Lonely Integer

Given an integer array where each number occurs twice except for one, find the unique (to rule them all?)

The point:

xor all elements together

Complexity :

Time	Space
O(nlog(n))	O(1)

O(nlog(n)) in time because for each integer from 0 to n, counting the number of bits set is in log(n) => n log(n)
O(1) in space because in place

About Rust :

Basic implementation
YES : tested on the Rust Playground

fn lonely_integer(nums : &[i32]) -> i32{
    let mut res = 0;
    // xor each element of the array
    // duplicate values cancel each other out (x^x=0)
    for num in nums{
        res ^=num;
    }
    // res store the lonely i32 because it is not canceled by any duplicate
    res
}

fn main() { // no main() if this code runs in a Jupyter cell 
    let nums = vec![1,3,3,2,1];
    println!("{}", lonely_integer(&nums)); // 2
} // end of local scope OR end of main()