Problem Description

You have an array of integers nums (0-indexed) and a positive integer k. Your goal is to make the XOR of all elements in the array equal to k by performing bit flip operations.

A bit flip operation allows you to:

• Select any element from the array
• Choose any bit position in that element's binary representation
• Flip that bit (change 0 to 1 or 1 to 0)

You can perform this operation as many times as needed on any elements. Leading zeros in binary representations can also be flipped - for example, the number (101)₂ can have its fourth bit flipped to become (1101)₂.

The task is to find the minimum number of operations needed to achieve the target XOR value of k.

Example understanding:

• If you have an array like [2, 1, 3] and k = 5
• The current XOR is 2 ⊕ 1 ⊕ 3 = 0
• You need to flip certain bits in the array elements so that the final XOR equals 5
• The solution counts how many individual bit flips are needed

The key insight is that XORing all elements gives you a result, and comparing this result with k bit by bit tells you exactly which bits need to be different - each differing bit requires exactly one flip operation somewhere in the array.

Intuition

Let's think about what happens when we XOR all elements in the array. If we have nums = [a, b, c], then the current XOR is a ⊕ b ⊕ c = current_result.

We want this current_result to equal k. The question becomes: what's the minimum number of bit flips needed to transform current_result into k?

Here's the key observation: when we flip a bit in any element of the array, that bit flip changes exactly one bit in the final XOR result. Why? Because of the properties of XOR:

• If we flip bit position i in element a, the new XOR becomes (a with bit i flipped) ⊕ b ⊕ c
• This is equivalent to a ⊕ b ⊕ c ⊕ (1 << i) due to XOR properties
• So flipping one bit in any element flips exactly that bit position in the final XOR result

This means the problem reduces to: "How many bits are different between current_result and k?"

To find which bits are different, we can XOR current_result with k. This gives us a number where:

• Each 1 bit represents a position where current_result and k differ
• Each 0 bit represents a position where they're already the same

For example, if current_result = 6 (110₂) and k = 3 (011₂), then 6 ⊕ 3 = 5 (101₂). This tells us bits at positions 0 and 2 are different, so we need 2 flips.

The minimum number of operations is simply the count of 1 bits in current_result ⊕ k. Since current_result = nums[0] ⊕ nums[1] ⊕ ... ⊕ nums[n-1], we can directly compute nums[0] ⊕ nums[1] ⊕ ... ⊕ nums[n-1] ⊕ k and count the 1 bits.

Solution Approach

The implementation is elegantly simple once we understand the underlying principle. We need to:

1. Calculate the XOR of all elements with k: This gives us a value where each 1 bit represents a position that needs to be flipped.

2. Count the number of 1 bits: This count is our answer.

The Python solution uses the reduce function with the XOR operator to accomplish this in one line:

class Solution:
    def minOperations(self, nums: List[int], k: int) -> int:
        return reduce(xor, nums, k).bit_count()

Let's break down how this works:

Step 1: XOR Reduction

• reduce(xor, nums, k) starts with initial value k
• It then XORs each element in nums with the accumulated result
• This is equivalent to: k ⊕ nums[0] ⊕ nums[1] ⊕ ... ⊕ nums[n-1]
• Due to XOR's associative and commutative properties, this equals (nums[0] ⊕ nums[1] ⊕ ... ⊕ nums[n-1]) ⊕ k

Step 2: Bit Counting

• .bit_count() is a Python method that counts the number of 1 bits in the binary representation
• This gives us the exact number of bit positions where the current XOR differs from k

Example walkthrough:

• If nums = [2, 1, 3, 4] and k = 1
• Current XOR: 2 ⊕ 1 ⊕ 3 ⊕ 4 = 4
• Difference: 4 ⊕ 1 = 5 = (101)₂
• Number of 1 bits in 5 is 2
• Therefore, we need 2 bit flip operations

The time complexity is O(n) where n is the length of the array (we process each element once). The space complexity is O(1) as we only use a constant amount of extra space.

Example Walkthrough

Let's walk through a concrete example with nums = [2, 7, 1] and k = 6.

