LeetCode-in-Java
1310. XOR Queries of a Subarray
Medium
You are given an array arr of positive integers. You are also given the array queries where queries[i] = [lefti, righti].
For each query i compute the XOR of elements from lefti to righti (that is, arr[lefti] XOR arr[lefti + 1] XOR ... XOR arr[righti] ).
Return an array answer where answer[i] is the answer to the ith query.
Example 1:
Input: arr = [1,3,4,8], queries = [[0,1],[1,2],[0,3],[3,3]]
Output: [2,7,14,8]
Explanation:
The binary representation of the elements in the array are:
1 = 0001
3 = 0011
4 = 0100
8 = 1000
The XOR values for queries are:
[0,1] = 1 xor 3 = 2
[1,2] = 3 xor 4 = 7
[0,3] = 1 xor 3 xor 4 xor 8 = 14
[3,3] = 8
Example 2:
Input: arr = [4,8,2,10], queries = [[2,3],[1,3],[0,0],[0,3]]
Output: [8,0,4,4]
Constraints:
1 <= arr.length, queries.length <= 3 * 1041 <= arr[i] <= 109queries[i].length == 20 <= lefti <= righti < arr.length
Solution
public class Solution {
public int[] xorQueries(int[] a, int[][] queries) {
int[] res = new int[queries.length];
for (int i = 1; i < a.length; i++) {
a[i] = a[i - 1] ^ a[i];
}
for (int i = 0; i < queries.length; i++) {
int[] query = queries[i];
res[i] = (query[0] == 0 ? a[query[1]] : a[query[0] - 1] ^ a[query[1]]);
}
return res;
}
}