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"question": {
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"categoryTitle": "Algorithms",
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"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>] = "aaa"</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> </p>\n<p><strong>Example :</strong></p>\n\n<pre>\n<strong>Input:</strong> s = "abcda", 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 = "d", is palidrome.\nqueries[1]: substring = "bc", is not palidrome.\nqueries[2]: substring = "abcd", is not palidrome after replacing only 1 character.\nqueries[3]: substring = "abcd", could be changed to "abba" which is palidrome. Also this can be changed to "baab" first rearrange it "bacd" then replace "cd" with "ab".\nqueries[4]: substring = "abcda", could be changed to "abcba" which is palidrome.\n</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = "lyb", queries = [[0,1,0],[2,2,1]]\n<strong>Output:</strong> [false,true]\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 <= s.length, queries.length <= 10<sup>5</sup></code></li>\n\t<li><code>0 <= left<sub>i</sub> <= right<sub>i</sub> < s.length</code></li>\n\t<li><code>0 <= k<sub>i</sub> <= 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 <code>s</code>，请你对 <code>s</code> 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 <code>queries[i] = [left, right, k]</code>。我们可以 <strong>重新排列</strong> 子串 <code>s[left], ..., s[right]</code>，并从中选择 <strong>最多</strong> <code>k</code> 项替换成任何小写英文字母。 </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 <code>true</code>，否则结果为 <code>false</code>。</p>\n\n<p>返回答案数组 <code>answer[]</code>，其中 <code>answer[i]</code> 是第 <code>i</code> 个待检子串 <code>queries[i]</code> 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 <code>s[left..right] = "aaa"</code> 且 <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> </p>\n\n<p><strong>示例：</strong></p>\n\n<pre><strong>输入：</strong>s = "abcda", 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 = "d"，回文。\nqueries[1] : 子串 = "bc"，不是回文。\nqueries[2] : 子串 = "abcd"，只替换 1 个字符是变不成回文串的。\nqueries[3] : 子串 = "abcd"，可以变成回文的 "abba"。 也可以变成 "baab"，先重新排序变成 "bacd"，然后把 "cd" 替换为 "ab"。\nqueries[4] : 子串 = "abcda"，可以变成回文的 "abcba"。\n</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 <= s.length, queries.length <= 10^5</code></li>\n\t<li><code>0 <= queries[i][0] <= queries[i][1] < s.length</code></li>\n\t<li><code>0 <= queries[i][2] <= 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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