LeetCode 1863. Sum of All Subset XOR Totals

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

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这题是要求一个数组的所有sublist里所有元素的XOR之和。第一反应就是先把所有的sublist求出来，然后再遍历一次求出XOR和sum，应该是有点排列组合那味儿？（不确定，以后再回来revisit。）反正也是不会做，就看了答案，其实是可以一边backtrack一边求XOR的。具体看的是这篇，讲的还是挺直观的但是我懒得理解那么深入了（可能也是智商限制无法研究那么深入）：- LeetCode

大意呢就是对于每个元素，我们都有取它和不取它的两种情况。我们在生成sublist的时候就可以进行XOR的计算了，我们计算取当前元素和不取当前元素两种情况的sublist XOR值，然后再把两个加起来，递归调用这个backtrack函数，直到当前已经遍历到了数组里的最后一个元素就可以停下了。这里的backtrack写的还是有返回值的，这样就可以每次都更新返回值，不需要用mutable variable in parameter了。

大概算是默写了一下答案：

class Solution {
    public int subsetXORSum(int[] nums) {
        return backtrack(nums, 0, 0);
    }

    private int backtrack(int[] nums, int index, int currXOR) {
        if (index == nums.length) {
            return currXOR;
        }
        int withElement = backtrack(nums, index + 1, currXOR ^ nums[index]);
        int withoutElement = backtrack(nums, index + 1, currXOR);
        return withElement + withoutElement;
    }
}

有人在底下评论说这个看起来不是那么显然的backtrack，其实在计算with/withoutElement的时候把nums[index]加入和移除currXOR就相当于是在currXORArray（需要进行XOR操作的数组）里add和remove了。

还找了一个看起来比较backtrack的做法，等哪天真正要来深入理解backtrack的时候再回来看看吧：- LeetCode

太难了，我的智商不足以支撑我掌握backtrack……