# Comments on the solution

The problem requires finding the longest segment of an array that contains no more than `k` different values. We can use a sliding window approach. We maintain a hashmap to store the count of each value in the current segment. We start with the left and right indices at 0 and keep incrementing the right index until the number of different values in the segment exceeds `k`. At this point, we increment the left index until the number of different values in the segment is less than or equal to `k`. We keep track of the maximum length of the segment seen so far and return the left and right indices of this segment.

### Complexity analysis

* **Time complexity**: $O(n)$ since we only iterate through the array once and perform constant time operations on each element.

* **Space complexity**: $O(k)$ since we store the count of each value in the hashmap.

---

### Rust Solution

This version reads the input from stdin and writes the output to stdout (as required by the problem statement)

```rust
fn main() {
    let mut input = String::new();
    std::io::stdin().read_line(&mut input).unwrap();
    let mut iter = input.split_whitespace();
    let n: usize = iter.next().unwrap().parse().unwrap();
    let k: usize = iter.next().unwrap().parse().unwrap();

    let mut input = String::new();
    std::io::stdin().read_line(&mut input).unwrap();
    let mut iter = input.split_whitespace();
    let mut a: Vec<usize> = Vec::with_capacity(n);
    for _ in 0..n {
        a.push(iter.next().unwrap().parse().unwrap());
    }

    let mut l = 0;
    let mut r = 0;
    let mut max = 0;
    let mut map: std::collections::HashMap<usize, usize> = std::collections::HashMap::new();
    while r < n {
        let count = map.entry(a[r]).or_insert(0);
        *count += 1;
        while map.len() > k {
            let count = map.entry(a[l]).or_insert(0);
            *count -= 1;
            if *count == 0 {
                map.remove(&a[l]);
            }
            l += 1;
        }
        if r - l + 1 > max {
            max = r - l + 1;
        }
        r += 1;
    }
    println!("{} {}", l + 1, r);
}
```

Here is a version that takes the input as arguments and returns the output as a tuple.


```rust
fn find_k_good_segment(n: usize, k: usize, a: &[usize]) -> (usize, usize) {
    let mut l = 0;
    let mut r = 0;
    let mut max = 0;
    let mut map: std::collections::HashMap<usize, usize> = std::collections::HashMap::new();
    while r < n {
        let count = map.entry(a[r]).or_insert(0);
        *count += 1;
        while map.len() > k {
            let count = map.entry(a[l]).or_insert(0);
            *count -= 1;
            if *count == 0 {
                map.remove(&a[l]);
            }
            l += 1;
        }
        if r - l + 1 > max {
            max = r - l + 1;
        }
        r += 1;
    }
    (l + 1, r)
}
```