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leetcode-problemset/leetcode-cn/originData/can-make-palindrome-from-substring.json
2022-05-02 23:44:12 +08:00

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{
"data": {
"question": {
"questionId": "1281",
"questionFrontendId": "1177",
"categoryTitle": "Algorithms",
"boundTopicId": 24022,
"title": "Can Make Palindrome from Substring",
"titleSlug": "can-make-palindrome-from-substring",
"content": "<p>You are given a string <code>s</code> and array <code>queries</code> where <code>queries[i] = [left<sub>i</sub>, right<sub>i</sub>, k<sub>i</sub>]</code>. We may rearrange the substring <code>s[left<sub>i</sub>...right<sub>i</sub>]</code> for each query and then choose up to <code>k<sub>i</sub></code> of them to replace with any lowercase English letter.</p>\n\n<p>If the substring is possible to be a palindrome string after the operations above, the result of the query is <code>true</code>. Otherwise, the result is <code>false</code>.</p>\n\n<p>Return a boolean array <code>answer</code> where <code>answer[i]</code> is the result of the <code>i<sup>th</sup></code> query <code>queries[i]</code>.</p>\n\n<p>Note that each letter is counted individually for replacement, so if, for example <code>s[left<sub>i</sub>...right<sub>i</sub>] = &quot;aaa&quot;</code>, and <code>k<sub>i</sub> = 2</code>, we can only replace two of the letters. Also, note that no query modifies the initial string <code>s</code>.</p>\n\n<p>&nbsp;</p>\n<p><strong>Example :</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;abcda&quot;, queries = [[3,3,0],[1,2,0],[0,3,1],[0,3,2],[0,4,1]]\n<strong>Output:</strong> [true,false,false,true,true]\n<strong>Explanation:</strong>\nqueries[0]: substring = &quot;d&quot;, is palidrome.\nqueries[1]: substring = &quot;bc&quot;, is not palidrome.\nqueries[2]: substring = &quot;abcd&quot;, is not palidrome after replacing only 1 character.\nqueries[3]: substring = &quot;abcd&quot;, could be changed to &quot;abba&quot; which is palidrome. Also this can be changed to &quot;baab&quot; first rearrange it &quot;bacd&quot; then replace &quot;cd&quot; with &quot;ab&quot;.\nqueries[4]: substring = &quot;abcda&quot;, could be changed to &quot;abcba&quot; which is palidrome.\n</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;lyb&quot;, queries = [[0,1,0],[2,2,1]]\n<strong>Output:</strong> [false,true]\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= s.length, queries.length &lt;= 10<sup>5</sup></code></li>\n\t<li><code>0 &lt;= left<sub>i</sub> &lt;= right<sub>i</sub> &lt; s.length</code></li>\n\t<li><code>0 &lt;= k<sub>i</sub> &lt;= s.length</code></li>\n\t<li><code>s</code> consists of lowercase English letters.</li>\n</ul>\n",
"translatedTitle": "构建回文串检测",
"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",
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"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 ?",
"As a preprocess, calculate the accumulate frequency of all characters for all prefixes of the string.",
"How to check if a substring can be changed to a palindrome given its characters frequency ?",
"Count the number of odd frequencies, there can be at most one odd frequency in a palindrome."
],
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