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Java-based LeetCode algorithm problem solutions, regularly updated

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package g2501_2600.s2564_substring_xor_queries;

// #Medium #Array #String #Hash_Table #Bit_Manipulation
// #2023_08_21_Time_26_ms_(95.83%)_Space_84.1_MB_(80.56%)

import java.util.HashMap;
import java.util.Map;

/**
 * 2564 - Substring XOR Queries\.
 *
 * Medium
 *
 * 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 <= s.length <= 10⁴
 * *   `s[i]` is either `'0'` or `'1'`.
 * *   1 <= queries.length <= 10⁵
 * *   0 <= first_i, second_i <= 10⁹
**/
public class Solution {
    public int[][] substringXorQueries(String s, int[][] queries) {
        int[][] ans = new int[queries.length][2];
        Map map = new HashMap<>();
        for (int i = 0; i < s.length(); i++) {
            if (s.charAt(i) == '0') {
                map.putIfAbsent(0, new int[] {i, i});
                continue;
            }
            int value = 0;
            for (int j = i; j <= Math.min(i + 30, s.length() - 1); j++) {
                value = (value << 1) + (s.charAt(j) - '0');
                map.putIfAbsent(value, new int[] {i, j});
            }
        }
        int i = 0;
        for (int[] q : queries) {
            ans[i++] = map.getOrDefault(q[0] ^ q[1], new int[] {-1, -1});
        }
        return ans;
    }
}