README.md

1177. Can Make Palindrome from Substring

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 1:

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.

Solutions (Rust)

1. Solution

impl Solution {
    pub fn can_make_pali_queries(s: String, queries: Vec<Vec<i32>>) -> Vec<bool> {
        let mut prefix = vec![0_i32; s.len() + 1];
        let mut answer = vec![true; queries.len()];

        for (i, c) in s.bytes().enumerate() {
            prefix[i + 1] = prefix[i] ^ (1 << (c - b'a'));
        }

        for i in 0..queries.len() {
            let left = queries[i][0] as usize;
            let right = queries[i][1] as usize;
            let k = queries[i][2] as u32;

            answer[i] = (prefix[left] ^ prefix[right + 1]).count_ones() / 2 <= k;
        }

        answer
    }
}