;, target = &quot;aaaaa&quot;\n<strong>Output:</strong> 1\n<strong>Explanation:</strong>\nWe can make one copy of &quot;aaaaa&quot; by taking the letters at indices 0, 3, 6, 9, and 12.\nWe can make at most one copy of &quot;aaaaa&quot;, so we return 1.\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;= 100</code></li>\n\t<li><code>1 &lt;= target.length &lt;= 10</code></li>\n\t<li><code>s</code> and <code>target</code> consist of lowercase English letters.</li>\n</ul>\n",
            "translatedTitle": "重排字符形成目标字符串",
            "translatedContent": "<p>给你两个下标从 <strong>0</strong> 开始的字符串 <code>s</code> 和 <code>target</code> 。你可以从 <code>s</code> 取出一些字符并将其重排，得到若干新的字符串。</p>\n\n<p>从 <code>s</code> 中取出字符并重新排列，返回可以形成 <code>target</code> 的 <strong>最大</strong> 副本数。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre><strong>输入：</strong>s = \"ilovecodingonleetcode\", target = \"code\"\n<strong>输出：</strong>2\n<strong>解释：</strong>\n对于 \"code\" 的第 1 个副本，选取下标为 4 、5 、6 和 7 的字符。\n对于 \"code\" 的第 2 个副本，选取下标为 17 、18 、19 和 20 的字符。\n形成的字符串分别是 \"ecod\" 和 \"code\" ，都可以重排为 \"code\" 。\n可以形成最多 2 个 \"code\" 的副本，所以返回 2 。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre><strong>输入：</strong>s = \"abcba\", target = \"abc\"\n<strong>输出：</strong>1\n<strong>解释：</strong>\n选取下标为 0 、1 和 2 的字符，可以形成 \"abc\" 的 1 个副本。 \n可以形成最多 1 个 \"abc\" 的副本，所以返回 1 。\n注意，尽管下标 3 和 4 分别有额外的 'a' 和 'b' ，但不能重用下标 2 处的 'c' ，所以无法形成 \"abc\" 的第 2 个副本。\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre><strong>输入：</strong>s = \"abbaccaddaeea\", target = \"aaaaa\"\n<strong>输出：</strong>1\n<strong>解释：</strong>\n选取下标为 0 、3 、6 、9 和 12 的字符，可以形成 \"aaaaa\" 的 1 个副本。\n可以形成最多 1 个 \"aaaaa\" 的副本，所以返回 1 。\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 &lt;= 100</code></li>\n\t<li><code>1 &lt;= target.length &lt;= 10</code></li>\n\t<li><code>s</code> 和 <code>target</code> 由小写英文字母组成</li>\n</ul>\n",
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            "hints": [
                "Count the frequency of each character in s and target.",
                "Consider each letter one at a time. If there are x occurrences of a letter in s and y occurrences of the same letter in target, how many copies of this letter can we make?",
                "We can make floor(x / y) copies of the letter."
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