update

leetcode-cn/originData/can-make-palindrome-from-substring.json

@@ -12,7 +12,7 @@
             "translatedContent": "<p>给你一个字符串&nbsp;<code>s</code>，请你对&nbsp;<code>s</code>&nbsp;的子串进行检测。</p>\n\n<p>每次检测，待检子串都可以表示为&nbsp;<code>queries[i] = [left, right, k]</code>。我们可以 <strong>重新排列</strong> 子串&nbsp;<code>s[left], ..., s[right]</code>，并从中选择 <strong>最多</strong> <code>k</code>&nbsp;项替换成任何小写英文字母。&nbsp;</p>\n\n<p>如果在上述检测过程中，子串可以变成回文形式的字符串，那么检测结果为&nbsp;<code>true</code>，否则结果为&nbsp;<code>false</code>。</p>\n\n<p>返回答案数组&nbsp;<code>answer[]</code>，其中&nbsp;<code>answer[i]</code>&nbsp;是第&nbsp;<code>i</code>&nbsp;个待检子串&nbsp;<code>queries[i]</code>&nbsp;的检测结果。</p>\n\n<p>注意：在替换时，子串中的每个字母都必须作为 <strong>独立的</strong> 项进行计数，也就是说，如果&nbsp;<code>s[left..right] = &quot;aaa&quot;</code>&nbsp;且&nbsp;<code>k = 2</code>，我们只能替换其中的两个字母。（另外，任何检测都不会修改原始字符串 <code>s</code>，可以认为每次检测都是独立的）</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例：</strong></p>\n\n<pre><strong>输入：</strong>s = &quot;abcda&quot;, queries = [[3,3,0],[1,2,0],[0,3,1],[0,3,2],[0,4,1]]\n<strong>输出：</strong>[true,false,false,true,true]\n<strong>解释：</strong>\nqueries[0] : 子串 = &quot;d&quot;，回文。\nqueries[1] :&nbsp;子串 = &quot;bc&quot;，不是回文。\nqueries[2] :&nbsp;子串 = &quot;abcd&quot;，只替换 1 个字符是变不成回文串的。\nqueries[3] :&nbsp;子串 = &quot;abcd&quot;，可以变成回文的 &quot;abba&quot;。 也可以变成 &quot;baab&quot;，先重新排序变成 &quot;bacd&quot;，然后把 &quot;cd&quot; 替换为 &quot;ab&quot;。\nqueries[4] :&nbsp;子串 = &quot;abcda&quot;，可以变成回文的 &quot;abcba&quot;。\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,&nbsp;queries.length&nbsp;&lt;= 10^5</code></li>\n\t<li><code>0 &lt;= queries[i][0] &lt;= queries[i][1] &lt;&nbsp;s.length</code></li>\n\t<li><code>0 &lt;= queries[i][2] &lt;= s.length</code></li>\n\t<li><code>s</code> 中只有小写英文字母</li>\n</ul>\n",
             "isPaidOnly": false,
             "difficulty": "Medium",
-            "likes": 58,
+            "likes": 60,
             "dislikes": 0,
             "isLiked": null,
             "similarQuestions": "[]",
@@ -155,7 +155,7 @@
                     "__typename": "CodeSnippetNode"
                 }
             ],
-            "stats": "{\"totalAccepted\": \"6.8K\", \"totalSubmission\": \"24.6K\", \"totalAcceptedRaw\": 6763, \"totalSubmissionRaw\": 24578, \"acRate\": \"27.5%\"}",
+            "stats": "{\"totalAccepted\": \"6.9K\", \"totalSubmission\": \"24.9K\", \"totalAcceptedRaw\": 6889, \"totalSubmissionRaw\": 24885, \"acRate\": \"27.7%\"}",
             "hints": [
                 "Since we can rearrange the substring, all we care about is the frequency of each character in that substring.",
                 "How to find the character frequencies efficiently ?",