Substring XOR Queries

You are given a binary string  s , and a 2D integer array queries where  queries[i] = [first_i, second_i] .

For the i^th query, find the shortest substring of s whose decimal value, val , yields second_i when bitwise XORed with first_i . In other words, val ^ first_i == second_i .

The answer to the i^th query is the endpoints (0-indexed) of the substring [left_i, right_i] or [-1, -1] if no such substring exists. If there are multiple answers, choose the one with the minimum  left_i .

Return an array ans where ans[i] = [left_i, right_i] is the answer to the i^th query.

A substring is a contiguous non-empty sequence of characters within a string.

Example 1:

Input: s = "101101", queries = [[0,5],[1,2]]
Output: [[0,2],[2,3]]
Explanation: For the first query the substring in range [0,2] is "101" which has a decimal value of 5, and 5 ^ 0 = 5, hence the answer to the first query is [0,2]. In the second query, the substring in range [2,3] is "11", and has a decimal value of 3, and 3 ^ 1 = 2. So, [2,3] is returned for the second query. 

Example 2:

Input: s = "0101", queries = [[12,8]]
Output: [[-1,-1]]
Explanation: In this example there is no substring that answers the query, hence [-1,-1] is returned.

Example 3:

Input: s = "1", queries = [[4,5]]
Output: [[0,0]]
Explanation: For this example, the substring in range [0,0] has a decimal value of 1, and 1 ^ 4 = 5. So, the answer is [0,0].

Constraints:

• $ 1 \leq \text{s.length} \leq {10}^4$
• $s[i]$ is either $0$ or $1$.
• $1 \leq \text{queries.length} \leq {10}^5$
• $0 \leq \text{first}_\text{i}, \text{second}_\text{i} \leq {10}^9$

解题思路

　　这场lc周赛跟爆零差不多，lc都打不动了好似喵。

　　对于每个询问本质上就是问$s$中是否存在$\text{first}_\text{i} \wedge \text{second}_\text{i}$的二进制表示连续子串，所以想到预处理出来$s$的所有连续子串。当时没注意到询问的二进制串长度最大是$30$，一直想着这么把$s$所有的连续子串全部预处理出来。其实只用预处理出来长度不超过$30$的所有连续子串就可以了。

　　比赛就是没想到询问的二进制串只用$30$，硬是不会做，好似喵。

　　AC代码如下，时间复杂度为$O({30}^3 + n)$，可以优化到$O({30}^2 + n)$，懒得优化了能过就行：

 1 class Solution {
 2 public:
 3     vector<vector<int>> substringXorQueries(string s, vector<vector<int>>& queries) {
 4         int n = s.size();
 5         unordered_map<int, vector<int>> mp;
 6         for (int i = 1; i <= 30; i++) {
 7             for (int j = 0; j + i - 1 < n; j++) {
 8                 int t = 0;
 9                 for (int k = j; k <= j + i - 1; k++) {
10                     t = t << 1 | s[k] - '0';
11                 }
12                 if (!mp.count(t)) mp[t] = vector<int>({j, j + i - 1});
13             }
14         }
15         vector<vector<int>> ans(queries.size(), vector<int>(2, -1));
16         for (int i = 0; i < queries.size(); i++) {
17             int t = queries[i][0] ^ queries[i][1];
18             if (mp.count(t)) ans[i] = mp[t];
19         }
20         return ans;
21     }
22 };

参考资料

　　枚举：https://leetcode.cn/problems/substring-xor-queries/solution/mei-ju-by-tsreaper-h2xh/