Description#
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 <= 121 <= nums[i] <= 20
Solutions#
Solution 1: Binary Enumeration#
We can use binary enumeration to enumerate all subsets, and then calculate the XOR sum of each subset.
Specifically, we enumerate $i$ in the range $[0, 2^n)$, where $n$ is the length of the array $nums$. If the $j$th bit of the binary representation of $i$ is $1$, it means that the $j$th element of $nums$ is in the current subset; if the $j$th bit is $0$, it means that the $j$th element of $nums$ is not in the current subset. We can get the XOR sum of the current subset according to the binary representation of $i$, and add it to the answer.
The time complexity is $O(n \times 2^n)$, where $n$ is the length of the array $nums$. The space complexity is $O(1)$.
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| 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
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| 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;
}
}
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| class Solution {
public:
int subsetXORSum(vector<int>& nums) {
int n = nums.size();
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) {
s ^= nums[j];
}
}
ans += s;
}
return ans;
}
};
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| 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
}
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| 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;
}
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| /**
* @param {number[]} nums
* @return {number}
*/
var subsetXORSum = function (nums) {
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;
};
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Solution 2: DFS (Depth-First Search)#
We can also use depth-first search to enumerate all subsets, and then calculate the XOR sum of each subset.
We design a function $dfs(i, s)$, where $i$ represents the current search to the $i$th element of the array $nums$, and $s$ represents the XOR sum of the current subset. Initially, $i=0$, $s=0$. During the search, we have two choices each time:
- Add the $i$th element of $nums$ to the current subset, i.e., $dfs(i+1, s \oplus nums[i])$;
- Do not add the $i$th element of $nums$ to the current subset, i.e., $dfs(i+1, s)$.
When we have searched all elements of the array $nums$, i.e., $i=n$, the XOR sum of the current subset is $s$, and we can add it to the answer.
The time complexity is $O(2^n)$, and the space complexity is $O(n)$. Where $n$ is the length of the array $nums$.
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| class Solution:
def subsetXORSum(self, nums: List[int]) -> int:
def dfs(i: int, s: int):
nonlocal ans
if i >= len(nums):
ans += s
return
dfs(i + 1, s)
dfs(i + 1, s ^ nums[i])
ans = 0
dfs(0, 0)
return ans
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| class Solution {
private int ans;
private int[] nums;
public int subsetXORSum(int[] nums) {
this.nums = nums;
dfs(0, 0);
return ans;
}
private void dfs(int i, int s) {
if (i >= nums.length) {
ans += s;
return;
}
dfs(i + 1, s);
dfs(i + 1, s ^ nums[i]);
}
}
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| class Solution {
public:
int subsetXORSum(vector<int>& nums) {
int n = nums.size();
int ans = 0;
function<void(int, int)> dfs = [&](int i, int s) {
if (i >= n) {
ans += s;
return;
}
dfs(i + 1, s);
dfs(i + 1, s ^ nums[i]);
};
dfs(0, 0);
return ans;
}
};
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| func subsetXORSum(nums []int) (ans int) {
n := len(nums)
var dfs func(int, int)
dfs = func(i, s int) {
if i >= n {
ans += s
return
}
dfs(i+1, s)
dfs(i+1, s^nums[i])
}
dfs(0, 0)
return
}
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| function subsetXORSum(nums: number[]): number {
let ans = 0;
const n = nums.length;
const dfs = (i: number, s: number) => {
if (i >= n) {
ans += s;
return;
}
dfs(i + 1, s);
dfs(i + 1, s ^ nums[i]);
};
dfs(0, 0);
return ans;
}
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| /**
* @param {number[]} nums
* @return {number}
*/
var subsetXORSum = function (nums) {
let ans = 0;
const n = nums.length;
const dfs = (i, s) => {
if (i >= n) {
ans += s;
return;
}
dfs(i + 1, s);
dfs(i + 1, s ^ nums[i]);
};
dfs(0, 0);
return ans;
};
|
