Problem Description

You need to find the sum of XOR values of all possible subsets of a given array.

The XOR total of an array is calculated by taking the bitwise XOR of all elements in that array. If the array is empty, the XOR total is 0.

For example, if you have the array [2, 5, 6], the XOR total would be 2 XOR 5 XOR 6 = 1.

Given an array nums, you must:

1. Find all possible subsets of the array (including the empty subset and the array itself)
2. Calculate the XOR total for each subset
3. Return the sum of all these XOR totals

A subset is formed by selecting zero or more elements from the original array. The order doesn't matter, and you can obtain a subset by removing some (or no) elements from the original array.

For instance, if nums = [1, 3]:

• The subsets are: [], [1], [3], [1, 3]
• The XOR totals are: 0, 1, 3, 1 XOR 3 = 2
• The sum of all XOR totals is: 0 + 1 + 3 + 2 = 6

The solution uses binary enumeration to generate all subsets. For an array of length n, there are 2^n possible subsets. Each number from 0 to 2^n - 1 represents a unique subset, where the binary representation indicates which elements to include: if bit j is 1, include nums[j]; if bit j is 0, exclude nums[j].

Flowchart Walkthrough

First, let's pin down the algorithm using the Flowchart. Here's a step-by-step walkthrough:

Is it a graph?

• No: The problem involves an array and its subsets, not nodes and edges as in graph problems.

Need to solve for kth smallest/largest?

• No: We're not looking for the kth smallest or largest element. We need to find the sum of XOR totals for all subsets.

Involves Linked Lists?

• No: The problem works with arrays, not linked lists.

Does the problem have small constraints?

• Yes: Although not explicitly stated in the problem, subset generation problems typically have small constraints since the number of subsets grows exponentially (2^n). For an array of size n, we have 2^n subsets, which becomes computationally intensive for large n.

Brute force / Backtracking?

• Yes: We need to explore all possible subsets of the array. This is a classic enumeration problem where we must generate every subset, calculate its XOR total, and sum all the results. Backtracking is the natural approach to systematically generate all subsets by making decisions to include or exclude each element.

Conclusion: The flowchart correctly leads us to use a Backtracking approach. We need to enumerate all possible subsets (which can be done through backtracking or binary enumeration), calculate the XOR total for each subset, and sum all these totals together.

Intuition

When we need to find all subsets of an array, we're essentially making a series of binary decisions: for each element, we either include it in our subset or we don't. This is fundamentally a backtracking problem where we explore all possible combinations.

Think of it like this: if we have an array [a, b, c], for each element we have 2 choices (include/exclude). This gives us 2 × 2 × 2 = 2^3 = 8 total subsets. We can visualize this as a decision tree where at each level we decide whether to include the current element.

The key insight is that we can represent each subset using a binary number. For an array of length n, we can use numbers from 0 to 2^n - 1 where each bit position tells us whether to include that element. For example, with [a, b, c]:

• 000 (0 in decimal) = [] (empty subset)
• 001 (1 in decimal) = [c] (only last element)
• 010 (2 in decimal) = [b] (only middle element)
• 101 (5 in decimal) = [a, c] (first and last elements)
• 111 (7 in decimal) = [a, b, c] (all elements)

This binary representation naturally maps to our backtracking approach. We iterate through all possible binary numbers, and for each number, we check each bit position. If bit j is set to 1, we include nums[j] in our current subset. We then calculate the XOR of all included elements and add it to our running sum.

The elegance of this approach is that it systematically generates all subsets without explicitly using recursion, making it both intuitive and efficient for problems with small constraints.

Learn more about Math, Backtracking and Combinatorics patterns.

Solution Approach

The solution uses binary enumeration to generate all possible subsets. Here's how the implementation works:

1. Iterate through all possible subsets: We use a loop from 0 to 2^n - 1 (written as 1 << n in the code), where n is the length of the array. Each number i in this range represents a unique subset.

for i in range(1 << n):  # Loop from 0 to 2^n - 1

2. For each subset representation i:

• Initialize a variable s to store the XOR sum of the current subset
• Check each bit position j from 0 to n-1

3. Determine which elements to include: For each bit position j, we check if the j-th bit of i is set using the expression i >> j & 1:

• i >> j shifts i right by j positions
• & 1 checks if the least significant bit is 1
• If the result is 1, we include nums[j] in our subset

for j in range(n):
    if i >> j & 1:  # Check if j-th bit is set
        s ^= nums[j]  # XOR with current element

4. Accumulate the result: After calculating the XOR sum s for the current subset, we add it to our answer:

ans += s

Example walkthrough with nums = [1, 3]:

• i = 0 (binary: 00): subset = [], XOR = 0
• i = 1 (binary: 01): subset = [1], XOR = 1
• i = 2 (binary: 10): subset = [3], XOR = 3
• i = 3 (binary: 11): subset = [1, 3], XOR = 1 ^ 3 = 2
• Total sum = 0 + 1 + 3 + 2 = 6

The time complexity is O(2^n × n) where we iterate through 2^n subsets and for each subset, we check n bit positions. The space complexity is O(1) as we only use a constant amount of extra space.

