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2023-12-09 19:57:46 +08:00
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leetcode-cn/originData/replace-the-substring-for-balanced-string.json
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@@ -12,7 +12,7 @@
"translatedContent": "<p>有一个只含有&nbsp;<code>'Q', 'W', 'E',&nbsp;'R'</code>&nbsp;四种字符，且长度为 <code>n</code>&nbsp;的字符串。</p>\n\n<p>假如在该字符串中，这四个字符都恰好出现&nbsp;<code>n/4</code>&nbsp;次，那么它就是一个「平衡字符串」。</p>\n\n<p>&nbsp;</p>\n\n<p>给你一个这样的字符串 <code>s</code>，请通过「替换一个子串」的方式，使原字符串 <code>s</code> 变成一个「平衡字符串」。</p>\n\n<p>你可以用和「待替换子串」长度相同的&nbsp;<strong>任何</strong> 其他字符串来完成替换。</p>\n\n<p>请返回待替换子串的最小可能长度。</p>\n\n<p>如果原字符串自身就是一个平衡字符串，则返回 <code>0</code>。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>s = \"QWER\"\n<strong>输出：</strong>0\n<strong>解释：</strong>s 已经是平衡的了。</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>s = \"QQWE\"\n<strong>输出：</strong>1\n<strong>解释：</strong>我们需要把一个 'Q' 替换成 'R'，这样得到的 \"RQWE\" (或 \"QRWE\") 是平衡的。\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong>s = \"QQQW\"\n<strong>输出：</strong>2\n<strong>解释：</strong>我们可以把前面的 \"QQ\" 替换成 \"ER\"。 \n</pre>\n\n<p><strong>示例 4：</strong></p>\n\n<pre>\n<strong>输入：</strong>s = \"QQQQ\"\n<strong>输出：</strong>3\n<strong>解释：</strong>我们可以替换后 3 个 'Q'，使 s = \"QWER\"。\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;= 10^5</code></li>\n\t<li><code>s.length</code>&nbsp;是&nbsp;<code>4</code>&nbsp;的倍数</li>\n\t<li><code>s</code>&nbsp;中只含有&nbsp;<code>'Q'</code>, <code>'W'</code>, <code>'E'</code>,&nbsp;<code>'R'</code>&nbsp;四种字符</li>\n</ul>\n",
"isPaidOnly": false,
"difficulty": "Medium",
"likes": 266,
"likes": 267,
"dislikes": 0,
"isLiked": null,
"similarQuestions": "[]",
@@ -149,7 +149,7 @@
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"stats": "{\"totalAccepted\": \"32.6K\", \"totalSubmission\": \"72.1K\", \"totalAcceptedRaw\": 32568, \"totalSubmissionRaw\": 72141, \"acRate\": \"45.1%\"}",
"stats": "{\"totalAccepted\": \"32.6K\", \"totalSubmission\": \"72.1K\", \"totalAcceptedRaw\": 32569, \"totalSubmissionRaw\": 72142, \"acRate\": \"45.1%\"}",
"hints": [
"Use 2-pointers algorithm to make sure all amount of characters outside the 2 pointers are smaller or equal to n/4.",
"That means you need to count the amount of each letter and make sure the amount is enough."