Today I have done leetcode's September LeetCoding Challenge with cpp.

September LeetCoding Challenge 13

Description

XOR Queries of a Subarray

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 * 104
• 1 <= arr[i] <= 109
• queries[i].length == 2
• 0 <= lefti <= righti < arr.length

Solution

class Solution {
public:
  vector<int> xorQueries(vector<int>& arr, vector<vector<int>>& queries) {
    arr.insert(arr.begin(), 0);
    for(int i = 1; i < arr.size(); ++i) {
      arr[i] = arr[i] ^ arr[i - 1];
    }
    vector<int> answer;
    answer.reserve(queries.size());
    for(const auto &query : queries) {
      answer.push_back(arr[query[1] + 1] ^ arr[query[0]]);
    }
    return answer;
  }
};