LeetCode in Kotlin

1835. Find XOR Sum of All Pairs Bitwise AND

Hard

The XOR sum of a list is the bitwise XOR of all its elements. If the list only contains one element, then its XOR sum will be equal to this element.

• For example, the XOR sum of [1,2,3,4] is equal to 1 XOR 2 XOR 3 XOR 4 = 4, and the XOR sum of [3] is equal to 3.

You are given two 0-indexed arrays arr1 and arr2 that consist only of non-negative integers.

Consider the list containing the result of arr1[i] AND arr2[j] (bitwise AND) for every (i, j) pair where 0 <= i < arr1.length and 0 <= j < arr2.length.

Return the XOR sum of the aforementioned list.

Example 1:

Input: arr1 = [1,2,3], arr2 = [6,5]

Output: 0

Explanation: The list = [1 AND 6, 1 AND 5, 2 AND 6, 2 AND 5, 3 AND 6, 3 AND 5] = [0,1,2,0,2,1]. The XOR sum = 0 XOR 1 XOR 2 XOR 0 XOR 2 XOR 1 = 0.

Example 2:

Input: arr1 = [12], arr2 = [4]

Output: 4

Explanation: The list = [12 AND 4] = [4]. The XOR sum = 4.

Constraints:

• 1 <= arr1.length, arr2.length <= 105
• 0 <= arr1[i], arr2[j] <= 109

Solution

class Solution {
    fun getXORSum(arr1: IntArray, arr2: IntArray): Int {
        var xor1 = 0
        var xor2 = 0
        for (i in arr1) {
            xor1 = xor1 xor i
        }
        for (j in arr2) {
            xor2 = xor2 xor j
        }
        return xor1 and xor2
    }
}

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