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This approach involves directly computing the XOR for each range specified in the queries. For each query, iterate over the specified subarray range and compute the XOR.
Time Complexity: O(n * q), where n is the number of elements in the largest range and q is the number of queries.
Space Complexity: O(q) for storing the query results.
1using System;
2using System.Collections.Generic;
3
4class XorQueries {
5 public static int[] XorQueries(int[] arr, int[][] queries) {
6 int[] result = new int[queries.Length];
7 for (int i = 0; i < queries.Length; i++) {
8 int xor_result = 0;
9 for (int j = queries[i][0]; j <= queries[i][1]; j++) {
10 xor_result ^= arr[j];
11 }
12 result[i] = xor_result;
13 }
14 return result;
15 }
16
17 static void Main() {
18 int[] arr = {1, 3, 4, 8};
19 int[][] queries = new int[][] {
20 new int[] {0, 1},
21 new int[] {1, 2},
22 new int[] {0, 3},
23 new int[] {3, 3}
24 };
25 int[] result = XorQueries(arr, queries);
26 Console.WriteLine(string.Join(" ", result));
27 }
28}In C#, the solution uses a main function to call XorQueries, which iterates over queries and computes the XOR for each specified range, storing the results in an array.
To optimize the XOR computation, we can use a prefix XOR array. Compute the XOR of all elements up to each position in the array. Then to find the XOR for a range simply subtract the prefix value before the start of the range from the prefix at the end of the range.
Time Complexity: O(n + q), where n is the number of elements and q is the number of queries.
Space Complexity: O(n) for the prefix XOR array.
1
This Java implementation computes the prefix XOR, allowing the XOR of any subarray to be quickly retrieved using two prefix XOR values. This makes the query resolution step constant time for each query.