Off-by-one on range boundaries
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.
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.
[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 <= 1050 <= arr1[i], arr2[j] <= 109Problem summary: 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.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array · Math · Bit Manipulation
[1,2,3] [6,5]
[12] [4]
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #1835: Find XOR Sum of All Pairs Bitwise AND
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;
}
}
// Accepted solution for LeetCode #1835: Find XOR Sum of All Pairs Bitwise AND
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
}
# Accepted solution for LeetCode #1835: Find XOR Sum of All Pairs Bitwise AND
class Solution:
def getXORSum(self, arr1: List[int], arr2: List[int]) -> int:
a = reduce(xor, arr1)
b = reduce(xor, arr2)
return a & b
// Accepted solution for LeetCode #1835: Find XOR Sum of All Pairs Bitwise AND
/**
* [1835] Find XOR Sum of All Pairs Bitwise AND
*
* 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 <= 10^5
* 0 <= arr1[i], arr2[j] <= 10^9
*
*/
pub struct Solution {}
// problem: https://leetcode.com/problems/find-xor-sum-of-all-pairs-bitwise-and/
// discuss: https://leetcode.com/problems/find-xor-sum-of-all-pairs-bitwise-and/discuss/?currentPage=1&orderBy=most_votes&query=
// submission codes start here
impl Solution {
pub fn get_xor_sum(arr1: Vec<i32>, arr2: Vec<i32>) -> i32 {
0
}
}
// submission codes end
#[cfg(test)]
mod tests {
use super::*;
#[test]
#[ignore]
fn test_1835_example_1() {
let arr1 = vec![1, 2, 3];
let arr2 = vec![6, 5];
let result = 0;
assert_eq!(Solution::get_xor_sum(arr1, arr2), result);
}
#[test]
#[ignore]
fn test_1835_example_2() {
let arr1 = vec![12];
let arr2 = vec![4];
let result = 4;
assert_eq!(Solution::get_xor_sum(arr1, arr2), result);
}
}
// Accepted solution for LeetCode #1835: Find XOR Sum of All Pairs Bitwise AND
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;
}
Use this to step through a reusable interview workflow for this problem.
Sort the array in O(n log n), then scan for the missing or unique element by comparing adjacent pairs. Sorting requires O(n) auxiliary space (or O(1) with in-place sort but O(n log n) time remains). The sort step dominates.
Bitwise operations (AND, OR, XOR, shifts) are O(1) per operation on fixed-width integers. A single pass through the input with bit operations gives O(n) time. The key insight: XOR of a number with itself is 0, which eliminates duplicates without extra space.
Review these before coding to avoid predictable interview regressions.
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Wrong move: Temporary multiplications exceed integer bounds.
Usually fails on: Large inputs wrap around unexpectedly.
Fix: Use wider types, modular arithmetic, or rearranged operations.