【leetcode】1177. Can Make Palindrome from Substring

题目如下：

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.

解题思路：对于给定一个query = [left,right,k]，很容易能求出这个区间内每个字符出现的次数，如果某个字符出现了偶数次，那说明不需要经过任何改变，这个字符就能组成回文。所以这里只需要计算有多少个字符出现的次数是奇数，假设有x个字符出现的次数为奇数，那么至少就需要经过x/2次改变，才能形成回文。这里有一种情况例外，那就是只有一个字符出现的次数为奇数，那么可以不需要做任何改变。

代码如下：

class Solution(object):
    def canMakePaliQueries(self, s, queries):
        """
        :type s: str
        :type queries: List[List[int]]
        :rtype: List[bool]
        """
        grid = [[0] * len(s) for _ in range(26)]
        count = [0] * 26

        for i,v in enumerate(s):
            for j in range(26):
                grid[j][i] = grid[j][i-1]
            inx = ord(v) - ord('a')
            count[inx] += 1
            grid[inx][i] = count[inx]

        res = []

        for left,right,k in queries:
            diff = 0
            for i in range(26):
                if left > 0 and (grid[i][right] - grid[i][left-1]) % 2 != 0:
                    diff += 1
                elif left == 0 and grid[i][right] % 2 != 0:
                    diff += 1
            if diff == 1 or diff / 2 <= k:
                res.append(True)
            else:
                res.append(False)

        return res