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.
Solution:
class Solution {
public List<Boolean> canMakePaliQueries(String s, int[][] queries) {
List<Boolean> res = new ArrayList();
int[][] count = new int[s.length() + 1][26];
for (int i = 0; i < s.length(); i ++) {
count[i + 1] = count[i].clone(); // copy previous sum.
count[i + 1][s.charAt(i) - 'a'] ++;
}
for (int[] query : queries) {
int start = query[0], end = query[1], k = query[2];
int odd = 0;
for (int i = 0; i < 26; i ++) {
if ((count[end + 1][i] - count[start][i]) % 2 == 1) {
odd ++;
}
}
res.add(odd / 2 <= k);
}
return res;
}
}