标签:nbsp turn round hat har pre info pull pos
Given a string s
, we make queries on substrings of s
.
For each query queries[i] = [left, right, k]
, we may rearrange the substring s[left], ..., s[right]
, and then choose up to k
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 an array answer[]
, where answer[i]
is the result of the i
-th query queries[i]
.
Note that: Each letter is counted individually for replacement so if for example s[left..right] = "aaa"
, and k = 2
, we can only replace two of the letters. (Also, note that the initial string s
is never modified by any query.)
Example :
Input: s = "abcda", queries = [[3,3,0],[1,2,0],[0,3,1],[0,3,2],[0,4,1]] Output: [true,false,false,true,true] Explanation: queries[0] : substring = "d", is palidrome. queries[1] : substring = "bc", is not palidrome. queries[2] : substring = "abcd", is not palidrome after replacing only 1 character. queries[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". queries[4] : substring = "abcda", could be changed to "abcba" which is palidrome.
Constraints:
1 <= s.length, queries.length <= 10^5
0 <= queries[i][0] <= queries[i][1] < s.length
0 <= queries[i][2] <= s.length
s
only contains lowercase English letters.Contest-LeetCode5175. Can Make Palindrome from Substring(unfinished)
标签:nbsp turn round hat har pre info pull pos
原文地址:https://www.cnblogs.com/Marigolci/p/11442610.html