Question

1863. Sum of All Subset XOR Totals

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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 <= 12
• 1 <= 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 $O(1)$.

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

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;
  }
}

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;
  }
};

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
}

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;
}

/**
 * @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;
};

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 $nums$.

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

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]);
  }
}

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;
  }
};

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
}

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;
}

/**
 * @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;
};