# 【leetcode】1177. Can Make Palindrome from Substring

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 : substring = "d", is palidrome.
queries : substring = "bc", is not palidrome.
queries : substring = "abcd", is not palidrome after replacing only 1 character.
queries : 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 : substring = "abcda", could be changed to "abcba" which is palidrome.


Constraints:

• 1 <= s.length, queries.length <= 10^5
• 0 <= queries[i] <= queries[i] < s.length
• 0 <= queries[i] <= s.length
• s only contains lowercase English letters.

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

posted @ 2019-09-06 15:59  seyjs  阅读(...)  评论(...编辑  收藏