LeetCode-in-Java

1177. Can Make Palindrome from Substring

Medium

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. q

ueries[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

import java.util.ArrayList;
import java.util.List;

public class Solution {
    public List<Boolean> canMakePaliQueries(String s, int[][] queries) {
        return this.canMakeP(s, queries);
    }

    private List<Boolean> canMakeP(String s, int[][] qs) {
        int n = s.length();
        int[] counts = new int[n];
        for (int i = 0; i < n; i++) {
            int m = 0;
            if (i > 0) {
                m = counts[i - 1];
            }
            char c = s.charAt(i);
            m = m ^ (1 << (c - 'a'));
            counts[i] = m;
        }
        List<Boolean> ans = new ArrayList<>();
        for (int[] q : qs) {
            ans.add(check(q, counts));
        }
        return ans;
    }

    private boolean check(int[] q, int[] counts) {
        int l = q[0];
        int r = q[1];
        int k = q[2];
        int prev = l > 0 ? counts[l - 1] : 0;
        int kk = Integer.bitCount(prev ^ counts[r]);
        return (kk / 2) <= k;
    }
}

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