competitive-programming/2023_09_18/missing_number

Comments on the solution

We are using a simple mathematical approach that it's pretty much self-explanatory. We can express the sum of all numbers from 0 to n as n * (n + 1) / 2. We can then subtract the sum of the array from the sum of all numbers from 0 to n to get the missing number.

Complexity

• Time Complexity: O(n) - We are iterating over the array once.
• Space Complexity: O(1) - We are not using any extra space.

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Solution in Rust

fn missing_number(nums: Vec<i32>) -> i32 {
    (1..=nums.len() as i32).sum::<i32>() - nums.iter().sum::<i32>()
}

where we use the ..= range operator to create a range from 1 to the length of the input vector, and then sum the values in that range using the sum method

Leetcode result