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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.
1import java.util.*;
2
3public class XorQueries {
4 public static int[] xorQueries(int[] arr, int[][] queries) {
5 int[] result = new int[queries.length];
6 for (int i = 0; i < queries.length; i++) {
7 int xor_result = 0;
8 for (int j = queries[i][0]; j <= queries[i][1]; j++) {
9 xor_result ^= arr[j];
10 }
11 result[i] = xor_result;
12 }
13 return result;
14 }
15
16 public static void main(String[] args) {
17 int[] arr = {1, 3, 4, 8};
18 int[][] queries = {{0, 1}, {1, 2}, {0, 3}, {3, 3}};
19 int[] result = xorQueries(arr, queries);
20 System.out.println(Arrays.toString(result));
21 }
22}In this Java implementation, we iterate over each query, then use a nested loop to compute the XOR for the specified range. The results are stored in an array and printed using Arrays.toString().
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 C implementation computes the prefix XOR array, which is then used to quickly compute the XOR for any subrange as the difference between the prefix XOR at right + 1 and left indices.