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268. Missing Number share

Problem Statement

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.

 

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.

 

Constraints:

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

 

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

Solution:

rs
impl Solution {
    pub fn missing_number(nums: Vec<i32>) -> i32 {
        let len = nums.len();
        let mut sum = len * (len + 1) / 2;

        for num in nums {
            sum -= num as usize;
        }

        sum as i32
    }
}

...


Released under the MIT License.