LeetCode in Kotlin

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

Solution

class Solution {
    fun xorQueries(arr: IntArray, queries: Array<IntArray>): IntArray {
        val res = IntArray(queries.size)
        for (i in 1 until arr.size) {
            arr[i] = arr[i - 1] xor arr[i]
        }
        for (i in queries.indices) {
            val query = queries[i]
            res[i] = if (query[0] == 0) arr[query[1]] else arr[query[0] - 1] xor arr[query[1]]
        }
        return res
    }
}

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