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.
class ValidPalindrome {
public boolean isPalindrome(String s) {
if (s == null || s.length() < 2) return true;
int left = 0;
int right = s.length() - 1;
while (left < right) {
if (s.charAt(left) != s.charAt(right)) return false;
left++;
right--;
}
return true;
}
}
关键点分析
前缀和数组计算前i个字母,j字母出现的次数
每一次query,扫描计算出次数为奇数的字母的个数,与k比较
代码 (Java/Python3)
Java code
public class CanMakePalindromeFromSubstring {
public List<Boolean> canMakePaliQueries(String s, int[][] queries) {
int n = s.length();
int[][] prefixCount = new int[n + 1][26];
// prefixCount[i][j] matrix, calculate number of letter from range [0,i] for j letter.
for (int i = 0; i < n; i++) {
prefixCount[i + 1] = prefixCount[i].clone();
prefixCount[i + 1][s.charAt(i) - 'a']++;
}
List<Boolean> res = new ArrayList<>();
for (int[] q : queries) {
int start = q[0];
int end = q[1];
int k = q[2];
int count = 0;
// count odd letters number from range [start, end]
for (int i = 0; i < 26; i++) {
count += (prefixCount[end + 1][i] - prefixCount[start][i]) % 2;
}
res.add(count / 2 <= k);
}
return res;
}
}
Python3 code (slow :( )
class Solution:
def canMakePaliQueries(self, s: str, queries: List[List[int]]) -> List[bool]:
prefix_sum = [[0] * 26]
for i, char in enumerate(s):
prefix_sum.append(prefix_sum[i][:])
prefix_sum[i + 1][ord(char) - ord('a')] += 1
res = [True] * len(queries)
for i, (start, end, k) in enumerate(queries):
odd_count = 0
for j in range(26):
odd_count += (prefix_sum[end + 1][j] - prefix_sum[start][j]) % 2
res[i] = (odd_count // 2 <= k)
return res