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85 lines
2.8 KiB
Markdown
85 lines
2.8 KiB
Markdown
1 year ago
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# Comments on the solution
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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.
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### Complexity analysis
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* **Time complexity**: $O(n)$ since we only iterate through the array once and perform constant time operations on each element.
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* **Space complexity**: $O(k)$ since we store the count of each value in the hashmap.
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---
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### Rust Solution
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This version reads the input from stdin and writes the output to stdout (as required by the problem statement)
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```rust
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fn main() {
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let mut input = String::new();
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std::io::stdin().read_line(&mut input).unwrap();
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let mut iter = input.split_whitespace();
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let n: usize = iter.next().unwrap().parse().unwrap();
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let k: usize = iter.next().unwrap().parse().unwrap();
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let mut input = String::new();
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std::io::stdin().read_line(&mut input).unwrap();
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let mut iter = input.split_whitespace();
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let mut a: Vec<usize> = Vec::with_capacity(n);
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for _ in 0..n {
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a.push(iter.next().unwrap().parse().unwrap());
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}
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let mut l = 0;
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let mut r = 0;
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let mut max = 0;
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let mut map: std::collections::HashMap<usize, usize> = std::collections::HashMap::new();
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while r < n {
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let count = map.entry(a[r]).or_insert(0);
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*count += 1;
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while map.len() > k {
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let count = map.entry(a[l]).or_insert(0);
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*count -= 1;
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if *count == 0 {
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map.remove(&a[l]);
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}
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l += 1;
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}
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if r - l + 1 > max {
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max = r - l + 1;
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}
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r += 1;
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}
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println!("{} {}", l + 1, r);
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}
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```
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Here is a version that takes the input as arguments and returns the output as a tuple.
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```rust
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fn find_k_good_segment(n: usize, k: usize, a: &[usize]) -> (usize, usize) {
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let mut l = 0;
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let mut r = 0;
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let mut max = 0;
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let mut map: std::collections::HashMap<usize, usize> = std::collections::HashMap::new();
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while r < n {
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let count = map.entry(a[r]).or_insert(0);
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*count += 1;
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while map.len() > k {
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let count = map.entry(a[l]).or_insert(0);
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*count -= 1;
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if *count == 0 {
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map.remove(&a[l]);
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}
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l += 1;
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}
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if r - l + 1 > max {
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max = r - l + 1;
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}
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r += 1;
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}
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(l + 1, r)
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}
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```
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