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25 lines
829 B
Markdown
25 lines
829 B
Markdown
# Comments on the solution
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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.
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### Complexity
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- **Time Complexity:** `O(n)` - We are iterating over the array once.
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- **Space Complexity:** `O(1)` - We are not using any extra space.
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---
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### Solution in Rust
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```rust
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fn missing_number(nums: Vec<i32>) -> i32 {
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(1..=nums.len() as i32).sum::<i32>() - nums.iter().sum::<i32>()
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}
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```
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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
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### Leetcode result
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