XOR queries of a subarray

Time: O(N); Space: O(1); medium

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 <= len(arr) <= 3 * 10^4

  • 1 <= arr[i] <= 10^9

  • 1 <= len(queries) <= 3 * 10^4

  • len(queries[i]) == 2

  • 0 <= queries[i][0] <= queries[i][1] < len(arr)

Hints:

  1. What is the result of x ^ y ^ x ?

  2. Compute the prefix sum for XOR.

  3. Process the queries with the prefix sum values.

[1]:

class Solution1(object):
    """
    Time: O(N)
    Space: O(1)
    """
    def xorQueries(self, arr, queries):
        """
        :type arr: List[int]
        :type queries: List[List[int]]
        :rtype: List[int]
        """
        for i in range(1, len(arr)):
            arr[i] ^= arr[i-1]

        return [arr[right] ^ arr[left-1] if left else arr[right] for left, right in queries]

[2]:

s = Solution1()

arr = [1,3,4,8]
queries = [[0,1],[1,2],[0,3],[3,3]]
assert s.xorQueries(arr, queries) == [2,7,14,8]

arr = [4,8,2,10]
queries = [[2,3],[1,3],[0,0],[0,3]]
assert s.xorQueries(arr, queries) == [8,0,4,4]