1177

1177. Can Make Palindrome from Substring (Medium)

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.

Related Topics:
Array, String

Solution 1.

// OJ: https://leetcode.com/problems/can-make-palindrome-from-substring/
// Author: github.com/lzl124631x
// Time: O(N + Q)
// Space: O(N)
class Solution {
public:
    vector<bool> canMakePaliQueries(string s, vector<vector<int>>& Q) {
        int N = s.size();
        vector<array<int, 26>> cnts(N);
        for (int i = 0; i < N; ++i) {
            if (i > 0) cnts[i] = cnts[i - 1];
            cnts[i][s[i] - 'a']++;
        }
        vector<bool> ans;
        for (auto &q : Q) {
            int from = q[0], to = q[1], k = q[2], odd = 0;
            auto cnt = cnts[to];
            if (from != 0) {
                for (int i = 0; i < 26; ++i) {
                    cnt[i] -= cnts[from - 1][i];
                }
            }
            for (int i = 0; i < 26; ++i) {
                odd += cnt[i] % 2;
            }
            ans.push_back(odd - 2 * k <= 1);
        }
        return ans;
    }
};

Or use bit mask

// OJ: https://leetcode.com/problems/can-make-palindrome-from-substring/
// Author: github.com/lzl124631x
// Time: O(N + Q)
// Space: O(N)
class Solution {
public:
    vector<bool> canMakePaliQueries(string s, vector<vector<int>>& Q) {
        int N = s.size(), mask = 0;
        vector<int> odds(1);
        for (char c : s) {
            odds.push_back(mask ^= (1 << (c - 'a')));
        }
        vector<bool> ans;
        for (auto &q : Q) {
            int from = q[0], to = q[1], k = q[2], odd = odds[to + 1] ^ odds[from], cnt = __builtin_popcount(odd);
            ans.push_back(cnt - 2 * k <= 1);
        }
        return ans;
    }
};