leetcode1863: Sum of All Subset XOR Totals
carloscn opened this issue · comments
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
Analysis
pub fn subset_xor_sum(nums: Vec<i32>) -> i32
{
let len:usize = nums.len();
if len < 1 {
return 0;
}
let mut ret:i32 = 0;
let (mut num, mut n, mut j, mut k) =
(1 << len, 0, 0, 0);
for i in 0..num {
j = i;
k = 0 as usize;
n = 0 as usize;
let mut t_buf:Vec<i32> = vec![0; i];
while j != 0 {
if j & 1 == 1 {
t_buf[n] = nums[k];
n += 1;
}
j >>= 1;
k += 1;
}
let mut xor_sum:i32 = 0;
for e in t_buf {
xor_sum ^= e;
}
ret += xor_sum;
}
return ret;
}