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829 B
829 B
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.
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
