Problem Description

You are given two arrays arr1 and arr2 that contain only non-negative integers. The problem asks you to find the XOR sum of all possible bitwise AND operations between pairs of elements from the two arrays.

Specifically:

1. For every pair (i, j) where i is an index in arr1 and j is an index in arr2, calculate arr1[i] AND arr2[j] (bitwise AND operation)
2. Collect all these AND results into a list
3. Find the XOR sum of this list (XOR all elements together)

XOR sum means applying the bitwise XOR operation to all elements in sequence. For example:

• XOR sum of [1,2,3,4] = 1 XOR 2 XOR 3 XOR 4 = 4
• XOR sum of [3] = 3

The mathematical insight here is that due to the distributive property of AND over XOR in Boolean algebra, the result can be simplified to:

• First, find the XOR sum of all elements in arr1 (let's call it a)
• Then, find the XOR sum of all elements in arr2 (let's call it b)
• The final answer is a AND b

This works because XOR behaves like addition and AND behaves like multiplication in Boolean algebra, allowing us to factor out the expression:

(a₁ AND b₁) XOR (a₁ AND b₂) XOR ... XOR (aₙ AND bₘ) = (a₁ XOR a₂ XOR ... XOR aₙ) AND (b₁ XOR b₂ XOR ... XOR bₘ)

Intuition

Let's think about what happens when we compute all possible AND operations and then XOR them together. If we were to naively compute this, we'd need to calculate n * m AND operations (where n and m are the lengths of the two arrays), store all results, and then XOR them together.

But there's a beautiful mathematical property we can exploit here. In Boolean algebra, XOR and AND operations have a special relationship - they follow distributive laws similar to addition and multiplication in regular algebra.

Consider a simple example with arr1 = [a, b] and arr2 = [c, d]. The naive approach would compute: (a AND c) XOR (a AND d) XOR (b AND c) XOR (b AND d)

Notice how we can group terms:

• Terms with a: (a AND c) XOR (a AND d) = a AND (c XOR d)
• Terms with b: (b AND c) XOR (b AND d) = b AND (c XOR d)

Now we have: (a AND (c XOR d)) XOR (b AND (c XOR d))

We can factor out (c XOR d): (a XOR b) AND (c XOR d)

This pattern generalizes! The key insight is that AND distributes over XOR in a way that allows us to separate the contributions from each array. Instead of computing all pairwise AND operations, we can:

1. First compute the XOR of all elements in arr1
2. Then compute the XOR of all elements in arr2
3. Finally, AND these two XOR results together

This reduces our time complexity from O(n * m) to O(n + m), which is a massive improvement. The mathematical elegance here is that the complex interaction of all pairs simplifies to a single AND operation between two XOR sums.

Learn more about Math patterns.

Solution Approach

Based on the mathematical insight that the XOR sum of all AND operations can be simplified to (XOR of arr1) AND (XOR of arr2), the implementation becomes remarkably straightforward.

Step 1: Calculate XOR sum of arr1 We iterate through all elements in arr1 and XOR them together. This can be done using Python's reduce function with the xor operator:

a = reduce(xor, arr1)

This computes arr1[0] XOR arr1[1] XOR ... XOR arr1[n-1]

Step 2: Calculate XOR sum of arr2 Similarly, we compute the XOR of all elements in arr2:

b = reduce(xor, arr2)

This computes arr2[0] XOR arr2[1] XOR ... XOR arr2[m-1]

Step 3: Apply bitwise AND Finally, we perform a single AND operation between the two XOR results:

return a & b

The complete solution uses Python's reduce function from the functools module along with the xor function from the operator module. The reduce function applies the XOR operation cumulatively to the elements of each array, effectively computing the XOR sum.

Time Complexity: O(n + m) where n and m are the lengths of arr1 and arr2 respectively. We only need to traverse each array once.

Space Complexity: O(1) as we only use two variables to store the XOR sums, regardless of input size.

This elegant solution transforms what could have been an O(n * m) problem with potentially massive memory requirements into a linear-time algorithm with constant space usage, all thanks to the distributive property of Boolean operations.

Example Walkthrough

Let's walk through a concrete example with arr1 = [1, 6, 3] and arr2 = [5, 2].

