not completed, still need to be fixes. Not fully working/tested

2023_10_09/check-if-all-the-integers-in-a-range-are-covered/Cargo.toml

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+[package]
+name = "check-if-all-the-integers-in-a-range-are-covered"
+version = "0.1.0"
+edition = "2021"
+
+# See more keys and their definitions at https://doc.rust-lang.org/cargo/reference/manifest.html
+
+[dependencies]

2023_10_09/check-if-all-the-integers-in-a-range-are-covered/src/main.rs

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+fn is_covered(ranges: Vec<Vec<i32>>, left: i32, right: i32) -> bool {
+    let mut line = vec![0; 52];  //
+    for r in ranges {
+        line[r[0] as usize] += 1;
+        line[r[1] as usize + 1] -= 1;
+    }
+    let mut overlaps = 0;
+    for i in 1..=right {
+        overlaps += line[i as usize];
+        if i >= left && overlaps == 0 {
+            return false;
+        }
+    }
+    true
+}
+
+fn main() {
+    let ranges = vec![vec![1, 2], vec![3, 4], vec![5, 6]];
+    let left = 2;
+    let right = 7;
+    println!("{}", is_covered(ranges, left, right));
+}

2023_10_09/longest-k-good-segment/Cargo.toml

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+[package]
+name = "longest-k-good-segment"
+version = "0.1.0"
+edition = "2021"
+
+# See more keys and their definitions at https://doc.rust-lang.org/cargo/reference/manifest.html
+
+[dependencies]

2023_10_09/longest-k-good-segment/README.md

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+# 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)
+}
+```

2023_10_09/longest-k-good-segment/src/main.rs

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+// https://codeforces.com/contest/616/problem/D?locale=en
+
+use std::collections::HashMap;
+use std::io::{self, Read};
+
+fn main() {
+    let mut buffer = String::new();
+    io::stdin().read_to_string(&mut buffer).unwrap();
+
+    let mut lines = buffer.lines();
+    let _ = lines.next();
+    let a: Vec<i32> = lines
+        .next()
+        .unwrap()
+        .split_whitespace()
+        .map(|x| x.parse().unwrap())
+        .collect();
+    let k: i32 = lines.next().unwrap().parse().unwrap();
+
+    let mut map = HashMap::new();
+    let mut max = 0;
+    let mut max_l = 0;
+    let mut max_r = 0;
+    let mut l = 0;
+    let mut r = 0;
+    while r < a.len() {
+        let count = map.entry(a[r]).or_insert(0);
+        *count += 1;
+        while map.len() > k as usize {
+            let count = map.get_mut(&a[l]).unwrap();
+            *count -= 1;
+            if *count == 0 {
+                map.remove(&a[l]);
+            }
+            l += 1;
+        }
+        if r - l + 1 > max {
+            max = r - l + 1;
+            max_l = l;
+            max_r = r;
+        }
+        r += 1;
+    }
+    println!("{} {}", max_l + 1, max_r + 1);
+}