Sum of All Subset XOR Totals - Practice Coding Problems

Sum of All Subset XOR Totals - Problem

Array Math Backtracking Bit Manipulation Combinatorics Enumeration Easy

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.

Input & Output

Example 1 — Basic Case

$ Input: nums = [1,3]

› Output: 6

💡 Note: Subsets: [] (XOR=0), [1] (XOR=1), [3] (XOR=3), [1,3] (XOR=1⊕3=2). Sum: 0+1+3+2 = 6

Example 2 — Three Elements

$ Input: nums = [5,1,6]

› Output: 28

💡 Note: 8 subsets: [] (0), [5] (5), [1] (1), [6] (6), [5,1] (4), [5,6] (3), [1,6] (7), [5,1,6] (2). Sum: 0+5+1+6+4+3+7+2 = 28

Example 3 — Single Element

$ Input: nums = [3]

› Output: 3

💡 Note: Only 2 subsets: [] (XOR=0) and [3] (XOR=3). Sum: 0+3 = 3

Constraints

1 ≤ nums.length ≤ 12
1 ≤ nums[i] ≤ 20

Visualization

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Understanding the Visualization

Input

Array [5,1,6] with 3 elements

Generate Subsets

Create all 2³ = 8 possible subsets

Calculate XOR

Find XOR total for each subset

Sum All

Add all XOR totals: 0+5+1+6+4+3+7+2 = 28

Key Takeaway

🎯 Key Insight: Generate all possible subsets and calculate the XOR total for each, then sum them all up

Asked in

a Amazon 15 M Microsoft 10

The key insight is that we need to calculate the XOR total for every possible subset and sum them up. The brute force approach generates all 2^n subsets explicitly. The backtracking approach uses recursion to explore include/exclude decisions. Best approach for understanding is backtracking with Time: O(2^n), Space: O(n).

Common Approaches

Approach	Time	Space	Notes
✓ Backtracking Recursion	O(2^n)	O(n)	Use recursion to explore include/exclude decisions for each element
Bit Manipulation Optimization	O(n)	O(1)	Each element appears in exactly 2^(n-1) subsets, contributing predictably to the sum
Brute Force - Generate All Subsets	O(n × 2^n)	O(1)	Generate all 2^n subsets, calculate XOR total for each, sum them up

Backtracking Recursion — Algorithm Steps

For each element, make include/exclude decision
Recursively process remaining elements
Base case: no more elements, add current XOR to sum

Visualization

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Step-by-Step Walkthrough

Decision

For each element: include it or exclude it

Recurse

Make recursive calls for remaining elements

Sum

Add up all XOR totals from leaf nodes

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int backtrack(int* nums, int n, int index, int currentXor) {
    if (index == n) {
        return currentXor;
    }
    
    // Include current element
    int include = backtrack(nums, n, index + 1, currentXor ^ nums[index]);
    // Exclude current element
    int exclude = backtrack(nums, n, index + 1, currentXor);
    
    return include + exclude;
}

int solution(int* nums, int n) {
    return backtrack(nums, n, 0, 0);
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    
    // Parse array manually
    int nums[20];
    int n = 0;
    char* token = strtok(line, "[,]");
    while (token != NULL) {
        nums[n++] = atoi(token);
        token = strtok(NULL, "[,]");
    }
    
    int result = solution(nums, n);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(2^n)

Each element has 2 choices (include/exclude), creating 2^n subsets

⚠ Quadratic Growth

Space Complexity

O(n)

Recursion call stack depth is n levels deep

⚡ Linearithmic Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Backtracking Recursion — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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