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title: Bitwise XOR of All Pairings
slug: bitwise-xor-of-all-pairings
difficulty: medium
leetcode_id: 2425
leetcode_url: https://leetcode.com/problems/bitwise-xor-of-all-pairings/
categories:
- arrays
- math
patterns:
- greedy
description: |
You are given two **0-indexed** arrays, `nums1` and `nums2`, consisting of non-negative integers.
Let there be another array, `nums3`, which contains the bitwise XOR of **all pairings** of integers between `nums1` and `nums2` (every integer in `nums1` is paired with every integer in `nums2` **exactly once**).
Return *the **bitwise XOR** of all integers in* `nums3`.
constraints: |
- `1 <= nums1.length, nums2.length <= 10^5`
- `0 <= nums1[i], nums2[j] <= 10^9`
examples:
- input: "nums1 = [2,1,3], nums2 = [10,2,5,0]"
output: "13"
explanation: "A possible nums3 array is [8,0,7,2,11,3,4,1,9,1,6,3]. The bitwise XOR of all these numbers is 13."
- input: "nums1 = [1,2], nums2 = [3,4]"
output: "0"
explanation: "All possible pairs of bitwise XORs are nums1[0] ^ nums2[0], nums1[0] ^ nums2[1], nums1[1] ^ nums2[0], and nums1[1] ^ nums2[1]. Thus, one possible nums3 array is [2,5,1,6]. 2 ^ 5 ^ 1 ^ 6 = 0, so we return 0."
explanation:
intuition: |
At first glance, this problem seems to require generating all `m * n` pairwise XORs (where `m = len(nums1)` and `n = len(nums2)`), then XORing them all together. With constraints up to `10^5` for both arrays, that's potentially `10^10` operations — far too slow.
The key insight comes from a fundamental property of XOR: **a value XORed with itself cancels out** (`x ^ x = 0`). This means if any value appears an **even number of times** in the final XOR, it contributes nothing to the result.
Think of it like this: imagine laying out all pairwise XORs in a grid. Each element `nums1[i]` appears in **exactly `n` pairs** (once with each element of `nums2`). Similarly, each element `nums2[j]` appears in **exactly `m` pairs**.
When we XOR all pairs together, we're effectively XORing each `nums1[i]` a total of `n` times, and each `nums2[j]` a total of `m` times. If `n` is even, all `nums1` elements cancel out. If `m` is even, all `nums2` elements cancel out.
This reduces the problem to simple parity checks on the array lengths.
approach: |
We solve this using **XOR properties and parity analysis**:
**Step 1: Understand the contribution of each element**
- Each element in `nums1` is XORed with every element in `nums2`, so it appears in `len(nums2)` pairs
- Each element in `nums2` is XORed with every element in `nums1`, so it appears in `len(nums1)` pairs
- When we XOR all pairs, each `nums1[i]` is XORed `len(nums2)` times total
&nbsp;
**Step 2: Apply the cancellation property**
- If `len(nums2)` is even, each `nums1[i]` appears an even number of times and cancels out
- If `len(nums2)` is odd, each `nums1[i]` appears an odd number of times and contributes once
- The same logic applies to `nums2` elements based on `len(nums1)`
&nbsp;
**Step 3: Compute the result based on parity**
- `result = 0`
- If `len(nums2)` is odd: XOR all elements of `nums1` into result
- If `len(nums1)` is odd: XOR all elements of `nums2` into result
- Return the final result
&nbsp;
This elegant solution avoids generating any pairs at all, working in O(m + n) time.
common_pitfalls:
- title: The Brute Force Trap
description: |
The naive approach generates all `m * n` pairwise XORs:
```python
for a in nums1:
for b in nums2:
result ^= (a ^ b)
```
With `m, n <= 10^5`, this results in up to `10^10` operations — a guaranteed **Time Limit Exceeded**. The mathematical insight about parity is essential to solve this efficiently.