Example Walkthrough

Let's walk through the solution with nums = [5, 1, 6]:

Step 1: Setup

• Array length n = 3
• Total subsets = 2^3 = 8 (represented by numbers 0 to 7)
• Initialize ans = 0 to accumulate our result

Step 2: Process each subset

For i = 0 (binary: 000):

• No bits are set, so we include no elements
• Subset: []
• XOR total: 0
• Update: ans = 0 + 0 = 0

For i = 1 (binary: 001):

• Bit 0 is set → include nums[0] = 5
• Subset: [5]
• XOR total: 5
• Update: ans = 0 + 5 = 5

For i = 2 (binary: 010):

• Bit 1 is set → include nums[1] = 1
• Subset: [1]
• XOR total: 1
• Update: ans = 5 + 1 = 6

For i = 3 (binary: 011):

• Bits 0 and 1 are set → include nums[0] = 5 and nums[1] = 1
• Subset: [5, 1]
• XOR total: 5 ^ 1 = 4
• Update: ans = 6 + 4 = 10

For i = 4 (binary: 100):

• Bit 2 is set → include nums[2] = 6
• Subset: [6]
• XOR total: 6
• Update: ans = 10 + 6 = 16

For i = 5 (binary: 101):

• Bits 0 and 2 are set → include nums[0] = 5 and nums[2] = 6
• Subset: [5, 6]
• XOR total: 5 ^ 6 = 3
• Update: ans = 16 + 3 = 19

For i = 6 (binary: 110):

• Bits 1 and 2 are set → include nums[1] = 1 and nums[2] = 6
• Subset: [1, 6]
• XOR total: 1 ^ 6 = 7
• Update: ans = 19 + 7 = 26

For i = 7 (binary: 111):

• All bits are set → include all elements
• Subset: [5, 1, 6]
• XOR total: 5 ^ 1 ^ 6 = 2
• Update: ans = 26 + 2 = 28

Step 3: Return the result

• Final answer: 28

The binary representation directly maps to subset selection: each bit position corresponds to an array index, and a 1 means "include this element." This systematic approach ensures we generate all possible subsets exactly once.

Solution Implementation

Time and Space Complexity

The time complexity is O(n × 2^n), where n is the length of the array nums. This is because:

• The outer loop iterates through all possible subsets, which is 2^n iterations (from 0 to 2^n - 1)
• For each subset, the inner loop iterates through all n bits to check which elements should be included in the current subset
• Therefore, the total operations are 2^n × n

The space complexity is O(1). The algorithm only uses a constant amount of extra space:

• Variable ans to store the running sum
• Variable n to store the array length
• Variable i for the outer loop counter
• Variable s to store the XOR sum of each subset
• Variable j for the inner loop counter

These variables use constant space regardless of the input size, hence the space complexity is O(1).

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

Common Pitfalls

1. Incorrect Bit Manipulation Order of Operations

A common mistake is writing the bit check condition incorrectly due to operator precedence confusion:

Incorrect:

if subset_mask >> bit_position & 1:  # This works but can be confusing

Why it's problematic: While this works in Python because >> has higher precedence than &, it can lead to bugs when translating to other languages or when developers add additional operations. In some languages, the precedence might differ, leading to unexpected behavior.

Better approach:

if (subset_mask >> bit_position) & 1:  # Explicit parentheses for clarity

2. Off-by-One Error in Subset Generation

Developers might mistakenly iterate through wrong range:

Incorrect:

for subset_mask in range(2 ** n):  # Using 2**n instead of 1<<n

Why it's problematic: While 2**n and 1<<n produce the same result, using power operator ** is less efficient for large values and inconsistent with the bit manipulation theme of the solution.

Correct approach:

for subset_mask in range(1 << n):  # Consistent bit operation

3. XOR Initialization Error

Incorrect:

xor_value = None  # or not initializing at all

Why it's problematic: XOR operations require a starting value. The identity element for XOR is 0 (since x ^ 0 = x).

Correct approach:

xor_value = 0  # Proper initialization for XOR operations

4. Misunderstanding Empty Subset Handling

Some might try to skip the empty subset:

Incorrect:

for subset_mask in range(1, 1 << n):  # Starting from 1 instead of 0

Why it's problematic: The problem explicitly states that the empty subset should be included. When subset_mask = 0, no bits are set, representing the empty subset with XOR value of 0.

Correct approach:

for subset_mask in range(1 << n):  # Include 0 for empty subset

5. Integer Overflow in Other Languages

While not an issue in Python, when implementing in languages like Java or C++:

Problematic in C++/Java:

for (int subset_mask = 0; subset_mask < (1 << n); subset_mask++)

Why it's problematic: If n = 32 or larger, 1 << n causes integer overflow.

Solution for other languages:

// Check n bounds first
if (n >= 31) {
    // Handle large n separately or use long long
}
// Or use condition differently
for (int subset_mask = 0; subset_mask < (1 << min(n, 30)); subset_mask++)