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Formatted question description: https://leetcode.ca/all/1835.html
1835. Find XOR Sum of All Pairs Bitwise AND
Level
Hard
Description
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 to1 XOR 2 XOR 3 XOR 4 = 4
, and the XOR sum of[3]
is equal to3
.
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 <= 10^5
0 <= arr1[i], arr2[j] <= 10^9
Solution
Use the distributive law. For integers a
, b
and c
, there is (a ^ b) & c == (a & c) ^ (b & c)
and a & (b ^ c) = (a & b) ^ (a & c)
.
For 0 <= i < arr1.length
, the XOR sum of arr1[i] & arr2[j]
where 0 <= j < arr2.length
is equal to arr1[i] & (arr2[0] ^ arr2[1] ^ ... ^ arr2[arr2.length - 1])
.
Therefore, the XOR sum of the list is calculated as follows.
(arr1[0] ^ arr1[1] ^ ... ^ arr1[arr1.length - 1]) & (arr2[0] ^ arr2[1] ^ ... ^ arr2[arr2.length - 1])
-
class Solution { public int getXORSum(int[] arr1, int[] arr2) { int xor1 = 0, xor2 = 0; for (int num : arr1) xor1 ^= num; for (int num : arr2) xor2 ^= num; return xor1 & xor2; } } ############ class Solution { public int getXORSum(int[] arr1, int[] arr2) { int a = 0, b = 0; for (int v : arr1) { a ^= v; } for (int v : arr2) { b ^= v; } return a & b; } }
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// OJ: https://leetcode.com/problems/find-xor-sum-of-all-pairs-bitwise-and/ // Time: O(A + B) // Space: O(1) class Solution { public: int getXORSum(vector<int>& A, vector<int>& B) { int cnt[31] = {}; for (int b : B) { for (int i = 0; i < 31; ++i) { cnt[i] += (b >> i & 1); } } int ans = 0; for (int a : A) { int x = 0; for (int i = 0; i < 31; ++i) { if ((a >> i & 1) == 0 || cnt[i] % 2 == 0) continue; x |= 1 << i; } ans ^= x; } return ans; } };
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class Solution: def getXORSum(self, arr1: List[int], arr2: List[int]) -> int: a = reduce(xor, arr1) b = reduce(xor, arr2) return a & b ############ # 1835. Find XOR Sum of All Pairs Bitwise AND # https://leetcode.com/problems/find-xor-sum-of-all-pairs-bitwise-and class Solution: def getXORSum(self, arr1: List[int], arr2: List[int]) -> int: xora = xorb = 0 for num in arr1: xora ^= num for num in arr2: xorb ^= num return xora & xorb
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func getXORSum(arr1 []int, arr2 []int) int { var a, b int for _, v := range arr1 { a ^= v } for _, v := range arr2 { b ^= v } return a & b }
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function getXORSum(arr1: number[], arr2: number[]): number { const a = arr1.reduce((acc, x) => acc ^ x); const b = arr2.reduce((acc, x) => acc ^ x); return a & b; }