RUST GYM Leetcode 268 Missing Number Rust Solution

268. Missing Number

Given an array nums containing n distinct numbers in the range [0, n], return the only number in the range that is missing from the array.

Follow up: Could you implement a solution using only O(1) extra space complexity and O(n) runtime complexity?

Example 1:

Input: nums = [3,0,1]
Output: 2
Explanation: n = 3 since there are 3 numbers, so all numbers are in the range [0,3]. 2 is the missing number in the range since it does not appear in nums.

Example 2:

Input: nums = [0,1]
Output: 2
Explanation: n = 2 since there are 2 numbers, so all numbers are in the range [0,2]. 2 is the missing number in the range since it does not appear in nums.

Example 3:

Input: nums = [9,6,4,2,3,5,7,0,1]
Output: 8
Explanation: n = 9 since there are 9 numbers, so all numbers are in the range [0,9]. 8 is the missing number in the range since it does not appear in nums.

Example 4:

Input: nums = [0]
Output: 1
Explanation: n = 1 since there is 1 number, so all numbers are in the range [0,1]. 1 is the missing number in the range since it does not appear in nums.

Constraints:

n == nums.length
1 <= n <= 10⁴
0 <= nums[i] <= n
All the numbers of nums are unique.

Rust Solution

struct Solution;

impl Solution {
    fn missing_number(nums: Vec<i32>) -> i32 {
        let mut xor: i32 = 0;
        let n = nums.len();
        for n in 0..=n {
            xor ^= n as i32;
        }
        for n in nums {
            xor ^= n;
        }
        xor
    }
}

#[test]
fn test() {
    let nums = vec![9, 6, 4, 2, 3, 5, 7, 0, 1];
    assert_eq!(Solution::missing_number(nums), 8);
}

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