子字符串异或查询

/*
 * Approach using Java Streams:
 * 1. Generate all possible substrings of the binary string s.
 * 2. Use streams to filter and find the shortest substring matching the query conditions.
 * 3. Return the indices of the substring or [-1, -1] if no such substring exists.
 */

import java.util.ArrayList;
import java.util.List;
import java.util.stream.IntStream;

public class Solution {
    public int[][] solveQueries(String s, int[][] queries) {
        int n = s.length();
        int[][] result = new int[queries.length][2];

        for (int i = 0; i < queries.length; i++) {
            int first = queries[i][0];
            int second = queries[i][1];
            result[i][0] = -1;
            result[i][1] = -1;
            int minLength = Integer.MAX_VALUE;

            for (int start = 0; start < n; start++) {
                int finalI = i;
                IntStream.range(start, n).forEach(end -> {
                    String subStr = s.substring(start, end + 1);
                    int val = Integer.parseInt(subStr, 2);
                    if ((val ^ first) == second && (end - start + 1) < minLength) {
                        result[finalI][0] = start;
                        result[finalI][1] = end;
                        minLength = end - start + 1;
                    }
                });
            }
        }
        return result;
    }

    public static void main(String[] args) {
        Solution solution = new Solution();
        String s = "101101";
        int[][] queries = {{0, 5}, {1, 2}};
        int[][] result = solution.solveQueries(s, queries);
        for (int[] res : result) {
            System.out.println("[" + res[0] + ", " + res[1] + "]");
        }
    }
}