wrong_approach: "Nested loops generating all pairs"
correct_approach: "Use parity to determine which elements contribute"
- title: Forgetting Both Arrays Can Contribute
description: |
It's easy to focus on just one array's contribution. Remember:
- `nums1` contributes if `len(nums2)` is **odd**
- `nums2` contributes if `len(nums1)` is **odd**
Both conditions can be true simultaneously! For example, if both arrays have odd lengths, elements from **both** arrays contribute to the final XOR.
wrong_approach: "Only checking one array's length parity"
correct_approach: "Check both len(nums1) and len(nums2) independently"
- title: Confusing Which Length Matters
description: |
A subtle error is checking the wrong length for each array:
- `nums1` elements appear `len(nums2)` times (not `len(nums1)` times)
- `nums2` elements appear `len(nums1)` times (not `len(nums2)` times)
If you check `len(nums1) % 2` to decide whether to include `nums1`, you'll get the wrong answer. The **other** array's length determines the contribution.
wrong_approach: "Checking len(nums1) % 2 for nums1's contribution"
correct_approach: "Check len(nums2) % 2 for nums1's contribution"
key_takeaways:
- "**XOR cancellation**: `x ^ x = 0` — values appearing an even number of times cancel out entirely"
- "**Parity analysis**: When elements repeat based on array lengths, check if the count is odd or even to determine contribution"
- "**Avoid unnecessary computation**: Mathematical insight can reduce O(n^2) or worse problems to O(n)"
- "**Bit manipulation patterns**: Many XOR problems have elegant solutions based on fundamental properties like `x ^ x = 0` and `x ^ 0 = x`"
time_complexity: "O(m + n). We iterate through each array at most once, where `m = len(nums1)` and `n = len(nums2)`."
space_complexity: "O(1). We only use a single variable to accumulate the XOR result."
solutions:
- approach_name: Parity-Based XOR
is_optimal: true
code: |
def xor_all_nums(nums1: list[int], nums2: list[int]) -> int:
result = 0
len1, len2 = len(nums1), len(nums2)
# Each nums1 element appears len2 times in all pairs
# If len2 is odd, it contributes to the final XOR
if len2 % 2 == 1:
for num in nums1:
result ^= num
# Each nums2 element appears len1 times in all pairs
# If len1 is odd, it contributes to the final XOR
if len1 % 2 == 1:
for num in nums2:
result ^= num
return result
explanation: |
**Time Complexity:** O(m + n) — Single pass through each array at most once.
**Space Complexity:** O(1) — Only one variable used.
We leverage the XOR cancellation property: elements appearing an even number of times contribute nothing. Each element in `nums1` appears `len(nums2)` times, and vice versa. We only XOR elements whose appearance count is odd.
- approach_name: Pythonic One-Liner
is_optimal: true
code: |
from functools import reduce
from operator import xor
def xor_all_nums(nums1: list[int], nums2: list[int]) -> int:
# XOR nums1 if len(nums2) is odd, XOR nums2 if len(nums1) is odd
return (
(reduce(xor, nums1, 0) if len(nums2) % 2 else 0) ^
(reduce(xor, nums2, 0) if len(nums1) % 2 else 0)
)
explanation: |
**Time Complexity:** O(m + n) — Same as the explicit version.
**Space Complexity:** O(1) — Constant extra space.
A more compact version using `reduce` to XOR all elements. The logic is identical: include `nums1`'s XOR if `len(nums2)` is odd, include `nums2`'s XOR if `len(nums1)` is odd.
- approach_name: Brute Force
is_optimal: false
code: |
def xor_all_nums(nums1: list[int], nums2: list[int]) -> int:
result = 0
# XOR every possible pair
for a in nums1:
for b in nums2:
result ^= (a ^ b)
return result
explanation: |
**Time Complexity:** O(m * n) — Nested loops over both arrays.
**Space Complexity:** O(1) — Only tracking result.
This directly computes the XOR of all pairs. While correct, it's far too slow for large inputs (up to `10^10` operations). Included to illustrate why the mathematical insight is necessary. This will cause TLE on LeetCode.