leetcode 1310. XOR Queries of a Subarray

Given the array arr of positive integers and the array queries where queries[i] = [Li, Ri], for each query i compute the XOR of elements from Li to Ri (that is, arr[Li] xor arr[Li+1] xor ... xor arr[Ri] ). Return an array containing the result for the given queries.

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 <= 3 * 10^4
• 1 <= arr[i] <= 10^9
• 1 <= queries.length <= 3 * 10^4
• queries[i].length == 2
• 0 <= queries[i][0] <= queries[i][1] < arr.length

思路：A[i] 表示[0,i]的所有值的异或，[i, j]的异或为A[j] ^ A[i - 1], 因为x ^ x = 0.

class Solution {
public:
    vector<int> xorQueries(vector<int>& arr, vector<vector<int>>& queries) {
        for (int i = 1; i < arr.size(); ++i) {
            arr[i] ^= arr[i - 1];
        }
        vector<int> ans(queries.size());
        for (int i = 0; i < queries.size(); ++i) {
            ans[i] = queries[i][0] > 0 ? arr[queries[i][1]] ^ arr[queries[i][0] - 1] : arr[queries[i][1]];
        }
        return ans;
    }
};

python:

class Solution:
    def xorQueries(self, arr: List[int], queries: List[List[int]]) -> List[int]:
        for i in range(len(arr) - 1):
            arr[i + 1] ^= arr[i]
        return [arr[j] ^ arr[i - 1] if i else arr[j] for i, j in queries]