{ "data": { "question": { "questionId": "1281", "questionFrontendId": "1177", "categoryTitle": "Algorithms", "boundTopicId": 24022, "title": "Can Make Palindrome from Substring", "titleSlug": "can-make-palindrome-from-substring", "content": "
You are given a string s
and array queries
where queries[i] = [lefti, righti, ki]
. We may rearrange the substring s[lefti...righti]
for each query and then choose up to ki
of them to replace with any lowercase English letter.
If the substring is possible to be a palindrome string after the operations above, the result of the query is true
. Otherwise, the result is false
.
Return a boolean array answer
where answer[i]
is the result of the ith
query queries[i]
.
Note that each letter is counted individually for replacement, so if, for example s[lefti...righti] = "aaa"
, and ki = 2
, we can only replace two of the letters. Also, note that no query modifies the initial string s
.
\n
Example :
\n\n\nInput: s = "abcda", queries = [[3,3,0],[1,2,0],[0,3,1],[0,3,2],[0,4,1]]\nOutput: [true,false,false,true,true]\nExplanation:\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\n\n
Example 2:
\n\n\nInput: s = "lyb", queries = [[0,1,0],[2,2,1]]\nOutput: [false,true]\n\n\n
\n
Constraints:
\n\n1 <= s.length, queries.length <= 105
0 <= lefti <= righti < s.length
0 <= ki <= s.length
s
consists of lowercase English letters.给你一个字符串 s
,请你对 s
的子串进行检测。
每次检测,待检子串都可以表示为 queries[i] = [left, right, k]
。我们可以 重新排列 子串 s[left], ..., s[right]
,并从中选择 最多 k
项替换成任何小写英文字母。
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 true
,否则结果为 false
。
返回答案数组 answer[]
,其中 answer[i]
是第 i
个待检子串 queries[i]
的检测结果。
注意:在替换时,子串中的每个字母都必须作为 独立的 项进行计数,也就是说,如果 s[left..right] = "aaa"
且 k = 2
,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 s
,可以认为每次检测都是独立的)
\n\n
示例:
\n\n输入:s = "abcda", queries = [[3,3,0],[1,2,0],[0,3,1],[0,3,2],[0,4,1]]\n输出:[true,false,false,true,true]\n解释:\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\n\n
\n\n
提示:
\n\n1 <= s.length, queries.length <= 10^5
0 <= queries[i][0] <= queries[i][1] < s.length
0 <= queries[i][2] <= s.length
s
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out-of-bounds<\\/code>\\u548c
use-after-free<\\/code>\\u9519\\u8bef\\u3002<\\/p>\\r\\n\\r\\n
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out-of-bounds<\\/code>\\u548c
use-after-free<\\/code>\\u9519\\u8bef\\u3002<\\/p>\\r\\n\\r\\n
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Node.js 16.13.2<\\/code><\\/p>\\r\\n\\r\\n
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Ruby 3.1<\\/code>\\u6267\\u884c<\\/p>\\r\\n\\r\\n
Swift 5.5.2<\\/code><\\/p>\\r\\n\\r\\n
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Python 3.10<\\/code><\\/p>\\r\\n\\r\\n
Scala 2.13<\\/code><\\/p>\"],\"kotlin\":[\"Kotlin\",\"
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rust 1.58.1<\\/code><\\/p>\\r\\n\\r\\n
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