Naive Approach (for comparison): First, let's see what we're actually computing by listing all pairwise AND operations:

• 1 AND 5 = 1 (binary: 001 AND 101 = 001)
• 1 AND 2 = 0 (binary: 001 AND 010 = 000)
• 6 AND 5 = 4 (binary: 110 AND 101 = 100)
• 6 AND 2 = 2 (binary: 110 AND 010 = 010)
• 3 AND 5 = 1 (binary: 011 AND 101 = 001)
• 3 AND 2 = 2 (binary: 011 AND 010 = 010)

XOR sum of all results: 1 XOR 0 XOR 4 XOR 2 XOR 1 XOR 2

• 1 XOR 0 = 1
• 1 XOR 4 = 5
• 5 XOR 2 = 7
• 7 XOR 1 = 6
• 6 XOR 2 = 4

Result: 4

Optimized Approach: Now let's use our mathematical insight:

Step 1: Calculate XOR sum of arr1

• 1 XOR 6 XOR 3
• 1 XOR 6 = 7 (binary: 001 XOR 110 = 111)
• 7 XOR 3 = 4 (binary: 111 XOR 011 = 100)
• XOR sum of arr1 = 4

Step 2: Calculate XOR sum of arr2

• 5 XOR 2
• 5 XOR 2 = 7 (binary: 101 XOR 010 = 111)
• XOR sum of arr2 = 7

Step 3: Apply bitwise AND

• 4 AND 7 = 4 (binary: 100 AND 111 = 100)
• Final result: 4

Both approaches give us the same answer of 4, confirming our mathematical optimization works! The key difference is that the naive approach required 6 AND operations and 5 XOR operations, while the optimized approach only needed 2 XOR operations and 1 AND operation.

Solution Implementation

Time and Space Complexity

The time complexity is O(n + m), where n is the length of arr1 and m is the length of arr2. This is because:

• The first reduce(xor, arr1) operation iterates through all n elements of arr1, performing XOR operations in O(n) time
• The second reduce(xor, arr2) operation iterates through all m elements of arr2, performing XOR operations in O(m) time
• The final AND operation between a and b takes O(1) time
• Total: O(n) + O(m) + O(1) = O(n + m)

The space complexity is O(1). The algorithm only uses two additional variables a and b to store the XOR results, regardless of the input size. The reduce function operates iteratively without creating additional data structures proportional to the input size.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Attempting the Brute Force Approach

Many developers initially try to solve this problem by computing all possible AND operations and then finding their XOR sum:

# Inefficient approach - DON'T DO THIS
def getXORSum_bruteforce(arr1, arr2):
    result = 0
    for num1 in arr1:
        for num2 in arr2:
            result ^= (num1 & num2)
    return result

Why it's problematic:

• Time complexity becomes O(n × m), which can be extremely slow for large arrays
• With constraints like arrays having up to 10^5 elements, this leads to 10^10 operations
• May cause Time Limit Exceeded (TLE) errors on platforms like LeetCode

Solution: Use the mathematical property that allows simplification to (XOR of arr1) & (XOR of arr2), reducing complexity to O(n + m).

2. Misunderstanding the Order of Operations

Some might incorrectly compute AND first, then XOR:

# WRONG - This doesn't give the correct result
def getXORSum_wrong(arr1, arr2):
    and_arr1 = reduce(lambda x, y: x & y, arr1)  # AND all elements
    and_arr2 = reduce(lambda x, y: x & y, arr2)  # AND all elements
    return and_arr1 ^ and_arr2  # XOR the results

Why it's wrong: The problem asks for XOR of all AND pairs, not AND of all elements then XOR. The correct approach requires XOR-ing within each array first, then AND-ing the results.

Solution: Remember the correct formula: (a1 XOR a2 XOR ... an) AND (b1 XOR b2 XOR ... bm)

3. Edge Case Handling

Not considering edge cases with single-element arrays or when XOR results in 0:

# Potential issue with empty array handling
def getXORSum_edge_case(arr1, arr2):
    if not arr1 or not arr2:  # This check might not be needed
        return 0
    return reduce(xor, arr1) & reduce(xor, arr2)

Why it matters: While the problem states arrays contain non-negative integers (implying non-empty), reduce without an initial value will fail on empty sequences.

Solution: Either validate input constraints or provide an initial value to reduce:

xor_arr1 = reduce(xor, arr1, 0)  # 0 is the identity element for XOR
xor_arr2 = reduce(xor, arr2, 0)

4. Not Recognizing the Mathematical Pattern

Trying to memorize the solution without understanding why it works can lead to errors in similar problems. The key insight is the distributive property: a & (b XOR c) = (a & b) XOR (a & c).

Solution: Understand that this property allows us to factor out the expression, similar to how multiplication distributes over addition in regular algebra. This pattern appears in other bit manipulation problems as well.