Problem Description

You are given a 0-indexed integer array nums.

For any three indices i, j, and k (where 0 <= i, j, k < n), you need to calculate an effective value using the formula: ((nums[i] | nums[j]) & nums[k]).

Here:

• | represents the bitwise OR operation
• & represents the bitwise AND operation

The xor-beauty of the array is calculated by taking the XOR of all possible effective values. This means you need to:

1. Generate all possible triplets (i, j, k) where each index can range from 0 to n-1
2. Calculate the effective value for each triplet
3. XOR all these effective values together

Your task is to return the final xor-beauty value.

For example, if nums = [a, b], you would have 8 possible triplets:

• (0,0,0), (0,0,1), (0,1,0), (0,1,1), (1,0,0), (1,0,1), (1,1,0), (1,1,1)

Each triplet produces an effective value, and the XOR of all these values gives you the xor-beauty.

The solution reveals an interesting mathematical property: despite the complex definition involving all triplets and bitwise operations, the xor-beauty simplifies to just the XOR of all elements in the array. This happens because when you XOR all the effective values, most terms cancel out due to the properties of XOR (where a XOR a = 0), leaving only the XOR of the original array elements.

Intuition

Let's think about what happens when we expand all possible triplets and their effective values.

For any triplet (i, j, k), we compute ((nums[i] | nums[j]) & nums[k]). If we consider all possible combinations:

1. When i = j = k, the effective value becomes ((nums[i] | nums[i]) & nums[i]) = (nums[i] & nums[i]) = nums[i].

2. When i = j ≠ k, the effective value is ((nums[i] | nums[i]) & nums[k]) = (nums[i] & nums[k]).

3. When i ≠ j but one of them equals k, we get expressions like ((nums[i] | nums[j]) & nums[i]) or ((nums[i] | nums[j]) & nums[j]).

4. When all three indices are different, we get ((nums[i] | nums[j]) & nums[k]).

The key insight comes from recognizing that XOR has a special property: a XOR a = 0. This means that any value that appears an even number of times in our XOR calculation will cancel out.

Let's analyze the frequency of different types of terms:

• Terms like (nums[i] & nums[j]) where i ≠ j appear multiple times across different triplet orderings. For instance, (i, i, j), (i, j, i), (j, i, i), (j, j, i), (j, i, j), and (i, j, j) all contribute similar terms.
• Due to the symmetry in how we generate triplets, most complex terms appear an even number of times.

When we XOR all these effective values together, the terms that appear an even number of times cancel out to 0. What remains are the terms where i = j = k, which gives us nums[i] for each index i.

Since each nums[i] (where all three indices are the same) appears exactly once in our XOR calculation, the final result is simply nums[0] XOR nums[1] XOR ... XOR nums[n-1].

This elegant simplification means that despite the seemingly complex definition involving all possible triplets and multiple bitwise operations, the xor-beauty reduces to just the XOR of all elements in the array.

Learn more about Math patterns.

Solution Approach

Based on our mathematical analysis that shows the xor-beauty simplifies to the XOR of all array elements, the implementation becomes remarkably straightforward.

The solution uses Python's reduce function from the functools module along with the xor operator from the operator module to compute the XOR of all elements in the array.

Here's how the implementation works:

1. Function Used: reduce(xor, nums)

   • reduce applies a function cumulatively to the items of an iterable, from left to right
   • xor is the bitwise XOR operation

2. Step-by-step execution:

   • Start with the first element of nums
   • XOR it with the second element
   • Take that result and XOR it with the third element
   • Continue this process until all elements are processed

For example, if nums = [a, b, c, d], the computation would be:

• Step 1: a XOR b → result1
• Step 2: result1 XOR c → result2
• Step 3: result2 XOR d → final result

This is equivalent to computing a XOR b XOR c XOR d.

Time Complexity: O(n) where n is the length of the array, as we iterate through each element exactly once.

Space Complexity: O(1) as we only use a constant amount of extra space for the reduction operation.

The elegance of this solution lies in recognizing that what appears to be a complex O(n³) problem (due to considering all triplets) can be reduced to a simple O(n) XOR operation through mathematical analysis and understanding of XOR properties.

Example Walkthrough

Let's walk through a small example with nums = [2, 3] to understand how the xor-beauty calculation works and why it simplifies.

