LeetCode 1835. Find XOR Sum of All Pairs Bitwise AND Solution in Java, C++, Python & More | Explanation + Code

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 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

Solutions

Solution 1: Bitwise Operation

Assume that the elements of array arr1 are a₁, a₂, ..., a_n, and the elements of array arr2 are b₁, b₂, ..., b_m. Then, the answer to the problem is:

\begin{aligned} ans &= (a₁ \wedge b₁) \oplus (a₁ \wedge b₂) ... (a₁ \wedge b_m) \ &\quad \oplus (a₂ \wedge b₁) \oplus (a₂ \wedge b₂) ... (a₂ \wedge b_m) \ &\quad \oplus … \ &\quad \oplus (a_n \wedge b₁) \oplus (a_n \wedge b₂) ... (a_n \wedge b_m) \ \end{aligned}

Since in Boolean algebra, the XOR operation is addition without carry, and the AND operation is multiplication, the above formula can be simplified as:

ans = (a₁ \oplus a₂ \oplus … \oplus a_n) \wedge (b₁ \oplus b₂ \oplus … \oplus b_m)

That is, the bitwise AND of the XOR sum of array arr1 and the XOR sum of array arr2.

The time complexity is O(n + m), where n and m are the lengths of arrays arr1 and arr2, respectively. The space complexity is O(1).

PythonJavaC++GoTypeScript

class Solution: def getXORSum(self, arr1: List[int], arr2: List[int]) -> int: a = reduce(xor, arr1) b = reduce(xor, arr2) return a & b(code-box)

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; } }(code-box)

class Solution { public: int getXORSum(vector<int>& arr1, vector<int>& arr2) { int a = accumulate(arr1.begin(), arr1.end(), 0, bit_xor<int>()); int b = accumulate(arr2.begin(), arr2.end(), 0, bit_xor<int>()); return a & b; } };(code-box)

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 }(code-box)

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; }(code-box)