Skip to content

1177. Can Make Palindrome from Substring

Difficulty Topics

Description

You are given a string s and array queries where queries[i] = [lefti, righti, ki]. We may rearrange the substring s[lefti...righti] for each query and then choose up to ki 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 a boolean array answer where answer[i] is the result of the ith query queries[i].

Note that each letter is counted individually for replacement, so if, for example s[lefti...righti] = "aaa", and ki = 2, we can only replace two of the letters. Also, note that no query modifies the initial string s.

 

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.

Example 2:

Input: s = "lyb", queries = [[0,1,0],[2,2,1]]
Output: [false,true]

 

Constraints:

  • 1 <= s.length, queries.length <= 105
  • 0 <= lefti <= righti < s.length
  • 0 <= ki <= s.length
  • s consists of lowercase English letters.

Solution

can-make-palindrome-from-substring.py
class Solution:
    def canMakePaliQueries(self, s: str, queries: List[List[int]]) -> List[bool]:
        n = len(s)
        res = []
        dp = [[0] * 26]

        for i in range(1, n + 1):
            t = dp[i - 1][:]
            ch = ord(s[i - 1]) - ord('a')
            t[ch] += 1
            dp.append(t)


        for l,r,k in queries:
            left = dp[l]
            right = dp[r + 1]

            cnt = 0
            for i in range(26):
                cnt += (right[i] - left[i]) & 1

            res.append(cnt // 2 <= k)

        return res