Step 1: List all possible triplets With 2 elements and indices 0 and 1, we have 2³ = 8 triplets:

• (0,0,0), (0,0,1), (0,1,0), (0,1,1)
• (1,0,0), (1,0,1), (1,1,0), (1,1,1)

Step 2: Calculate effective values for each triplet Remember the formula: ((nums[i] | nums[j]) & nums[k])

Let's compute each one (with nums[0]=2=10₂ and nums[1]=3=11₂):

• (0,0,0): ((2 | 2) & 2) = (2 & 2) = 2
• (0,0,1): ((2 | 2) & 3) = (2 & 3) = 2
• (0,1,0): ((2 | 3) & 2) = (3 & 2) = 2
• (0,1,1): ((2 | 3) & 3) = (3 & 3) = 3
• (1,0,0): ((3 | 2) & 2) = (3 & 2) = 2
• (1,0,1): ((3 | 2) & 3) = (3 & 3) = 3
• (1,1,0): ((3 | 3) & 2) = (3 & 2) = 2
• (1,1,1): ((3 | 3) & 3) = (3 & 3) = 3

Step 3: XOR all effective values We have: 2, 2, 2, 3, 2, 3, 2, 3

Let's XOR them step by step:

• 2 XOR 2 = 0
• 0 XOR 2 = 2
• 2 XOR 3 = 1
• 1 XOR 2 = 3
• 3 XOR 3 = 0
• 0 XOR 2 = 2
• 2 XOR 3 = 1

Result: 1

Step 4: Verify with the simplified formula According to our analysis, the xor-beauty should equal nums[0] XOR nums[1]:

• 2 XOR 3 = 1 ✓

The result matches! This demonstrates how all the complex triplet calculations ultimately cancel out, leaving us with just the XOR of the original array elements.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n) where n is the length of the input array nums. The reduce function with xor operation iterates through all elements in the array exactly once, performing a constant-time XOR operation for each element.

Space Complexity: O(1). The reduce function only uses a constant amount of extra space to maintain the accumulator value during the XOR operations. No additional data structures are created that depend on the input size.

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

Common Pitfalls

1. Attempting Brute Force Implementation

The most common pitfall is trying to implement the problem exactly as described - calculating all n³ triplets and their effective values. This leads to unnecessary complexity and performance issues.

Incorrect Approach:

def xorBeauty(self, nums: List[int]) -> int:
    n = len(nums)
    result = 0
    # This works but is O(n³) - unnecessarily slow!
    for i in range(n):
        for j in range(n):
            for k in range(n):
                effective = (nums[i] | nums[j]) & nums[k]
                result ^= effective
    return result

Why it's problematic: While this produces the correct answer, it has O(n³) time complexity, which will cause Time Limit Exceeded (TLE) for large inputs when the optimal solution is O(n).

2. Misunderstanding the Mathematical Simplification

Some might think the simplification only applies to specific cases (like when i=j=k) and try to optimize by only considering those cases.

Incorrect Approach:

def xorBeauty(self, nums: List[int]) -> int:
    result = 0
    # Wrong: Only considering diagonal elements
    for i in range(len(nums)):
        result ^= nums[i]  # Only when i=j=k
    # Missing the mathematical proof that OTHER terms cancel out
    return result

Why it's problematic: While this gives the correct answer, the reasoning is incomplete. Without understanding WHY other terms cancel out, you might doubt the solution or apply it incorrectly to similar problems.

3. Edge Case Handling

Forgetting to handle edge cases or assuming the array has specific properties.

Potential Issues:

• Empty array handling (though constraints typically guarantee non-empty arrays)
• Single element arrays
• Arrays with duplicate values

Solution: The reduce approach naturally handles all these cases:

def xorBeauty(self, nums: List[int]) -> int:
    if not nums:  # Handle empty array if needed
        return 0
    return reduce(xor, nums)

4. Import Statement Omission

Forgetting to import necessary modules when using reduce and xor.

Incorrect:

class Solution:
    def xorBeauty(self, nums: List[int]) -> int:
        # This will cause NameError without imports!
        return reduce(xor, nums)

Correct:

from functools import reduce
from operator import xor

class Solution:
    def xorBeauty(self, nums: List[int]) -> int:
        return reduce(xor, nums)

5. Manual XOR Implementation Errors

When implementing XOR manually without reduce, common mistakes include:

Incorrect:

def xorBeauty(self, nums: List[int]) -> int:
    result = nums[0]  # Wrong: crashes on empty array
    for i in range(1, len(nums)):
        result ^= nums[i]
    return result

Correct:

def xorBeauty(self, nums: List[int]) -> int:
    result = 0  # Start with 0 (XOR identity)
    for num in nums:
        result ^= num
    return result

The key insight is recognizing that 0 is the identity element for XOR (a ^ 0 = a), so we should initialize with 0, not the first element.