Mastering XOR Operations: Solving Leetcode 1863 - Sum of All Subset XOR Totals

Although listed as an easy problem, LeetCode's problem 1863, "Sum of All Subset XOR Totals," was, in my opinion, the hardest problem I've tackled to date. The challenge requires a deep understanding of binary operations and XOR totals. There is prerequisite knowledge expected, such as familiarity with binary numbers, XOR operations, and the XOR truth table. In this article, we’ll explore the problem, understand the concept of XOR, and walk through a detailed solution using JavaScript. We'll also cover the underlying data structures and analyze the solution's time complexity.

Problem Description:

The problem requires us to find the sum of XOR totals for all subsets of a given array. The XOR total of an array is defined as the bitwise XOR of all its elements, or 0 if the array is empty. For example, the XOR total of the array [2, 5, 6] is 2 XOR 5 XOR 6 = 1. Given an array nums, we need to return the sum of all XOR totals for every subset of nums.

Prerequisite Knowledge: To fully understand this problem, you need to be familiar with the following concepts:

1. Subsets: A subset is a portion of a set. For example, the subsets of [1, 2] include [], [1], [2], [1, 2].
2. Bitwise XOR: XOR (exclusive OR) is a bitwise operation that outputs true or 1 only when the input bits are different. For instance, 5 XOR 1 calculates as follows:

• 5 in binary: 101
• 1 in binary: 001
• Performing XOR: 101 XOR 001 = 100 (which is 4 in decimal).

XOR Total Explanation: The XOR total of an array is the result of performing the XOR operation on all its elements. For example:

1. XOR total of [5, 1, 6]:

• 5 in binary: 101
• 1 in binary: 001
• 6 in binary: 110
• Performing XOR: 101 XOR 001 XOR 110 = 010 (which is 2 in decimal).

Solution Approach

To solve this problem, we use Depth-First Search (DFS) to explore all possible subsets of the array and calculate their XOR totals. Here’s a detailed breakdown of the solution:

Fully Commented JavaScript Solution

/**
 * @param {number[]} nums
 * @return {number}
 */
var subsetXORSum = function(nums) {
    // Initialize a variable to keep track of the total XOR sum of all subsets
    let totalXORSum = 0;

    // Helper function to perform DFS and calculate the XOR sum of subsets
    const dfs = (index, currentXOR) => {
        // Base case: if the index is out of bounds, add the current XOR to the totalXORSum
        if (index === nums.length) {
            totalXORSum += currentXOR;
            return;
        }
        
        // Recursive case 1: Include the current element in the subset and calculate XOR
        dfs(index + 1, currentXOR ^ nums[index]);
        
        // Recursive case 2: Exclude the current element from the subset
        dfs(index + 1, currentXOR);
    };

    // Start the DFS traversal from the 0th index with an initial XOR of 0
    dfs(0, 0);
    
    // Return the total XOR sum of all subsets
    return totalXORSum;
};        

Detailed Explanation of the Solution

Initialization

First, we initialize a variable totalXORSum to keep track of the sum of XOR totals for all subsets.

let totalXORSum = 0;        

DFS Helper Function

We define a helper function dfs that performs DFS on the array nums. This function takes two parameters: index and currentXOR.

• index: Tracks the current position in the array.
• currentXOR: Stores the XOR total of the current subset being explored.

const dfs = (index, currentXOR) => {
    if (index === nums.length) {
        totalXORSum += currentXOR;
        return;
    }
    
    dfs(index + 1, currentXOR ^ nums[index]);
    dfs(index + 1, currentXOR);
};        

1. Base Case: If index equals the length of the array, it means we have processed all elements. Add currentXOR to totalXORSum and return.
2. Recursive Case 1: Include the current element in the subset and calculate the new currentXOR.
3. Recursive Case 2: Exclude the current element from the subset and keep the currentXOR unchanged.

Start DFS Traversal

We start the DFS traversal from the 0th index with an initial currentXOR of 0.

dfs(0, 0);        

Return the Result

Finally, return the totalXORSum, which now contains the sum of XOR totals for all subsets.

return totalXORSum;        

Example Explanation

Consider the array [5, 1, 6]:

• Subsets: [], [5], [1], [6], [5, 1], [5, 6], [1, 6], [5, 1, 6]
• XOR Totals:[]: 0[5]: 5[1]: 1[6]: 6[5, 1]: 5 XOR 1 = 4[5, 6]: 5 XOR 6 = 3[1, 6]: 1 XOR 6 = 7[5, 1, 6]: 5 XOR 1 XOR 6 = 2
• Sum: 0 + 5 + 1 + 6 + 4 + 3 + 7 + 2 = 28

Underlying Data Structures

The main data structure used in this solution is the call stack due to the recursive DFS implementation. The array nums is used to store the input values.

Time Complexity

The time complexity of this solution is O(2n), where n is the number of elements in the array. This is because each element can either be included in or excluded from a subset, resulting in 2n possible subsets. The space complexity is also O(2n) due to the recursion stack.

This problem highlights the power of DFS in exploring all subsets of an array and calculating their XOR totals efficiently. By understanding and leveraging DFS, we can solve complex problems involving subsets and bitwise operations with ease.

Feel free to reach out with any questions or thoughts! Happy coding! ??