LeetCode #1863 — EASY

Sum of All Subset XOR Totals

Build confidence with an intuition-first walkthrough focused on array fundamentals.

Solve on LeetCode

The Problem

Problem Statement

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, return the sum of all XOR totals for every subset of nums.

Note: Subsets with the same elements should be counted multiple times.

An array a is a subset of an array b if a can be obtained from b by deleting some (possibly zero) elements of b.

Example 1:

Input: nums = [1,3]
Output: 6
Explanation: The 4 subsets of [1,3] are:
- The empty subset has an XOR total of 0.
- [1] has an XOR total of 1.
- [3] has an XOR total of 3.
- [1,3] has an XOR total of 1 XOR 3 = 2.
0 + 1 + 3 + 2 = 6

Example 2:

Input: nums = [5,1,6]
Output: 28
Explanation: The 8 subsets of [5,1,6] are:
- The empty subset has an XOR total of 0.
- [5] has an XOR total of 5.
- [1] has an XOR total of 1.
- [6] has an XOR total of 6.
- [5,1] has an XOR total of 5 XOR 1 = 4.
- [5,6] has an XOR total of 5 XOR 6 = 3.
- [1,6] has an XOR total of 1 XOR 6 = 7.
- [5,1,6] has an XOR total of 5 XOR 1 XOR 6 = 2.
0 + 5 + 1 + 6 + 4 + 3 + 7 + 2 = 28

Example 3:

Input: nums = [3,4,5,6,7,8]
Output: 480
Explanation: The sum of all XOR totals for every subset is 480.

Constraints:

1 <= nums.length <= 12
1 <= nums[i] <= 20

Step 01

Brute Force Baseline

Problem summary: 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, return the sum of all XOR totals for every subset of nums. Note: Subsets with the same elements should be counted multiple times. An array a is a subset of an array b if a can be obtained from b by deleting some (possibly zero) elements of b.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Math · Backtracking · Bit Manipulation

Example 1

[1,3]

Example 2

[5,1,6]

Example 3

[3,4,5,6,7,8]

Step 02

Core Insight

What unlocks the optimal approach

Is there a way to iterate through all the subsets of the array?
Can we use recursion to efficiently iterate through all the subsets?

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1863: Sum of All Subset XOR Totals
class Solution {
    public int subsetXORSum(int[] nums) {
        int n = nums.length;
        int ans = 0;
        for (int i = 0; i < 1 << n; ++i) {
            int s = 0;
            for (int j = 0; j < n; ++j) {
                if ((i >> j & 1) == 1) {
                    s ^= nums[j];
                }
            }
            ans += s;
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1863: Sum of All Subset XOR Totals
func subsetXORSum(nums []int) (ans int) {
	n := len(nums)
	for i := 0; i < 1<<n; i++ {
		s := 0
		for j, x := range nums {
			if i>>j&1 == 1 {
				s ^= x
			}
		}
		ans += s
	}
	return
}

# Accepted solution for LeetCode #1863: Sum of All Subset XOR Totals
class Solution:
    def subsetXORSum(self, nums: List[int]) -> int:
        ans, n = 0, len(nums)
        for i in range(1 << n):
            s = 0
            for j in range(n):
                if i >> j & 1:
                    s ^= nums[j]
            ans += s
        return ans

// Accepted solution for LeetCode #1863: Sum of All Subset XOR Totals
impl Solution {
    pub fn subset_xor_sum(nums: Vec<i32>) -> i32 {
        let n = nums.len();
        let mut ans = 0;

        for i in 0..(1 << n) {
            let mut s = 0;
            for j in 0..n {
                if ((i >> j) & 1) == 1 {
                    s ^= nums[j];
                }
            }
            ans += s;
        }

        ans
    }
}

// Accepted solution for LeetCode #1863: Sum of All Subset XOR Totals
function subsetXORSum(nums: number[]): number {
    let ans = 0;
    const n = nums.length;
    for (let i = 0; i < 1 << n; ++i) {
        let s = 0;
        for (let j = 0; j < n; ++j) {
            if ((i >> j) & 1) {
                s ^= nums[j];
            }
        }
        ans += s;
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(2^n)

Space

O(n)

Approach Breakdown

EXHAUSTIVE

O(nⁿ) time

O(n) space

Generate every possible combination without any filtering. At each of n positions we choose from up to n options, giving nⁿ total candidates. Each candidate takes O(n) to validate. No pruning means we waste time on clearly invalid partial solutions.

BACKTRACKING + PRUNING

O(n!) time

O(n) space

Backtracking explores a decision tree, but prunes branches that violate constraints early. Worst case is still factorial or exponential, but pruning dramatically reduces the constant factor in practice. Space is the recursion depth (usually O(n) for n-level decisions).

Shortcut: Backtracking time = size of the pruned search tree. Focus on proving your pruning eliminates most branches.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

Missing undo step on backtrack

Wrong move: Mutable state leaks between branches.

Usually fails on: Later branches inherit selections from earlier branches.

Fix: Always revert state changes immediately after recursive call.