Step 1: Calculate current XOR of all elements

• Binary representations: 2 = 010₂, 7 = 111₂, 1 = 001₂
• Current XOR: 2 ⊕ 7 ⊕ 1
  • 2 ⊕ 7 = 010 ⊕ 111 = 101 = 5
  • 5 ⊕ 1 = 101 ⊕ 001 = 100 = 4
• So current XOR = 4

Step 2: Find which bits need to change

• We have current XOR = 4 = 100₂
• We want XOR = k = 6 = 110₂
• To find differences: 4 ⊕ 6 = 100 ⊕ 110 = 010₂ = 2

Step 3: Count the '1' bits in the difference

• The difference 2 = 010₂ has exactly one '1' bit (at position 1)
• This means we need exactly 1 bit flip operation

Step 4: Verify with an actual flip

• We could flip bit 1 of element 2 (changing 010₂ to 000₂ = 0)
• New array would be [0, 7, 1]
• New XOR: 0 ⊕ 7 ⊕ 1 = 111 ⊕ 001 = 110 = 6 ✓

Using the solution approach:

reduce(xor, [2, 7, 1], 6)
= 6 ⊕ 2 ⊕ 7 ⊕ 1
= ((6 ⊕ 2) ⊕ 7) ⊕ 1
= (4 ⊕ 7) ⊕ 1
= 3 ⊕ 1
= 2

2.bit_count() = 1  # Since 2 = 010₂ has one '1' bit

The answer is 1 operation.

Solution Implementation

Time and Space Complexity

The time complexity is O(n), where n is the length of the array nums. This is because the reduce function with xor operation iterates through all elements in the array exactly once, performing a constant-time XOR operation for each element. The bit_count() method then counts the number of set bits in the final XOR result, which takes O(1) time since it operates on a single integer with a fixed maximum number of bits.

The space complexity is O(1). The reduce function only maintains a single accumulator value throughout the iteration, using constant extra space regardless of the input size. The bit_count() operation also uses constant space.

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

Common Pitfalls

1. Misunderstanding the XOR Operation Direction

A common mistake is thinking you need to XOR the array first, then XOR with k separately, leading to code like:

# Incorrect approach
array_xor = reduce(xor, nums)
result = array_xor ^ k
return result.bit_count()

Why it's wrong: This performs unnecessary steps and can be confusing. The reduce(xor, nums, k) with k as the initial value is more elegant and direct.

Solution: Use reduce(xor, nums, k) which starts with k and XORs through all elements in one pass.

2. Manual Bit Counting Instead of Using Built-in Methods

Some might implement bit counting manually:

# Inefficient approach
xor_result = reduce(xor, nums, k)
count = 0
while xor_result:
    count += xor_result & 1
    xor_result >>= 1
return count

Why it's problematic: While correct, it's slower and more error-prone than using Python's optimized bit_count() method (available in Python 3.10+).

Solution: Use .bit_count() for cleaner, faster code. For older Python versions, use bin(xor_result).count('1').

3. Overthinking the Problem

Some might try to track which specific elements need bit flips:

# Overcomplication
operations = 0
current_xor = reduce(xor, nums)
diff = current_xor ^ k
for i in range(32):  # Assuming 32-bit integers
    if diff & (1 << i):
        # Try to find which element to flip...
        operations += 1

Why it's wrong: The problem only asks for the minimum number of operations, not which elements to modify. The XOR difference directly tells us how many bits differ.

Solution: Focus on the XOR difference between the current result and target - the number of differing bits is the answer.

4. Forgetting Edge Cases

Not handling when the array already has the desired XOR:

# Missing edge case consideration
def minOperations(self, nums: List[int], k: int) -> int:
    return reduce(xor, nums, k).bit_count()
    # This actually handles it correctly! When XOR equals k, 
    # the result is 0, and bit_count() returns 0

Why it matters: While the given solution handles this automatically (returns 0 when no operations needed), being aware of edge cases helps in understanding and testing.

Solution: The provided implementation naturally handles all cases, including when no operations are needed.