9. 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]

Approach:(V.IMP)

The XOR trick: If we have a sequence of XOR operations a ^ b ^ c ^ ..., then we can remove all pairs of duplicated values without affecting the result.

Commutativity allows us to re-order the applications of XOR so that the duplicated elements are next to each other. Since x ^ x = 0 and a ^ 0 = a, each pair of duplicated values has no effect on the outcome.

Let’s go through an example of this:

  a ^ b ^ c ^ a ^ b     # Commutativity
= a ^ a ^ b ^ b ^ c     # Using x ^ x = 0
= 0 ^ 0 ^ c             # Using x ^ 0 = x (and commutativity)
= c

Solution:

class Solution
{
public:
    vector<int> xorQueries(vector<int> &arr, vector<vector<int>> &queries)
    {

        vector<int> res;
        for (int i = 1; i < arr.size(); i++)
        {
            arr[i] = arr[i] ^ arr[i - 1];
        }

        for (int i = 0; i < queries.size(); i++)
        {
            int start = queries[i][0];
            int end = queries[i][1];
            int ans;
            if (start != 0)
            {
                ans = arr[end] ^ arr[start - 1];
            }
            else
            {
                ans = arr[end];
            }

            res.push_back(ans);
        }

        return res;
    }
};

Time Complexity: O(n)